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:
\[\begin{equation*}
g(a,b)= \left\{ \begin{array}{lr}
\sqrt{a^2+b^2} \angle arccos\left(\tfrac{a}{\sqrt{a^2+b^2}}\right) & \text{if } b >= 0,\\
\sqrt{a^2+b^2} \angle 2 \pi-arccos\left(\tfrac{a}{\sqrt{a^2+b^2}}\right) & \text{if } b < 0.
\end{array} \right.
\end{equation*}\]
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