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.
No comments:
Post a Comment