Complex Operations and MORE TRIGONOMETRY

Right, we have complex numbers, we can draw them on our plane. We can convert between the two forms of notation that we are using, Polar and Rectangular. We also saw that multiplying by \(i\) had the effect of rotating a number by \(90^{\circ}\), or as we are now referring to angles, \(\tfrac{\pi}{2}\).

How, more generally, do we go about adding and multiplying and raising to an exponent complex numbers?

Adding is actually pretty easy. Lets put our number in rectangular notation. So it looks like \(a+bi\). Lets make a new number, \(c+di\). Now lets add them together:

\[(a+bi)+(c+di)\]

There is no multiplication there, apart from the \(b\) and \(d\) both being multiplied by \(i\). So we could just add all the terms up:

\[a+bi+c+di\]

And tidy:

\[a+c+bi+di\]

Now if we look closely, we can see that we have two numbers multiplied by \(i\). If you remember our red and blue baskets with apples and oranges in them, we can just pull out the \(i\)'s leaving the \(b\) and \(d\) in a bracket multiplied by \(i\):

\[a+c+i(b+d)\]

Hang on, we now have two real numbers added together, and then two other real numbers added together and multiplied by \(i\). That is very nearly back to our rectangular form of one real part, and one imaginary part represented by a real number multiplied by \(i\). We just stick another set of brackets in like this:

\[(a+c)+(b+d)i\]

And we are done, that looks just like a rectangular number. Adding numbers in rectangular form is as simple as just adding the real parts together, and then adding the imaginary parts together.

OK, what about multiplication?

\[(a+bi)\cdot (c+di)\]

Again, remember our baskets. We can write this out like this:

\[a\cdot c + a\cdot di + bi\cdot c + bi\cdot di\]

So we have each number in the first bracket multiplied by each number in the second bracket. It tidies up to look like:

\[a\cdot c + a\cdot di + bi\cdot c + b\cdot d\cdot i^2\]
\[a\cdot c + a\cdot di + bi\cdot c - bd\]
(Remember that when we multiply \(i\) by itself we get negative one, so the last term loses its \(i^2\) and is subtracted instead of added at the same time.) We can take away the dots as well:

\[ac + adi + bci - bd\]

And then rearrange to the real and imaginary parts:

\[ac - bd + adi + bci\]

Again we can take out the \(i\)'s, and bracket off the real and imaginary parts:

\[(ac - bd) + (ad + bc)i\]

Well, it IS a rectangular form complex number, but it looks horrid. You get no real feel for what is happening on the complex plane. What happens if we deal with multiplication in the polar form instead? Let's look at what happens to the absolute value of the number. We'll call \(r_{\scriptsize 1}\) the absolute value of the first number and \(r_{\scriptsize 2}\) the absolute value of the second number, and \(r_{\scriptsize 12}\) the absolute value of the pair of them multiplied together. And remember that the square of the absolute value of a complex number is the real part squared plus the imaginary part squared. So we get this:

\[r_{\scriptsize 1}^2=a^2+b^2\]
\[r_{\scriptsize 2}^2=c^2+d^2\]
\[r_{\scriptsize 12}^2=(ac-bd)^2+(ad+bc)^2\]

Let's take a very deep breath and try to expand the squares in the last line:

\[r_{\scriptsize 12}^2=acac-acbd-bdac+bdbd+adad+adbc+bcad+bcbc\]

Oh, boy. Lets try to tidy:

\[r_{\scriptsize 12}^2=a^2c^2-abcd-abcd+b^2d^2+a^2d^2+abcd+abcd+b^2c^2\]

Hmm. Not a great deal better. It's hard to spot but notice something about the four terms which are just \(abcd\). Two of them are subtracted, and two are added, so it could be written:

\[r_{\scriptsize 12}^2=a^2c^2-2\cdot abcd+2\cdot abcd+b^2d^2+a^2d^2+b^2c^2\]

But, of course if you add two somethings, and take two somethings away, then you end up with no somethings, so we can remove those parts altogether:

\[r_{\scriptsize 12}^2=a^2c^2+b^2d^2+a^2d^2+b^2c^2\]

Right. What do we do with that mess? The answer is red and blue baskets again. You spot that there are two parts multiplied by \(a^2\), namely \(a^2c^2\) and \(a^2d^2\), so we can take the \(a\)'s out and putting what remains in a bracket:

\[r_{\scriptsize 12}^2=a^2(c^2+d^2)+b^2d^2+b^2c^2\]

It should be easier to spot this time, but you can then do exactly the same with the \(b^2\) terms:

\[r_{\scriptsize 12}^2=a^2(c^2+d^2)+b^2(c^2+d^2)\]

And if we remember about the red and blue baskets, and we just want to know how many of everything we have, we can put the \(a^2\) and \(b^2\) in brackets as well:

\[r_{\scriptsize 12}^2=(a^2+b^2)(c^2+d^2)\]

Remember our definitions of the absolute values of the numbers we are multiplying?

\[r_{\scriptsize 1}^2=a^2+b^2\]
\[r_{\scriptsize 2}^2=c^2+d^2\]

So we can replace the contents of the first bracket with \(r_{\scriptsize 1}^2\) and the second bracket with \(r_{\scriptsize 2}^2\)

\[r_{\scriptsize 12}^2=(r_{\scriptsize 1}^2)(r_{\scriptsize 2}^2)\]

That can just be written as:

\[r_{\scriptsize 12}^2=r_{\scriptsize 1}^2\cdot r_{\scriptsize 2}^2\]
\[r_{\scriptsize 12}^2=r_{\scriptsize 1}\cdot r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot r_{\scriptsize 2}\]
\[r_{\scriptsize 12}^2=r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot r_{\scriptsize 1}\cdot r_{\scriptsize 2}\]
\[r_{\scriptsize 12}^2=(r_{\scriptsize 1}\cdot r_{\scriptsize 2})\cdot (r_{\scriptsize 1}\cdot r_{\scriptsize 2})\]
\[r_{\scriptsize 12}^2=(r_{\scriptsize 1}\cdot r_{\scriptsize 2})^2\]

And if we now take the square roots of both sides we get:

\[r_{\scriptsize 12}=r_{\scriptsize 1}\cdot r_{\scriptsize 2}\]

This tells us that when you multiply two complex numbers together, the absolute value of the result is the absolute values of the original two numbers multiplied together. So if the absolute values of your starting numbers were two and three, the absolute value of the result will be six.

Where will the angle at the origin point to once the numbers have been multiplied? Well, our formula for working out the polar number from the rectangular number was a horrific mess. If we actually wanted to trail through that for the rectangular form \((ac - bd) + (ad + bc)i\) just to see what would happen, we would have to be nuts. Stark raving mad. Is there an easier way? Joyfully, yes. Remember our formula for working out the rectangular form of a polar number?

\[rcos\theta + risin\theta\]

That is a perfect halfway house between a rectangular form number and a polar form number. Why? Well it has a real part (distance along the real axis) and an imaginary part (distance along the imaginary axis), but it is actually written in terms of the size of the angle involved. So we can use this to see what happens to the angle when we multiply. The first thing to do is to spot that there are two entities multiplied by \(r\) and take the \(r\) out:

\[r(cos\theta + isin\theta)\]

Lets call that complex number one:

\[r_{\scriptsize 1}(cos\theta + isin\theta)\]

Lets make this next one complex number two:

\[r_{\scriptsize 2}(cos\alpha + isin\alpha)\]

If we multiply them together we get:

\[r_{\scriptsize 1}(cos\theta + isin\theta)\cdot r_{\scriptsize 2}(cos\alpha + isin\alpha)\]

Rearranging the \(r\)'s gives us:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \big(cos\theta + isin\theta\big)\cdot \big(cos\alpha + isin\alpha\big)\]

This should make sense to us, because look what we are doing with our absolute values: we are multiplying them together, just like we worked out above. If we multiply out all the stuff in the brackets, we will get this:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \big(
cos\theta cos\alpha+cos\theta isin\alpha+isin\theta cos\alpha+iisin\theta sin\alpha
\big)\]

Which simplifies (a bit), because the term at the end has \(i^2\) in it:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \big(
cos\theta cos\alpha+cos\theta isin\alpha+isin\theta cos\alpha-sin\theta sin\alpha
\big)\]

And then we can move the \(i\)'s out of the middle two terms, putting the remainder in brackets:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos\theta cos\alpha+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)-sin\theta sin\alpha
\Big)\]

I'll then do one further rearranging to get the real parts together:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos\theta cos\alpha-sin\theta sin\alpha+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)
\Big)\]

Now what in the name of sanity are we going to do with that mess in brackets? What we need to do is start drawing lines on the unit circle to see exactly what we are talking about. What IS the cosine of one angle multiplied by the cosine of another minus the sines of both multiplied together? To find out what this looks like, we are going to go back to the unit circle, and we are going to draw on the angle \(\theta\). I am marking the origin as point A, and the point where the radius hits the circle as B. We have seen all this before.

Forget for a moment that I am showing the angle down near the origin, or point A, rather than drawn onto the unit circle. This will make sense in due course. It doesn't really matter what size the angle is, because we are trying to come up with a general rule that will work for all angles. OK, next let's mark off the sine of the angle as a line going straight down from point B to the x axis:

We'll call the point where it meets the x axis C, and we will draw on a square to show that the angle at which it meets the x axis is a right angle:

And so we know what we are talking about, lets describe the line:

Good. Now what about the cosine of the angle \(\theta\)? That's just the x axis out to point C isn't it? So lets mark that on:

This is exactly the same picture that we have seen before, with a bit more writing on it for clarity's sake. For the purposes of this task I am going to zoom into the quarter circle between the positive x and y axes. That is the place where we are going to be playing, so we can just concentrate there:

OK, that's a bit clearer. Now what I am going to do is to take the point B and start to move it around the unit circle. But this time I am going to FIX the angle \(\theta\) so that does not change. That sounds very odd. What is that going to tell us? Well, wait and see. What is going to happen is that the whole triangle that we have just drawn, from A to B to C and back to A, is going to keep its shape but it is going to rotate about the point A. Lets see what we get:

So, you can see that we now have the triangle seemingly floating around. Point C is in empty space. The important fact is that I have ONLY rotated. The lengths of the sides have not changed. The angle at C is still a right angle. The length of A to B is still the radius of the unit circle, which is still one. What can I do now? Well I can show that I have rotated the first triangle by a new angle called \(\alpha\):

At this point, what I could do is draw a line all the way from the origin at A, through C and out to the unit circle to let us measure the sine and cosine of \(\alpha\). I am not going to do that though. What I am going to try and do is measure the sine and cosine of \(\alpha\) using the measurements we have already made for \(\theta\). You will see why shortly.

The first thing we need to do here is to draw on another line, to stand in for the sine of \(\alpha\). We want it to reach the x axis at right angles, just like a normal sine line. Because we are using the measurements of \(\theta\) to work out \(\alpha\) we need to use the whole of the line from A to C. So we will drop our "sine" of \(\alpha\) line down from C to the x axis, and we will mark the point where it hits as D, which is a right angle:

Right then, first BIG question. How do we work out the lengths of the line from A to D and from C to D? Just in case you are trying to figure this out I am talking about these distances:

If the line from A to C was actually the radius of a unit circle, then we could easily work out AD and CD. They would just be the cosine and sine of the angle \(\alpha\). So that's a start. Let's stick a circle on the diagram anyway, to see what that would look like:

So you can see that IF \(cos\theta\) was one, then the other two sides would just be the cosine and sine of \(\alpha\). What we are dealing with here is another problem of scaling. If you remember back to when we were first working out the rectangular form version of a complex number in polar form, what we did was to work out the sine and cosine on the unit circle and then we just multiplied them by the absolute value of the number. We did this because we had scaled the absolute value to one, so that we could work in the unit circle. Here we just need to do the same thing. Point C is not on the unit circle. It is like the complex number we are trying to find. The absolute value of that number is \(cos\theta\). So we just need to scale the ACTUAL sine and cosine of \(\alpha\) (whatever they are) by the factor \(cos\theta\), and that will give us the real and imaginary parts of the complex number C. Those real and imaginary parts are, of course, the lengths of the lines AD and CD that we are trying to find!

So we end up by saying that the length of AD is the cosine of \(\alpha\) MULTIPLIED by \(cos\theta\). Or \(cos\theta\cdot cos\alpha\). Hey! We've seen that before:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos\theta cos\alpha-sin\theta sin\alpha+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)
\Big)\]

It's the very first term of the mess in brackets that we are trying to sort out. Great, we are making progress! So what about the line CD? Well that is just going to be the sine of \(\alpha\) MULTIPLIED by \(cos\theta\). Or \(cos\theta\cdot sin\alpha\). And look, that's the third term in the mess in brackets. 50% done! Let's mark this up on our diagram:

I have taken the example unit circle away, because it was just making a mess, and we have finished with it. Right, we are halfway done now. We just need the other two terms of that mess so we can start to make sense of it. Looking at our diagram, we have now worked out the length of all our sides. What we need to make progress is ANOTHER triangle. No, I am not about to rotate the whole mess again, no more rotation. Instead I am going to look at the mess I am trying to make sense of and I am going to realise that to get the remaining two terms I need to start multiplying stuff by \(sin\theta\). I was able to multiply stuff by \(cos\theta\) because I made that the radius of a circle, and then scaled the other two sides by that factor. So lets make \(sin\theta\) the radius of a circle so we can play the same trick. How can we do that? Well if \(sin\theta\) is going to be the radius of a circle that makes either point B or point C the centre of the circle. Let's pick point C as the centre, and draw on the circle:

I am now going to twist the whole diagram through a quarter turn:

And I am going to draw on, in a dashed line because it is temporary, an x axis for this new circle. Now I can drop a line to this x axis from the point B, so as to meet the line at right angles at the new point E:

I will also draw in the line along the temporary x axis between C and E:

Now we can start working out the lengths of CE and BE. Hang on though, don't we need the size of the angle as well? Which angle? The angle marked \(\beta\):

So what size is this angle? The angle of a straight line is \(\pi\) radians. Is this right? Of course. If you take the unit circle, and you start at the normal place on the real axis, and you move along half the circle (\(\pi\) radians) you end up back on the real axis, on the negative side. You may as well have travelled in a straight line along the real axis and you would have ended up at the same place. Knowing what we know about angles on straight lines, what can we say about \(\beta\)?

Well, we can say that it is on a straight line, the line from D to E which passes through C. So the total angle on that straight line must be \(\pi\). We can see by looking that there are three angles that make up that total angle. They are the unmarked angle, the right angle (size \(\tfrac{\pi}{2}\)) and \(\beta\). So we can say:

\[\pi=\tfrac{\pi}{2}+\beta+unknown\]

Take away \(\tfrac{\pi}{2}\) from both sides and you get:

\[\tfrac{\pi}{2}=\beta+unknown\]

So what's the unknown angle? It is on the line, AND it is inside a right angled triangle. In case it is unclear it is a corner in this bold triangle:

If you add up all the angles in a right angled triangle in radian measure they total \(\pi\). Is this right? Yes, a right angled triangle is half of a rectangle. Each corner of the rectangle has an angle of \(\tfrac{\pi}{2}\), making a total of \(2\pi\). So one half of this shape will have angles that add up to one half of \(2\pi\) or \(\pi\). If they didn't two of them could not fit together to make a rectangle. (Actually this is true of any triangle drawn on a flat surface, but it is harder to explain why and involves Euclid and Parallel lines, which we do not need to concern ourselves with here).

The other two angles in that triangle are a right angle \(\tfrac{\pi}{2}\), and \(\alpha\), and all three angles must add up to \(\pi\). So we can also say that:

\[\pi=\tfrac{\pi}{2}+\alpha+unknown\]

Again, take away \(\tfrac{\pi}{2}\) from both sides:

\[\tfrac{\pi}{2}=\alpha+unknown\]

If we take \(\alpha\) away from both sides we get:

\[\tfrac{\pi}{2}-\alpha=unknown\]

So the unknown angle has size \(\tfrac{\pi}{2}-\alpha\). Let's put that back in our equation above, so:

\[\tfrac{\pi}{2}=\beta+unknown\]

Becomes:

\[\tfrac{\pi}{2}=\beta+\tfrac{\pi}{2}-\alpha\]

Now what can we do with that? First add \(\alpha\) to both sides:

\[\tfrac{\pi}{2}+\alpha=\beta+\tfrac{\pi}{2}\]

And finally, remove \(\tfrac{\pi}{2}\) from both sides:

\[\alpha=\beta\]

Hey presto, just like magic we have worked out that the angle \(\beta\) MUST be the same size as angle \(\alpha\). So lets mark that on our diagram:

Bingo. We can now deduce, using exactly the same logic we used for the lines AD and CD, the lengths of lines BE and CE. In particular, CE would be the cosine of \(\alpha\) IF you scaled it by the factor \(sin\theta\), so CE is \(sin\theta cos\alpha\). And in turn BE would be the sine of \(\alpha\) if you scaled it by the factor \(sin\theta\), so BE is \(sin\theta sin\alpha\). Lets draw those on:

Let's also get rid of the other circle, the temporary x axis, and rotate back to normal:

Now we have lines on our diagram that match all the parts of the mess in the brackets in this equation:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos\theta cos\alpha-sin\theta sin\alpha+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)
\Big)\]

Now we can try to answer the question what is the cosine of \(\theta\) multiplied by the cosine of \(\alpha\) minus the sine of \(\theta\) mutlipied by the sine of \(\alpha\). We know what \(cos\theta cos\alpha\) is, it is the line from A to D. We also know what \(sin\theta sin\alpha\) is, it is the line from B to E. So what do we get if we subtract the line BE from the line AD? First notice that the point E is directly above the point D. So if we start at D, and move back towards the origin at A until we are directly under B, the remaining distance to A will be AD-BE. The first thing we need to do is drop a line from B to the real axis so we can see exactly where that point will be:

I have moved the text down a bit showing the length of A to D. The angles are getting in the way a bit though, so I will shrink them down:

Can you see what has happened? The two angles we have been dealing with, \(\theta\) and \(\alpha\) can now be see as one big angle, which I can call \(\theta + \alpha\). What is the cosine of the angle \(\theta + \alpha\)? Well, first you would fix the point on the circle that corresponds to a journey around the circle of the size of the angle. That point is B isn't it? Remember that the line from A to B is the radius of the unit circle, so it is length one. That means that the cosine of the angle \(\theta + \alpha\) is the length from the origin along the real axis to the point directly under the point B. That's the point F we just drew isn't it? So \(cos(\theta + \alpha)\) is the length A to F. But how did we find the length A to F? We started with A to D, which was \(cos\theta cos\alpha\) and we subtracted the length from B to E, which was \(sin\theta sin\alpha\). So if AF=AD-BE, then:

\[cos(\theta + \alpha)=cos\theta cos\alpha-sin\theta sin\alpha\]

Fantastic! We can now go back to our horrible equation:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos\theta cos\alpha-sin\theta sin\alpha+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)
\Big)\]

and we can replace the first part of the mess with:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos(\theta + \alpha)+i\big(cos\theta sin\alpha+sin\theta cos\alpha\big)
\Big)\]

Well, that's better, and we are nearly finished. Let's tackle the second part. Back to the diagram, what is the sine of \(\theta + \alpha\)? Well given that the line from A to B is length one, the sine is just the height of the point B off the real axis, isn't it? Can you see a line that shows that height? Yes, it is the line from F to B. What OTHER line have we drawn that is the same height as F to B? Yes, D to E. So \(sin(\theta + \alpha)\) is the same as the length D to E. And look closely, what is D to E? It is \(cos\theta sin\alpha+sin\theta cos\alpha\). So:

\[sin(\theta + \alpha)=cos\theta sin\alpha+sin\theta cos\alpha\]

Fantastic times two! We can now finish clearing up our horrible equation by replacing the second mess in brackets with:

\[r_{\scriptsize 1}\cdot r_{\scriptsize 2}\cdot \Big(
cos(\theta + \alpha)+isin(\theta + \alpha)
\Big)\]

Just in case you think I fiddled the diagram with special angles, have a look at an animation of varying angles, and you'll set it holds for all of them:



What can we learn from this? Well, as we had already worked out, if you multiply two complex numbers in polar form, you multiply their absolute values together. That's the \(r_{\scriptsize 1}\cdot r_{\scriptsize 2}\) bit. You then just ADD the angles together - look you can see it happening in the brackets above. So the whole multiplication looks like this:

\[r_{\scriptsize 1}(cos\theta + isin\theta)\cdot r_{\scriptsize 2}(cos\alpha + isin\alpha)=
r_{\scriptsize 1}r_{\scriptsize 2}\Big(
cos(\theta + \alpha)+isin(\theta + \alpha)
\Big)\]

Formally then we can say that:

\[(a+bi)+(c+di)=(a+c)+(b+d)i\]

and

\[r_{\scriptsize 1}\angle \theta \cdot r_{\scriptsize 2}\angle \alpha = r_{\scriptsize 1}r_{\scriptsize 2}\angle (\theta + \alpha)\]

Hurrah! We've done multiplication and addition. Subtraction and division are essentially just the reverse. What about exponents?

Raising a complex number to a real exponent is just the same as repeated multiplications, so that's fine. What about raising a real number to a complex exponent though? What does that look like? Answer a) it looks fucking horrible, is what it looks like. Answer b) it is the secret at the heart of finding out why \(e^{i\pi}+1=0\), because look, there is a complex number as an exponent of \(e\)! To get into that in any detail, we will have to work out a different way of writing out \(e\), which is going to be pretty traumatic. So as a diversion, lets look at something shiny instead.

Polar Form, Rectangular Form and TRIGONOMETRY

So how have we finally come to a topic called trigonometry. Is this not just a hellish torture for school pupils who have to learn crap mnemonics? Well, not quite.

Last time we looked at the two forms of complex numbers. The rectangular form identified the location of our complex number on the complex plane by it’s real and imaginary parts. You mark them off on the relevant axes, and then draw a rectangle. The corner opposite the origin (the zero point) is your complex number. Fine. The other way to describe EXACTLY the same number, the polar form, was to note down which direction to point in at the origin and then how far to move to get to the number. So with the rectangular form you draw a rectangle, with the polar form you draw a line at some angle.

So we need to describe the angle when placing a value on our complex number. I picked a really easy complex number to get to last time, which was just at a \(45^{\circ}\) angle, or one eighth of the whole circle. I could work out the angle without any particular bother, because the real and imaginary parts of the number were the same. If we are dealing with any other complex numbers, we will need to deal with more complicated angles. How, then, would we go about converting any complex number in rectangular form to polar form and vice versa?

First of all I like to ask the question “Can we do that?” I mean, do we have enough information available to actual perform the task? For instance, I once struggled for quite some time trying to work out how to get the area of a piece of ground when the only information I had was the lengths of each of the four sides. I was staring idly into space trying to get an answer when my eyes focused on the mechanism use to automatically close my door. I am talking about the things described in this article. I realised that the two arms of the door closer, the door frame and the door itself made up four fixed length sides of an area. However I saw when the door opened and closed that the area enclosed by the sides went from very long and slim to very broad and fat. The area was obviously changing depending on the angles. I then stopped trying to find a solution to the area of the ground, because I realised that I could not find one without some more information, namely the angles between the sides of the area of ground.

(Second reasonable question, why are we doing this? It turns out, as we will see, that it is dead easy to add complex numbers in rectangular form, but hard to multiply them. Conversely it is easy to multiply numbers in polar form, but hard to add them. So if we can swap between the two forms, we get the best of both worlds. Also, it turns out that the process of converting between sides and angles is going to be key to the understanding of this whole project, so we will have to look at all this eventually.)

So, do we have enough information from the angle and length only to derive the length of the sides of the rectangle? Lets draw out roughly what I am talking about:

[pic of rect and polar]

See the polar form line? It has a fixed angle and length? What happens if we change either the angle or the length? The point that the line reaches will change won’t it? And if that point changes, then the lengths of the rectangle that reach that point will also have to change? Yes. And vice versa, if we change the lengths of either of the sides of the rectangle, that will force the length AND the angle to change as well? Yes. So one set of rectangle lengths is equal to one, and only one, set of line and angle sizes. So, if we have the rectangle lengths OR the angle and line length, we DO have enough information to work out the one from the other. Good stuff. At least this won’t be an afternoon wasted.

Let’s start with the number \(\sqrt{5}\angle 30^{\circ}\). How do we find the rectangular form of this number? Let’s see what it looks like first:

[pic of number]

Well, it looks like it ends up at roughly our old friend \(2+i\), but I may not have drawn the line with sufficient precision. Is it exactly \(2+i\)? The first thing to do is to test that the length of the line fits. When we convert to rectangular form, we are dealing with a right angled triangles. Why? Because a defining characteristic of rectangles is that they have four right angles in them. So with the rectangular form that we are converting to, the angle where the line from the number hits either the real or the imaginary axis will be a right angle. The absolute value line cuts through the middle of this rectangle creating two triangles. Pythagoras’ theorem tells us that the square of the length of the absolute value line must equal the square of the lengths of the other two sides added together. The other two sides are of course, the size of the real and imaginary parts of the number marked off on those axes.

We said that it looks like our candidate number is \(2+i\), so if we guess that the other two sides have lengths of two and one, then we get \(2^2=4\) and \(1^1=1\) and four plus one is five, and we know our line is \(\sqrt{5}\) long, so that fits. It does appear that \(2+i\) is correct. But, it only MIGHT be correct. There are lots of points which have a line length of \(\sqrt{5}\); they lie on a circle with that radius. That circle goes through the number \(2+i\). Because the circle goes through that number it also goes through other numbers that are very very close indeed to \(2+i\) (actually as close as you like). That means that just because our line length fits our guess, we cannot stop there. We need to confirm that the angle we have, \(30^{\circ}\) points directly at the number \(2+i\) and not to a number very close to it.

So how are we going to do this? Well, let's try to work out what the relationship is between the size of the angle, and the real and imaginary parts of the complex number. In other words, if you change the size of the angle, what happens to the size of the real and imaginary parts? Remember, though, we are looking only at changes in the real and imaginary parts and the angle, so we will be leaving the length of the line itself fixed so it cannot be a factor in what happens.

Right. If we have a fixed length, and a varying angle at one end, what is going to happen to the other end of that line? Glittering prizes for anyone who said it will mark out a circle. Remember that our definition of a circle was all points a fixed distance from a fixed point? Here we have that fixed point, the origin, and a fixed distance, the length of the line attached to the origin (otherwise known as the absolute value of our complex number). And we also just said above that there are lots of points with a length of \(sqrt{5}\) which lie on a circle with that radius about the origin.

As our angle increases, one point remains at the origin, and the other point will trace out the shape of a circle around the origin.

[pic] [pic] [pic]

This all looks a bit messy with the square root floating about there. Even if the absolute value was something nice like two or seven, rather than a square root, if we study this arrangement in detail we will only be able to draw conclusions about this arrangement. We want to be able to reach more general conclusions that will help us find out what angles do to the real and imaginary parts of complex numbers with ANY absolute value. We therefore want to get rid of the square root. To get rid of it, I am going to shrink the circle down so that it just has a radius of one. The units we use are unimportant. This is OK, as long as we remember that we need to scale all the other measurements to get answers. The good news is that we do not need to do any scaling on the angle. Angles do not change when the size of the things meeting at an angle change. Basically any rules that we discover about the behaviour of the line lengths and angles here, will be true for bigger circles, as long as we remember to scale them back up properly. So our circle now looks like this:

[pic]

Lets give the various things we are going to talk about some names. A circle with a radius of one, is known as a unit circle. We also have an angle here, which I am going to call Angie, because I am sick of one letter names for variables. Let’s call the line with unchanging length Abby. Abby is one unit long. Good. Now remember that the circumference of a circle is \(\pi\) times the diameter. The diameter here is two Abbies. So the whole circumference is \(2\cdot \pi \cdot Abby\). But we have already decided that Abby is one, so that makes the circle \(2\pi\) long.

Next we are going to name the point where Abby meets the circle, Polly. As Angie changes in size Polly is going to move around the circle. We can then draw two new lines along the real and imaginary axes. The line on the real axis corresponds the Polly’s real value, and the one on the imaginary axis corresponds to Polly’s imaginary part. The real axis line we will call Raul, and the imaginary axis line we will call Imogen. This all looks like this:

OK, starting places please! Where does Polly start if the angle is actually closed all the way? We agreed that we start measuring angles on positive side of the x axis, so Polly will be on the circle where it meets the positive real axis here:

Now, what can we say about our characters at this size of Angie? Well, Raul is one unit long, all the way to Polly. Imogen is actually zero, because Polly has not lifted off the real axis. So this number on the complex plane at Polly, has no Imaginary part – in other words it is just a real number. Now let’s open up the angle to one quarter of the circle. Polly is now right at the very top of the circle. It is now Raul's turn to be zero, and Imogen is now exactly one unit long because Polly is actually ON the imaginary axis, directly above the origin. Now the number has no REAL value.

You can see in the picture that the computer has written Imogen and Abby on top of each other because with this size of Angie, they are actually exactly the same size.

If we move Polly further round the circle, anticlockwise, Imogen is going to start shrinking, and Raul is going to start growing into a negative length:

So in that picture Raul is now negative because it is to the left of the zero on the real axis. We can keep picturing what happens to Raul and Imogen at different sizes of Angie, OR we could calculate what happens for each of the three hundred and sixty different whole angles, and then make an animation of the result. That would be handy.



So as the angle increases in size, Polly travels around the circle, Raul grows, while Imogen shrinks, until Raul is at maximum size, and Imogen at zero, and then the process reverses. At any point, either Raul or Imogen is growing and the other shrinking. The important fact to notice is that any point on the circle has a unique value of both Raul and Imogen. This means if you are given the lengths of Raul AND Imogen, you can work out the size of the angle. Can you do that if you are only given the size of Raul OR Imogen? No. Say you are told that Imogen is half a unit long. Well, Imogen is half a unit long twice - first on the way UP to Polly being at the top of the circle, and second on the way DOWN from Polly being at the top. So you also need to know something about Raul (whether it is positive or negative), to tell whether the Imogen in question is that on the left or right of the imaginary axis. In fact, if all you know is whether Raul and Imogen are positive or negative, you can tell which quarter of the circle Polly will be on. For instance, if both are positive, Polly will be in the first quarter. If Raul is negative, but Imogen is Positive, Polly will be in the second quarter. If both are negative, Polly is in the third quarter, and if Raul is positive but Imogen negative, Polly will be in the fourth and final quarter.

Let’s talk about Polly’s journey as well. When Polly reached the very top of the circle, we said that Angie was one quarter of a circle. How far has Polly traveled? Remember that Polly is stuck on the circle. So if we ask how far Polly has traveled to get to the top of the circle, what we are really asking is what is the length of that quarter of the circle. How do we find that out? Well, remember that we know, exactly, how long the circumference of our unit circle is. It is \(2\pi\) long. So a quarter of that total distance would be \(\tfrac{1}{2}\pi\), or \(\tfrac{\pi}{2}\). So the distance covered by Polly from the start to the very top of the circle is exactly \(\tfrac{\pi}{2}\) units. When Polly is all the way over at the left hand side, the total distance traveled is exactly \(\pi\) units. In fact, instead of talking about the amount of “turn” at the centre of the circle to define the angle, we could just talk about the distance covered by Polly. If we decided to do that, and I am going to do that from now on, we would be using something called radian measure. This just measures the distance around the circle in units called radians. There are exactly \(2\pi\) radians in the unit circle. So instead of saying \(180^{\circ}\), we would say \(\pi\) radians.

Let's quickly convert our polar form complex number into radians. It looks like this in degrees: \(\sqrt{5}\angle 30^{\circ}\). So what is thirty degrees in radians? It is going to be some fraction of \(\pi\). If \(180^{\circ}\) is \(\pi\) radians, then one degree is \(\tfrac{\pi}{180}\) radians. That means that thirty degrees is just thirty times that amount, which is \(\tfrac{30\cdot \pi}{180}\). That's a bit of a mouthful though, and we can simplify it by spotting that one hundred and eighty can be divided up into six parts, each of size thirty. In other words, or mathematical notation, \(\tfrac{30}{180}=\tfrac{1}{6}\). What we have done there is to say that the ratio of thirty to one hundred and eighty is the same as the ratio of one to six. So there are six thirties in one hundred and eighty and six ones in six, so the ratios represented by both of those fractions are the same. Instead of \(\tfrac{30\cdot \pi}{180}\) radians, we can say \(\tfrac{\pi}{6}\) radians. Our polar form number using radian measure looks like this: \(\sqrt{5}\angle \tfrac{\pi}{6}\). We do not typically need to write radians after the angle units. If an angle does not have the degree symbol, and involves \(\pi\) in any capacity, you are safe to assume it is in radians.

I like radians because they are a much more sensible form of measurement. You can see exactly what they are on the unit circle. You could try to measure the size of an angle in radians using only a bit of string (lay it along the line of the circle one unit from the origin life it up and lie it flat along a ruler, there's your angle size). You can't do that with degrees.

OK, let's get back to the relationship between Angie, Raul and Imogen. If you want to find the size of Imogen for any given size of Angie, you just use a function with Angie as the input variable. Good, that sounds simple. What does the function look like then? The function is not the same as functions we have looked at before. These had letters in place of variables, and algebraic notation. In other words they were algebraic functions. You get different kinds of functions. Remember our very first function hotbeverage of drink flavouring? No algebra there was there? So what is the function we use to find the size of Imogen for any specific value of Angie? It is a geometric function, meaning that it involves drawing pictures. Just like the pictures above in fact. The steps in the function are:

Draw a unit circle on the real and imaginary axes, centered at the origin.
Start at the point where the circle meets the real axis on the positive side.
Moving along the unit circle until you have traveled the distance equal to the angle in radians.
Measure the height you are above, or as the case may be below, the real axis.
The distance is the value of Imogen you want.

This function has a name. It is called the sine function. None of this was explained to me in school. Exactly what the sine function was, was a complete mystery. It is hardly difficult. It only involves the measurement of two lines, the circumference of the unit circle, and the distance away from the real axis. Anyway, we write thes function down like this \(sin(angle)\). If we use variables instead of precise numbers for the size of the angle, we leave out the brackets around the angle. Also, we don't tend to use letters of the latin alphabet (the one this is written in), we use the greek alphabet instead.

(I am only calling the horizontal axis the real axis here because we are dealing with complex numbers. The sine function works perfectly well if you change from real and imaginary to horizontal and vertical, or x and y, or any other labels you want to give the axes. What is important is that you are measuring the lengths of two lines on a two dimensional plane (a flat surface). So the sine function is not restricted to the complex plane, and indeed was discovered, or invented if you have that world view, millennia before the complex plane).

What about Raul? We've left it behind just now. There is another function for working out Raul. It looks almost exactly the same as the sine function:

Draw a unit circle on the real and imaginary axes, centered at the origin.
Start at the point where the circle meets the real axis on the positive side.
Moving along the unit circle until you have traveled the distance equal to the angle in radians.
Measure the distance you are to the right, or as the case may be left, of the imaginary axis.
The distance is the value of Raul you want.

This is called the cosine function. We write it as \(cos(angle)\), and the same rule apply if using variables as applied to the sine function.

So, back to our complex number. We want to find out what number we get with the angle \(\tfrac{\pi}{6}\) at length \(\sqrt{5}\) from the origin. First of all let's look at that angle on the unit circle:

You can see that I am now drawing the “angle” ONTO the unit circle itself instead of down about the origin. This shows us exactly what we are talking about in our geometric function above. We have completed the first three instructions. We now just need to measure the height we are above the real axis. Lets stick show this:

There's our Imogen! It's a bit difficult to see what value it has though, because I haven't drawn on any other markings. Let's zoom in a bit to the intersection between the dashed line and the imaginary axis:

If you think that Imogen looks like it is a half, then you are right. No matter how closely you measure Imogen, it is perfectly a half. Right, what about Raul?

OK, lets zoom in again to see if we can find it more accurately:

Well, that's not really helping this time. It is between eight and nine tenths. To save zooming time, I will say that no matter how carefully you measure Raul, it does not resolve into a nice fraction. It is actually an irrational number. I can actually prove what it is exactly! How? Look at the diagram. You have the green line, the dashed line and the real axis. We know the length of the green line, it is one unit. We also know the value of the dashed line – it's Imogen, and we just measured it in the last step – it is a half. The dashed line meets at a right angle, so we can use pythagoras to get the length of Raul. So the length of the green line squared has to equal the length of the dashed line squared plus the length of Raul. Lets algebra this sod:

\[Green^2=Imogen^2+Raul^2\]
\[Green^2-Imogen^2=Raul^2\]
\[(1)^2-(\tfrac{1}{2})^2=Raul^2\]
\[1-\tfrac{1}{4}=Raul^2\]
\[\tfrac{3}{4}=Raul^2\]
\[Raul=\sqrt{\tfrac{3}{4}}\]

What can we do with the square root of a fraction? Well, let's think of it as raising the fraction to a power instead. So:

\[Raul=(\tfrac{3}{4})^{\tfrac{1}{2}}\]

And if you raise a fraction to a power, that is the same as raising the top part by the power and the bottom part by the power. So:

\[Raul=\frac{3^{\tfrac{1}{2}}}{4^{\tfrac{1}{2}}}\]

We know what the square root of four is, that's just two. To save time I will just tell you that the square root of three is irrational, so we just leave it as the square root of three. That gives us:

\[Raul=\frac{\sqrt{3}}{2}\]

If you ask your calculator what the square root of three divided by two is, you will get a number which goes on for ever, but is somewhere between eight and nine tenths.

OK, so we have our sine and cosine functions of the angle. How do we turn these into the rectangular form of the complex number? There is actually only one stage left to go. We just need to scale the numbers back up. When we scaled down the number to the unit circle, we saw the absolute value of \(\sqrt{5}\) becoming one. To get from one back to \(\sqrt{5}\), we just multiply one by \(\sqrt{5}\). If we are multiplying one length by this scale factor, we need to multiply all lengths by the same factor.

So the real part of our complex number is going to be Raul (\(\tfrac{\sqrt{3}}{2}\)) multiplied by \(\sqrt{5}\). That will be \(\frac{\sqrt{5}\cdot \sqrt{3}}{2}\). To multiply two square roots together, you just multiply the two numbers being square rooted, and then take the square root of that. (Think about it: \(\sqrt{2}\cdot \sqrt{2}\) is the same as \(\sqrt{4}\)). So the real part of the complex number is \(\tfrac{\sqrt{15}}{2}\). If you think that looks like a horror, you are right. It certainly is not two, which was our guess. What about the imaginary part of the number? That's going to be Imogen (a half) multiplied by \(\sqrt{5}\), or just \(\tfrac{\sqrt{5}}{2}\). Again, that's a horror, and certainly is not one, which was our guess. This can be confirmed, if we measure the actual real and imaginary parts of the number using dashed lines:

As you can see when we plot this out properly, you can see that the real value is a little bit less than two, and the imaginary value is a little bit more than \(i\), or one. If you run the numbers we calculated above through a calculator to get a decimal representation, you will see they match this geometric one.

So we can say that the following polar form equals the following rectangular form:

\[\sqrt{5}\angle \tfrac{\pi}{6}=\tfrac{\sqrt{15}}{2}+\tfrac{\sqrt{5}}{2}i\]

We can say more generally that given any polar form number with an absolute value of \(r\) and an angle of \(\theta\) (\(r\angle \theta\)) the rectangular form of that number will be \(r\cdot cos\theta + r\cdot sin\theta i\). The bit at the end looks a little ambiguous. Do we mean \(sin(\theta \cdot i)\) or \(sin(\theta)\cdot i\)? To avoid this potential confusion, I am going to move the \(i\): \(r\cdot cos\theta + r\cdot i\cdot sin\theta\). We may as well lose the dots for multiplication as well: \(rcos\theta + risin\theta\)

What about going back the other way? You basically just do everything in reverse. Let's consider the number \(2+2i\) that we looked at before. What you do is simple Pythagoras to get the absolute value (we did this last time). So that absolute value is \(\sqrt{2^2+2^2}\) or \(\sqrt{8}\). Easy. Now, what about the angle, which we will call \(\theta\)?

To get the real and imaginary parts we multiplied the sine and cosine of the angle by the absolute value. So to GET the sine and cosine of the real and imaginary parts, we just DIVIDE those parts by the absolute value. That gets us to:

\[sin\theta=\frac{2}{\sqrt{8}}\]
\[cos\theta=\frac{2}{\sqrt{8}}\]

Those are the co-ordinates for the point on the unit circle we are interested in. So let's mark these off as values for Imogen and Raul:

Now let's find the point on the unit circle that corresponds to those values:

The two dashed lines that we have just projected represent the OPPOSITE of the sine and cosine functions. These are known as the arcsine or arccosine functions. They take a value on the axes and convert it into possible distances around the unit circle for those functions.

But, here's the thing, the function arcsin, doesn't know whether the point on the circle you are trying to find is to the left or right of the imaginary axis. Equally, the function arccosine doesn't know if the point is above or below the real axis. So the honest output of the function \(arcsine(imogen)\) should be either the distance to Polly OR Polly-A. Equally, the honest output of the function \(arccosine(raul)\) shoud be either the distance to Polly OR Polly-B. This is really the same as the honest answer to the square root of four being both plus and minus two. There are two equally good answers. Ask your calculator though, and it will only tell you about plus two. Ask your calculator about \(arcsine(imogen)\) and it will only tell you about distances on the right hand side of the imaginary axis. Ask about \(arccosine(raul)\) and it will only tell you about distances above the real axis.

We'll come back to this problem, but for now, we can see the only possible distance is where \(arcsine(imogen)=arccosine(raul)\). This is where the two dashed lines meet, and this must be the Polly we are after. Remember that if we only had one or the other of imogen or raul, we would be in trouble, because we wouldn't know which of the two possible Pollys was the correct one. Finally, lets work out how far around the circle you have to go from the starting point to get to Polly:

If you actually measure the distance around the circle to Polly, you will find it is exactly one eighth of the circle, or \(\tfrac{\pi}{4}\) radians. So that gives us our angle for the polar form:

\[2+2i=\sqrt{8}\angle \tfrac{\pi}{4}\]

Now, the problem at this stage is finding an easy way to express this as a general rule. Let's look at the function to turn a polar form complex number into a rectangular form. The polar form has two variables, the absolute value of the number, and the angle. We are using the variables \(r\) for the absolute value, and \(\theta\) for the angle. So the function to convert one to the other is going to have TWO variables going into it. We write that like this: \(f(r,\theta)\). And setting the whole thing out we get:

\[f(r,\theta)= rcos\theta + risin\theta\]

Now it makes perfect sense. Moving on, how do we write the same kind of function to show the relationship between Raul, Imogen and the distance around the unit circle to Polly? How do we, in otherwords, turn the step of “look for where the two lines cross, and that's your Polly” into mathematical notation? What we need is a function that takes a variable and then asks a question. Weird? Yes, but very useful. What am I talking about? Well, the function is going to have two variables going into it again. This time they are just the real and imaginary parts of the complex number. We'll call them \(a\) for the real part and \(b\) for the imaginary part. So our new function for converting a rectangular number to a polar one is going to be written like this: \(g(a,b)\). The function is going to start in any case with the absolute value of the complex number being set to the square root of \(a^2+b^2\), which is the same, whatever the angle.

But we cannot just write out a simple one line function can we? No, because we know that the exact length of \(\theta\) is going to depend on the reverse sine or cosine function of one part, and whether the other part is positive or negative. It is actually easier to work with the reverse of the cosine function for this, because it always produces a positive number. Basically the arccosine function always tells you how far you need to move away from the starting position, and then you just need to know which side of the real axis you are. You could work with the arcsine function, but it has the problem that it spits out negative distances. This is not a massive problem - a negative angle, just really means go round the whole circle and then back up that quantity, but it is an unnecessary hassle.

So, we can actually write out our function as follows:
\[\begin{equation*}
g(a,b)= \left\{ \begin{array}{lr}
\sqrt{a^2+b^2} \angle arccos\left(\tfrac{a}{\sqrt{a^2+b^2}}\right) & \text{if } b >= 0,\\
\sqrt{a^2+b^2} \angle 2 \pi-arccos\left(\tfrac{a}{\sqrt{a^2+b^2}}\right) & \text{if } b < 0.
\end{array} \right.
\end{equation*}\]

If that looks fucking horrendous, yes it is pretty much the worst thing we have seen in this adventure so far. Lets have a go at translating it into english shall we? It says that you have two options for finding the polar form of the rectangular number with real part \(a\) and imaginary part \(b\). No matter which option you choose, the absolute value is always the square root of \(a\) squared plus \(b\) squared. If the imaginary part is greater than or equal to zero, then the angle is the arccosine of the real part divided by the absolute value of the number. If the imaginary part is less than zero, then the angle is \(2\pi\) (or the full circle) minus the arccosine of the real part divided by the absolute value of the number.

Actually, lumping zero in with the positive imaginary parts is a red herring. It doesn't matter which club zero joins. If the imaginary part is zero, then Polly has to lie on the real number line. There are only two places on the real number line where polly can lie. So if the imaginary part is zero, then the value of the real part will tell you whether it is one or the other. In other words, \(arccosine(-1)=\pi\), and \(arccosine(1)=0\). So in either case it doesn't matter if you take the first option and ADD the result to 0, or start with \(2\pi\) and SUBTRACT the result, you end up in the same place. We are assuming, incidentally, that the position on the circle with a distance of \(0\) is the same as the position \(2\pi\). Which is a safe assumption to make for now.

So there you go, Polar to Recangular:
\[f(r,\theta)= r\cdot cos\theta + risin\theta\]
and Rectangular to Polar:
\[\begin{equation*}
g(a,b)= \left\{ \begin{array}{lr}
\sqrt{a^2+b^2} \angle arccos\left(\tfrac{a}{\sqrt{a^2+b^2}}\right) & \text{if } b >= 0,\\
\sqrt{a^2+b^2} \angle 2 \pi-arccos\left(\tfrac{a}{\sqrt{a^2+b^2}}\right) & \text{if } b < 0.
\end{array} \right.
\end{equation*}\]

Real or Imaginary Part 4

Last time we managed to graph the function \(f(x)=\frac{1}{1-x^2}\). First we calculated some values for the function around zero in intervals of a quarter. Then we drew a line and marked off these intervals to scale and then drew circles above each interval at a point that was to scale with the value of the function when we used that particular interval as the input to the function.

Now we are going to try that with the other function that we created. This is the other one that is exactly equal to the result of an infinite sum, but lets give it a different name to avoid confusion: \(g(x)=\frac{1}{1+x^2}\). The first job is to work out the values of the function at the same intervals as before, to get a rough idea about how this function behaves.

\[g(0)=1\]
\[g(\tfrac{1}{4})=\frac{16}{17}\]
\[g(\tfrac{1}{2})=\frac{4}{5}\]
\[g(\tfrac{3}{4})=\frac{16}{25}\]
\[g(1)=\frac{1}{2}\]

Hey look! This time the function is perfectly calculable at one. No breaking. What about a quarter more than one?

\[g(\tfrac{5}{4})=\frac{16}{41}\]
OK, so it is still going then. Let's do one more just to check.
\[g(\tfrac{3}{2})=\frac{4}{13}\]
Yep, still going strong, and it seems to be getting smaller and smaller. What about negative numbers as inputs to the function? Well again the only variable that we plug into this function is immediately squared, so it matters not a jot whether or not the input is positive or negative, the square is always going to be positive. So the function should be symmetrical around zero, because there is no difference in the output between a positive or negative input.

So let's set up the same graph as we did last time. This time we have calculated the values of the function at up to one and a half. We'll stop there for now, and make our line stretch from minus one and a half to plus one and a half. That makes our line look like this:

Now we have our line, lets mark off, using circles, the values of the function that we worked out above - and their negative mirrors which we know have the same value.

Those plotted points look very different to the set of points for the other function, don't they? Instead of a smiley face it is a moustache. There is a hump in the middle at \(x\) equal to zero and then the ends just seem to tail off. Let's fill in the gaps in-between our plotted values by getting the computer to calculate and draw the hundreds of points needed to form what looks like a line:

As we thought the line that appears bulges in the middle and then tails off at either end towards zero. Notice that it does not go nuts at plus and minus one by heading off to infinity, it just smoothly sails through those values.

As with the other function, we can construct estimates of our function from the first so many terms of the infinite sum. These should get closer and closer to our function just like when we first tested to see if we could get to \(\tfrac{4}{3})\) by adding up terms of the sum. Remember we saw that the more terms we added onto the infinite sum, the closer the total did indeed get to the answer we got from our function. We can do that again, this time starting by graphing the function \(g(x)=1-x^2\). That looks like this:

OK, that looks a bit different to our last one as well. Why? Well look what we do with our \(x^2\). We subtract it instead of adding it. That is why the line in this one disappears off below our number line instead of heading off to the top. You can see, as it was last time, that around the zero value the function we just drew is a pretty good fit but it all goes to hell shortly afterwards. Ok, let's add \(x^4\) onto our function and see where that takes us (in cyan):

Bloody hell, what's happened here then? Well the plus \(x^4\) drowns out the other parts of the function, so this time it stays positive. Again it looks like it is trying to match the real function around the zero point but then it heads off at high speed up the way. Lets now subtract the next term which is \(x^6\). We'll draw resulting line in green:

That is starting to look distinctly odd. The green one seems to flatten out along the real line and then it spears off down the way. Like adding the previous term, the subtraction of the sixth power of \(x\) is now dominating, so the function is only very briefly positive. Also, can you see that the lines at either end head down the way at a much more pronounced angle than the blue line? OK, lets add \(x^8\) to the mix to see what happens. We'll do this one in red:

It looks much like we have come to expect. Very similar to the \(x^4\) addition, but with the ends tilted in much more - like the previous one where we subtracted \(x^3\). What is going on with these line ends? They are acting much like the estimate lines in the previous function when they were getting closer and closer to the real line that went off to infinity. But HERE the real line doesn't go off to infinity at all. Just to make what seems to be happening clearer, I'll graph the functions that would result if we worked out all the terms to our sum up to subtracting \(x^{18}\) in green, and up to adding \(x^{20}\) in red. We'll take the other functions we've drawn out, apart from the real one, so we can see what is going on:

That's pretty clear. The more and more powers of \(x\) that you add on to the sum, the faster and more dramatically they head off to positive and negative infinity. And look where they are doing it! Right at plus and minus one. But why is this? The real function that short cuts to the answer is nice and smooth at plus and minus one. In fact it is nice and smooth all along the graph. Why on earth does the shortcut to the answer stay normal at plus and minus one, but the actual totals of the infinite sum go daft at plus and minus one?

The answer is that the number \(x\) that we plug into the function to give us our graph is, like every number, made up of a real and an imaginary part. What we are seeing at plus and minus one on that graph above, is the effect of the imaginary part of \(x\) sending the function off to infinity. I thought in school that imaginary numbers were just tacked onto real numbers to answer questions like what is the square root of minus one. In other words, that they were an artificial invention of mathematicians. But here you have no square roots of negative numbers at all. You just have plus and minus various powers of \(x\). Nevertheless though, the imaginary part of \(x\) is making itself known by sending the totals of the sums closer and closer to infinity at plus and minus one. So imaginary numbers are not imaginary at all - they exist and have an effect even when we don't invoke them, or expect them, at all.

Why does this happen? Well, very basically, remember that each number has a real and imaginary part. With the function \(f(x)=\frac{1}{1-x^2}\) when the REAL part of \(x\) is plus or minus the square root of one (i.e. either positive or negative one), you square it, getting one, which you then subtract from one getting zero on the bottom of the equation which breaks it. With \(g(x)=\frac{1}{1+x^2}\), when the IMAGINARY part of \(x\) is plus or minus the square root of minus one (i.e. plus or minus \(i\)) you then square the square root of minus one getting ... minus one, you cretin. When you add minus one to one and you get ... zero on the bottom of the equation which breaks it just as badly as when the REAL part of \(x\) is zero.

But hang on I hear you cry, the function goes nuts around the REAL numbers plus and minus one. As I was at pains to say this number line is for real numbers only. It does not show imaginary numbers. If the imaginary part of a number is responsible for sending the function off to infinity, then the whole number must square to negative one. If it does so then its real part must be zero. In other words why doesn't the function go daft at the real value zero?

The answer involves the absolute value of numbers and something called the circle of convergence. Which sounds like it should be drawn on a dusty basement floor in pigs blood and surrounded by hooded figures illuminated by large dribbly white candles. Both of these concepts will require pictures. More of which next time.

Real or Imaginary Part 3

Last time we took two infinitely long sums and we worked out general equations which allow us to jump straight to the answer without having to add up an infintely large amount of individual terms. We have expressed our series as a list of \(x\) variables to various powers, and unsurprisingly our formulas are also in terms of the \(x\) variable. Because we are working with formulas we can now call them functions which will be \(f(x)\) and \(g(x)\). This is because there is only one variable in the formula - the \(x\) bit.

Remember that an infinite sum can either add up to some specific number, or it can go nuts and head off to infinity. Now that we have these to functions, if you want to know what will happen with a specific infinite sum is to pick a number, any number, stick it in the function in place of the \(x\). The output of the function will tell you whether or not the infinite sum for that number adds up to any specific number, or disappears off to infinity. When I say "pick a number" this would be the number that would appear in place of the half in the brackets in this description of the sum:

\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\]

What we are interested in, for the purpose of thinking about imaginary numbers, is at what point the infinite sums we are looking at go from a finite to an infinite answer. When I say "what point" I really mean which numbers when plugged in in place of \(x\) give infinite answers and which give finite answers. What we could do to find this out is write down a list of the numbers we are interested in and, next to them, the output from the function we calculated last time. We could then consider the answers. We've already worked out \(f(x)=\frac{1}{1-x^2}\) is four thirds, so let's try numbers around a half to start with.

\[f(0)=1\]
\[f(\tfrac{1}{4})=\tfrac{16}{15}\]
\[f(\tfrac{1}{2})=\tfrac{4}{3}\]
\[f(\tfrac{3}{4})=\tfrac{16}{7}\]
\[f(1)=\]
Oh dear, it is broken. We cannot divide by zero, which is what one minus one squared is. We could consider this to be an infinite answer. And because we square the \(x\) component, we know that whether the \(x\) starts as positive or negative it will square to the same number. In other words we know that the result for \(f(-\tfrac{1}{4})\) is still going to be \(\tfrac{16}{15}\)

That all looks a bit boring though, and it hardly gives us a feeling for what is actually happening with the output of the function as the numbers that we put into it change. Lets make things a bit more dramatic by making them graphic! How do we do that then? Lets draw a line just like when we were making our number line:

Now lets mark the centre of the line as zero:

Now let's mark off the other values for which we have just calculated the function along the line which passes through zero, using the left of zero as negative numbers and the right of zero as positive numbers. Lets make these all proportionate, so the distance between a half and one is the same as negative three quarters and negative one quarter. In other words lets mark these off to scale:

Note that this is just a one dimensional line. It does not have an imaginary vertical axis like our number plain. Any dots that appear above the number line are NOT imaginary numbers. Instead we will mark off the \(f(x)\) values for each of the values on the number line. We will do so by just placing a dot the same distance above the line that the value turned out to be. So for instance, if you stuck a half into the function the output was four thirds. So our dot will be four thirds of a unit above the line. A "unit" here is just the distance from zero to one on the line. This is just going to be a simple graph to give us a visual idea of how the output of the function behaves depending on changes to the input. Let's start with \(f(\tfrac{1}{2})\):

See the dot? That represents our value for \(f(\tfrac{1}{2})\). It is four thirds of a unit directly above the marker for a half along the line. Lets fill in dots for all the other values that we know about:

Hey! We got a smiley face! Can we see what is happening now though? The dots start off high (technically infinite at negative one) and get lower as they get closer to zero. Then they start getting higher again. Of course we have only marked off seven values. Those seven values fall on an invisible line that would appear if we calculated ALL the infinitely many \(f(x)\)'s between negative one and one. We can still draw that line though, by telling the computer to calculate hundreds of different \(f(x)\)'s. We are still drawing dots, but now we are drawing so many that it will just look like a line. That line looks like this:

The black lines at the sides are just the computers way of trying to draw infinity. It just draws that as a line going directly up and down. But, do you see the rest of the curved line going through all the dots that we plotted? That line is called the plot of the function \(f(x)=\frac{1}{1-x^2}\). This gives you a much better idea of what is going on. As we calculate values closer and closer to plus and minus one, the function just takes off towards infinity. To give an example of this, lets calculate two more values of \(x\) - between the two extremes on the outside (plus and minus three quarters) and plus and minus one. If we make them mid point between those, we will end up with plus and minus seven eighths. Which is fine.

What we should see is plus and minus one divided by one minus seven eighths squared. Seven eighths squared is seven times seven (forty nine) divided by eight times eight (sixty four). Take that away from one and you get fifteen sixty fourths. One divided by fifteen sixty fourths is the same as sixty four fifteenths. So the answer at plus and minus seven eighths is sixty four fifteenths. That looks like:

The image is now much taller, to fit in the dots getting higher and higher. See how the gap between three quarters and seven eighths is much bigger than all the other gaps. I also had to make the text at the bottom a bit smaller so it would all fit in.

Just in case we are getting too far away from where we started, lets remember that the right most dot in that picture represents the total that you get if you add up:

\[1+(\tfrac{7}{8})^2+(\tfrac{7}{8})^4+(\tfrac{7}{8})^6+(\tfrac{7}{8})^8+\ldots\]

What we can also do is draw a plot on the graph that represents the first so many terms of the infinite sum. Think back to when we were first checking that we had successfully turned an infinite sum into a function. Remember the bit where we added up the first few terms of the infinite sum by working out what they were:

\[1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}=\frac{64+16+4+1}{64}=\tfrac{85}{64}\]

And then we asked if they were getting close to our prediction of \(\tfrac{4}{3}\). So we asked what \(\tfrac{4}{3}\) was in sixty-fourths. We chose sixty-fourths because they were the smallest fraction that we had reached so far in the infinite sum. We got the answer: \(\frac{85\tfrac{1}{3}}{64}\), so we confirmed that we were moving towards \(\tfrac{4}{3}\) as we added more terms to the infinite sum.

We can now do that using the graph, by plotting functions that gradually have more and more terms of the infinite sum in them, to see if we get close to the prediction. Last time we were doing it with individual numbers, this time it is whole functions. To start, we plot the function \(f(x)=1+x^2\). That looks like this:

I have drawn it in blue so it stands out. As you can see it is really close to the proper curve down by zero, but it gets less accurate quickly, the further you move out. What about adding another term in a different colour? Lets do \(f(x)=1+x^2+x^4\) in Cyan:

As we may have expected the line gets close again to the true total of the infinite sum. Lets add two more powers of \(x\): \(f(x)=1+x^2+x^4+x^6\) and \(f(x)=1+x^2+x^4+x^6+x^8\) in green and red respectively:

Again, much as we expected, the lines are getting progressively closer to the "true" line.

So what does this tell us? Well, the functions we are generating by adding more powers of \(x\) are getting closer and closer to infinity at plus and minus one. We know why. It is because the function for working out the actual answer to the infinite sum goes off to infinity at that very point. This is because you end up dividing by zero at plus and minus one. Because the "true" function goes to infinity, then all our estimates of that function get closer and closer to infinity at that point. All we are doing each time we add a new power of \(x\) to the estimate term is getting one step closer to the real answer, which is demonstrated by the real line.

So what does this have to do with imaginary numbers? To find out we will have to look at what happens when we try to graph our OTHER function: \(g(x)=\frac{1}{1+x^2}\)

Real or Imaginary Part 2

OK, last time we looked at these two formulas:
\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\]
and
\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\cdot-1^n\]
We worked out that the first one adds up to \(\tfrac{4}{3}\) and the second one \(\tfrac{4}{5}\).

Now lets replace the \(\tfrac{1}{2}\) in each one with \(x\). This makes both sums:
\[\sum_{n=0}^\infty x^{2n}\]
and
\[\sum_{n=0}^\infty x^{2n}\cdot-1^n\]
When we do the expansion they are going to look like this:
\[1+x^2+x^4+x^6+\ldots\]
and
\[1-x^2+x^4-x^6+\ldots\]
We can then set both sides equal to a variable. Best give them slightly different names so we do not get too confused doing this simultaneously. I think we can also dispense with the "ands".
\[S_a=1+x^2+x^4+x^6+\ldots\]
\[S_b=1-x^2+x^4-x^6+\ldots\]
OK. Now multiply both sides of the equation by \(x^2\). Hang on though - didn't we divide by four last time? Well, yes, but dividing by four is the same as multiplying by \(\tfrac{1}{4}\) and \(\tfrac{1}{2}^2=\tfrac{1}{4}\), so this step is actually the same as we did last time. It gives us:
\[x^2\cdot S_a=x^2\cdot (1+x^2+x^4+x^6+\ldots)\]
\[x^2\cdot S_b=x^2\cdot (1-x^2+x^4-x^6+\ldots)\]
If we then multiply out the right hand side we get:
\[x^2\cdot S_a=x^2+x^4+x^6+x^8+\ldots\]
and
\[x^2\cdot S_b=x^2-x^4+x^6-x^8+\ldots\]
Now we have defined our long series of fractions we can substitute into \(S_a\) and \(S_b\) above.
\[S_a=1+x^2\cdot S_a\]
\[S_b=1-x^2\cdot S_b\]
We then subtract, or as the case may be add the \(x^2\) multiple of \(S_{a/b}\) to both sides:
\[S_a-x^2\cdot S_a=1\]
\[S_b+x^2\cdot S_b=1\]
We then look at the left hand side of the equation and we spot that the \(S\) term appears twice. We can actually reform the left hand side by working out what we need to multiply by the \(S\) term. Here it looks like this:
\[S_a(1-x^2)=1\]
\[S_b(1+x^2)=1\]
We then divide both sides by the expression in the brackets. (Look carefully and if you multiply the stuff in brackets by the \(S\) term, you get the same left hand side as we had in the last step.
\[S_a=\frac{1}{1-x^2}\]
\[S_b=\frac{1}{1+x^2}\]
What we have now achieved are two functions of \(x\). You can see this because only the right hand side has an \(x\) in it. We have named these things already, but now they are functions, lets give them proper names:
\[f(x)=\frac{1}{1-x^2}\]
\[g(x)=\frac{1}{1+x^2}\]
OK. Lets check that we have done this right by sticking in \(\tfrac{1}{2}\) for \(x^2\) in each equation and seeing if we get the same results as last time:
\[f(\tfrac{1}{2})=\frac{1}{1-\tfrac{1}{2}^2}\]
\[g(\tfrac{1}{2})=\frac{1}{1+\tfrac{1}{2}^2}\]
Then:
\[f(\tfrac{1}{2})=\frac{1}{1-\tfrac{1}{4}}\]
\[g(\tfrac{1}{2})=\frac{1}{1+\tfrac{1}{4}}\]
Then:
\[f(\tfrac{1}{2})=\frac{1}{\tfrac{3}{4}}\]
\[g(\tfrac{1}{2})=\frac{1}{\tfrac{5}{4}}\]
And remember that dividing by a number is the same as multiplying by the reciprocal:
\[f(\tfrac{1}{2})=1\cdot \tfrac{4}{3}\]
\[g(\tfrac{1}{2})=1\cdot \tfrac{4}{5}\]
And finally:
\[f(\tfrac{1}{2})=\tfrac{4}{3}\]
\[g(\tfrac{1}{2})=\tfrac{4}{5}\]
Which is exactly what we worked out last time. So our formulas work just fine. Now that we have a formula for working out the solution to the infinite sum based on one variable, we can make a graph of that function and it will show us at what point our infinite sum goes from a sensible proper answer to an infinite answer. That fun will have to wait for next time. For now, lets sum up (haha) what we have established so far:
\[\sum_{n=0}^\infty x^{2n}=\frac{1}{1-x^2}=f(x)\]
and
\[\sum_{n=0}^\infty x^{2n}\cdot-1^n=\frac{1}{1+x^2}=g(x)\]

Real or Imaginary Part 1

OK, so we just made up a number \(i\) which if you multiplied it by itself you get negative one. That sounds like the laziest way of solving the problem of the square root of negative one - just make shit up. Does it actually exist (if any number can be said to exist)? let's see if I can convince you it does.

First of all remember our infinite sums. We have decided that some of them add up to infinity, and others add up to a finite number, like two. I want to consider two more infinite sums. Lets look at these:
\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\]
and
\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\cdot-1^n\]
Right. Those look a bit odd. First of all they do not have \(x\) in them, it has been replaced with \(n\). Why? Well, we'll come back to that. I want to do a general version of these two in due course, where I will replace \(\tfrac{1}{2}\) with \(x\), so I don't want to confuse things by having \(x\) appear now.

So what do these look like expanded out? Lets work through the first one. For \(n\) equal to zero, we get one, because the exponent is two times zero, which is zero and everything to the power of zero is one. So we have:
\[1+\ldots\]
Good. Now lets go to \(n\) as one. Now we have two times one as the exponent, which is just two. So we need to square a half. Half of a half is a quarter. So we are now up to:
\[1+\tfrac{1}{4}+\ldots\]
Great. OK, with \(n\) as two, we are going to have the exponent equal to two times two, or four. This means that we need to do a half of a half of a half of a half, or a sixteenth.
\[1+\tfrac{1}{4}+\tfrac{1}{16}+\ldots\]
Fantastic. Lets do one more with \(n\) equal to three. We get an exponent equal to six, and I will just tell you that a half to the power of six is a sixty-fourth.
\[1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+\ldots\]
This looks quite like our previous infinite sum, in that the amounts added are getting smaller very quickly. In fact each term is a quarter of the last term. We can work out what number this is going to approach. We do so in the same way as before. Lets pick a name for the series \(S\):
\[S=1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+\ldots\]
Now let us divide both sides of the equation by four:
\[\frac{S}{4}=\frac{1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+\ldots}{4}\]
Again, we can just divide each term on top of the large fraction by four. Just like we saw before what will happen is that the one drops off the end and all the rest of the infinite terms march one place to the left. But an infinitely long list of numbers less one is still an infinitely long list of numbers.
\[\frac{S}{4}=\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+\tfrac{1}{256}+\ldots\]
Again we can spot that this term \(\frac{S}{4}\) is what we add to one to get \(S\) above. I mean, you look up to the first place where we say what \(S\) is. \(S\) is one plus a whole long list of fractions. If you compare that long list of fractions to the list that we now know is equal to \(\frac{S}{4}\), you should see that they are identical. So we can just replace the long list of fractions with \(\frac{S}{4}\), which tidies things up dramatically.
\[S=1+\frac{S}{4}\]
If that is the case then to get rid of the fraction we can just multiply both sides by four:
\[4\cdot S=4\cdot(1+\frac{S}{4})\]
\[4S=4+S\]
We can then subtract an \(S\) from both sides.
\[3S=4\]
Finally we can divide both sides by three to find out what \(S\) actually is:
\[S=\tfrac{4}{3}\]
Does this make sense? Lets look at the results of adding up the terms that we worked out above:
\[1=1\]
\[1+\tfrac{1}{4}=\frac{4+1}{4}=\tfrac{5}{4}\]
\[1+\tfrac{1}{4}+\tfrac{1}{16}=\frac{16+4+1}{16}=\tfrac{21}{16}\]
\[1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}=\frac{64+16+4+1}{64}=\tfrac{85}{64}\]
What number are those results moving towards? Well let me put it this way. What is \(\tfrac{4}{3}\) in sixty-fourths?
\[\frac{85\tfrac{1}{3}}{64}\]
So, yes we are moving towards \(\tfrac{4}{3}\) as we add more terms. OK, so let's look at the second infinite sum I put up there.
\[\sum_{n=0}^\infty (\tfrac{1}{2})^{2n}\cdot-1^n\]
The only difference is that it multiplies each term by \(-1^n\). What effect does that have on proceedings? Well, remember that the zeroth term (where \(n=0\)) will be multiplied by \(-1^0\) and ANYTHING raised to the power of zero is one. So the term is multiplied by one, so remains the same. But what happens when \(n=1\)? Well negative one to the power one is just negative one. So whatever was our next term will be multiplied by negative one. That means we are going to have to subtract it instead of adding it. The next term will be multiplied by negative one squared, which is one (as we saw before), so it stays the same. The next term will be multiplied by negative one cubed. What will that be? Well, one way to think about it is negative one squared multiplied by negative one. We know what negative one squared is - it is just one. One multiplied by negative one is negative one. So we will end up subtracting this term instead of adding it.

Can you see what is going to happen here? This \(-1^n\) multiple that I have introduced is going to alternate between one and negative one all the way to infinity. The effect is that it is going to flip every other addition sign between the terms in our equation to a subtraction. But that is all. So instead of the term:
\[S=1+\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+\ldots\]
We are going to get:
\[S=1-\tfrac{1}{4}+\tfrac{1}{16}-\tfrac{1}{64}+\ldots\]
We can divide both sides by four again to get this:
\[\frac{S}{4}=\frac{1-\tfrac{1}{4}+\tfrac{1}{16}-\tfrac{1}{64}+\ldots}{4}\]
BUT look what happens when we divide through all the terms. Each term moves to the left BUT IT LEAVES ITS SIGN BEHIND!
\[\frac{S}{4}=\tfrac{1}{4}-\tfrac{1}{16}+\tfrac{1}{64}-\tfrac{1}{256}+\ldots\]
So this time when we look up, we cannot spot a duplicate of our long list of fractions because the signs have switched. What to do?

Well, lets look at what happens when you take a series of additions and subtractions and deduct them all from zero. Lets use letters to stand in for any numbers at all. I am going to use \(a,b,c,d\) to save confusion. Lets start with:
\[a-b+c-d=E\]
Now, lets take away both sides from zero:
\[0-(a-b+c-d)=0-E\]
Now get rid of the brackets:
\[(0-a)+(0--b)+(0-c)+(0--d)=0-E\]
\[-a+b-c+d=-E\]
We know what \(E\) is, so we can write:
\[-a+b-c+d=-(a-b+c-d)\]
So can you see what happens when you negate a whole string of numbers? Each number in the string flips its sign from positive to negative. Just like we saw with our long list of fractions in this case. So we can actually say that:
\[S=1-\frac{S}{4}\]
Remember that the FIRST term in \(\frac{S}{4}\) is \(+\tfrac{1}{4}\) so the NEGATIVE sign between the one and the \(\frac{S}{4}\) turns the \(+\tfrac{1}{4}\) into \(-\tfrac{1}{4}\), and vice versa for all the other entries in that string of fractions. Guess what? All the signs have now switched back again - hooray! So what next? Well lets multiply through by four again:
\[4\cdot S=4\cdot (1-\frac{S}{4})\]
\[4S=4-S\]
This time we have to ADD \(S\) to each side:
\[5S=4\]
And then divide both sides by five:
\[S=\frac{4}{5}\]
Again lets check that makes sense:
\[1=1\]
\[1-\tfrac{1}{4}=\frac{4-1}{4}=\tfrac{3}{4}\]
\[1-\tfrac{1}{4}+\tfrac{1}{16}=\frac{16-4+1}{16}=\tfrac{13}{16}\]
\[1-\tfrac{1}{4}+\tfrac{1}{16}-\tfrac{1}{64}=\frac{64-16+4-1}{64}=\tfrac{51}{64}\]
What number are those results moving towards? Well let me put it this way. What is \(\tfrac{4}{5}\) in sixty-fourths?
\[\frac{51\tfrac{2}{5}}{64}\]
Yay!
OK, next time we'll reintroduce the \(x\) variable in place of the \(\tfrac{1}{2}\) that we used this time.