The easy choice is: . This is monotonically increasing. and .

I needed to scale three such quantities and mush them together. I thought it’d be spiffy then to have three different functions that satisfy my criteria. The next logical choice was . It is everywhere positive and increasing. And, it has .

Now, I needed third function that was always positive, always increasing, had and . One choice was: . But, that seemed like overkill. It also meant that I really had to keep my tiny if I didn’t want to scale things into the stratosphere.

Playing with hyperbolas

So, I thought… why don’t I make a hyperbola, rotate it, and shift it so that the apex of one side of the hyperbola is at . And, I can adjust the parameters of the hyperbola so that . After a variety of false starts where I tried to keep the hyperbola general until the very end (, rotated by degrees, and shifted by ), I quickly got bogged down in six or seven incredibly ugly equations in eight or nine variables.

So, it was time to start trying to make it easy from the beginning. I noticed that if was going to rotate it by an angle in the clockwise direction, then I needed to be such that if my slope was going to work out right in the end. So, I’m looking at the basic triangle on the right then to determine all of my sines and cosines and such.

Based on that triangle, it was also obvious that the asymptote for my starting hyperbola had to have . I played around a bit then with making and . In the end, I found things simplified sooner if I started with and .

I also needed to be such that the point rotate up so that its -coordinate was . This meant that or .

So, my starting hyperbola then was: .

From there, I had to rotate the and by in the clockwise direction. This gave me:

A little multiplying out leads to:

From there, using , , and , we come to:

The only step remaining was to shift it all over so that when , we end up with . Plugging in , we see that . That equation balances when . We want it to balance when , so we’re going to substitute in place of the in the above equation to get:

We can easily verify that the point is on the curve. And, we can implicitly differentiate the above to get:

This is pretty good as it goes. The only step is to complete the square to find a nicer expression for in terms of . We start by adding to both sides to get:

This is easy enough to solve for by taking the square root of both sides and shuffling some things about:

Here are all three curves with . The exponential is in black, the hyperbolic tangent is in red, and the hyperbola is in blue:

The first image on the page here was made with Zach’s Vecto library with some post-processing in the GIMP. (Here is the source file: hyperbola.lisp.) The second image was made entirely within the GIMP. And, the last image was made using Foo Plot and the GIMP.

Many years back, I wrote some ambient music generation code. The basic structure of the code is this: Take one queen and twenty or so drones in a thirty-two dimensional space. Give them each random positions and velocities. Limit the velocity and acceleration of the queen more than you limit the same for the drones. Now, select some point at random for the queen to target. Have the queen accelerate toward that target. Have the drones accelerate toward the queen. Use the average distance from the drones to the queens in the -th dimension as the volume of the -th note where the notes are logarithmically spaced across one octave. Clip negative volumes to zero. Every so often, or when the queen gets close to the target, give the queen a new target.

It makes for some interesting ambient noise that sounds a bit like movie space noises where the lumbering enemy battleship is looming in orbit as its center portion spins to create artificial gravity within.

I started working on an iPhone application based on this code. The original code was in C++. The conversion to Objective C was fairly straightforward and fairly painless (as I used the opportunity to try to correct my own faults by breaking things out into separate functions more often).

Visualization troubles

The original code though chose random positions and velocities from uniform distributions. The iPhone app is going to involve visualization as well as auralization. The picture at the right here is a plot of five thousand points with each coordinate selected from a uniform distribution with range [-20,+20]. Because each axis value is chosen independently, it looks very unnatural.

What to do? The obvious answer is to use Gaussian random variables instead of uniform ones. The picture at the right here is five thousand points with each coordinate selected from a Gaussian distribution with a standard-deviation of 10. As you can see, this is much more natural looking.

How did I generate the Gaussians?

I have usually used the Box-Muller method of generating two Gaussian-distributed random variables given two uniformly-distributed random variables:

But, I found an article online that shows a more numerically stable version:

For a quick sanity check, I thought, let’s just make sure it looks like a Gaussian. Here, I showed the code in Lisp, but the original code was in Objective-C. I figured, If I just change the function declaration, I can plop this into a short C program, run a few thousand trials into some histogram buckets, and see what I get.

The trouble with zero

So, here comes the problem with zero. I had the following main loop:

I now present you with three different implementations of the val_to_bucket() function.

As you can probably guess, after years or reading trick questions, only the last one actually works as far as my main loop is concerned. Why? Every number between -1 and +1 becomes zero when you cast the double to an integer. That’s twice as big a range as any other integer gets. So, for the first implementation, the middle bucket has about twice as many things in it as it should. For the second implementation, the first bucket has more things in it than it should. For the final implementation, the non-existent bucket before the first one is the overloaded bucket. In the end, I used this implementation instead so that I wouldn’t even bias non-existent buckets:

• using complex numbers for rotating, translating, and scaling the plane
• using Clifford algebras for rotating, translating, and scaling the plane, and
• using quaternions for rotating, translating, and scaling three-dimensional space.

Today, it is time to tackle rotating, translating, and scaling three-dimensional space using Clifford algebras.

Three dimensions now instead of two

Back when we used Clifford algebras to rotate, translate, and scale the plane, we were using the two-dimesional Clifford algebra. With the two-dimensional Clifford algebra, we represented two-dimensional coordinates as . It shouldn’t surprise you then to find we’re going to represent three-dimensional coordinates as .

As before, we will have and . Similarly, we will have . In the two-dimesional case, we showed that . By the same logic as the two-dimensional case, we also find that and . We could potentially also end up multiplying , , and all together. This isn’t going to be equal to any combination of the other things we’ve seen so we’ll just leave it written .

Using the identities like for , we can shuffle any combination of , , and so that all of the indexes are in order. We may have to change the sign, but we won’t have to change anything else. For example:

Recall that was a bivector. Here, we have added one more vector: , two more bivectors: and , and a trivector . The general element of the three-dimensional Clifford algebra looks like:

For the two-dimensional Clifford algebra, we had terms in the general element—the Clifford algebra is an algebra on a four-dimensional vector space. For the three-dimensional Clifford algebra, we have terms in the general element. If we backtrack here to the one-dimensional Clifford algebra, we see that the general element looks like meaning there are terms in the general element of the one-dimensional Clifford algebra. If we backtrack even further, we see that for a zero-dimensional Clifford algebra, the general element looks like meaning there is term in the general element of the zero-dimensional Clifford algebra.

This relation is going to hold for Clifford algebras in general. If is a non-negative integer, then there are terms in the general element of the -dimensional Clifford algebra. The -dimensional Clifford algebra is an algebra over a -dimensional vector space. There are degrees of freedom, if you like. We won’t be using all of those degrees of freedom because we are restricting ourselves to scaling, rotating, and translating vectors.

Translating and Scaling

We are going to represent our points in three-dimensional space using the vector portion of our three-dimensional Clifford algebra . So, a point with coordinates will be .

Just as we did in the two-dimensional case, we can scale three-space away from the origin by a factor of just by multiplying each point by the real number . So, .

Similarly, we can translate the point by units in the x-direction, units in the y-direction and units in the z-direction by creating a translation vector and adding it to each point so that .

Rotating three-space

When we used Clifford algebras to rotate the xy-plane, we let and let . Then, to rotate a point we multiplied out .

Three-dimensional space contains the xy-plane. We want to be able to rotate things in the xy-plane in just the same way we did in the two-dimensional case. We even spent some time in another article verifying that these rotations in the xy-plane would leave coordinates on other axises alone.

Rotating space with the quaternions, we found that could rotate around an arbitrary axis. If we pick an axis that points from the origin to where , then we used the quaternion (where and just like we used the two-dimensional Clifford algebra rotation. We let and . Then, we could rotate a point by evaluating .

With Clifford algebras, we don’t mush together translations and rotations both into the vector portion. We keep translations in the vector portion, but we move rotations to the bivector portion. We’re going to rotate around the axis from the origin to (where by an angle by letting
and where and . Then, we can rotate a point just by multiplying on the left by and on the right by so that .

I wrote a little differently here than I did above for the general element. If I were to keep totally consistent with what I wrote above, I should write . Certainly, we can add in any order, so it is totally cosmetic to write it one way or the other. I chose to keep , , and in order rather than , , and . I also chose to keep them all positive at the expense of using instead of .

Before we get into why we used instead of for the , let’s look at the case of rotating around the z-axis. In that case, would be . This should degenerate to exactly what we had for rotating the xy-plane with Clifford algebras. Sure enough, if we plop those values in for , , and we find that exactly as we had in the two-dimensional case. Bonus!

Why should the end up with instead of ? The way that our axis system worked with the quaternions is that we had the x-axis going east, the y-axis going north, and the z-axis going up. If we are going to rotate around the y-axis, then we need to rotate the z-axis toward the x-axis as opposed to rotating the x-axis toward the z-axis. The rotation around the z-axis takes the x-axis toward the y-axis. This is represented with the . is the x-axis and it is moving toward the y-axis . If we reverse the order, , then we are moving the y-axis toward the x-axis. Using our transposing tricks, . Thus, rotating the y-axis toward the x-axis is just negatively rotating the x-axis toward the y-axis. In better English, that is: rotating the y-axis toward the x-axis is just rotating the x-axis away from the y-axis.

If we had chosen instead of , then we would have messed up how our axises fit together. To keep the axis of rotation obvious, we used . We can still rewrite the result in any form we like. If we were given , that would be a rotation around the axis from the origin to the point . I think it’s more intuitive to write as because then you can read the axis of rotation right off of the coefficients.

Next time

Next time, we’ll tackle rotating, scaling, and translating hyperspace (four-dimensional space).

What are the quaternions

Okay, remember how we got the complex numbers? We needed something that was the square root of negative one.

Now, imagine that you are Sir William Rowan Hamilton. The year is 1843. It is springtime. You know how to use the complex numbers to represent points in the plane. And, you know that when you do that, you can use complex numbers to rotate, scale, and translate the points. That’s all well and good, but you don’t live in a two-dimensional world. How are you going to do the same sort of thing with three-dimensional space? How are you going to multiply triples?

You spend months on this. If only you could say, How about I let there be another number that is different from (and from ) that has the same property that its square is negative one? You fight with this for months. You try to represent a point with coordinates as . But, nothing you come up with makes any sense.

Your kids are harassing you, Daddy, did you figure out how to multiply triples yet? You have to answer them every morning with a polite, No, not yet.

Then, you’re walking along the Royal Canal in Dublin. It’s mid-October already. My, how the year has flown by. Bam, it hits you. If you add a third number like and which is equal to , everything works out. You get so excited, that you carve your equations into a stone bridge over the canal:

Now, instead of trying to represent as , it is represented as . Immediately, this looks like an improvement. Before, things were not symmetric. Squaring something that was only in the x-direction was fundamentally different than squaring something that was only in the y- or z-direction because , but . Now, at least, there is no singled-out axis.

Let’s just explore the part for a minute. If we multiply both sides (on the right) by , we see that: . If, instead, we multiply by on the left, we see that . If we take that and multiply on the right by , we get . We’re going to show in a minute that . So, if take , , and in that order and take two in a row (wrapping around if needed), the product is the next one: , , and .

So, how do we show that ? Well, it’s pretty simple to show that . We already know that . So, . We know that , so . And, we know that , so . And, since we already know that , we have . We can use this, then to manipulate . . We can do a similar thing to show that .

This should look familiar from the the Clifford algebras. In the Clifford algebras, . Here, . It turns out that this is extra important in three-dimensions compared to two.

In two dimensions, if you rotate by an angle and then by and angle , it is the same thing as if you rotate by the angle and then by the angle . In three-dimensions, you have more options. In addition to picking the angle, you can also pick the axis of rotation. (Actually, for later on, it may be helpful to think of picking the plane parallel to which the rotation will take place rather than thinking of it as an axis.) In the picture at the right, the die in the back is the original position of a die. The die on the left is the result of rotating the original die a quarter turn on its left face (orange arrow) and then a quarter turn on its right face (blue arrow). The die on the right is the result of rotating the original die around its right face (blue arrow) and then its left face (orange arrow). As you can see, the resulting positions are different.

[Ugh... I just realized the free model of a die that I downloaded is not an actual, legal die. If it were, the opposite faces would sum to seven. Clearly, however, the two and the five are on adjacent faces of this die, as are the three and four, and thus also the one and six. Erf. So, you can't repeat this experiment on your desktop unless you fabricate your own die. Sorry.]

Like the Clifford algebras and the complex numbers, addition and subtraction are simply adding and subtracting (respectively, of course) the corresponding parts.

Armed with the above identities, we have enough information to multiply together quaternions. Rather than do the full general case of , I’m going to do a simpler case to give you the flavor without wasting a ton of space in this article beating a horse that needn’t be beaten that much. Let’s just try: . We start with the distributive law, twice, and then combine like terms.

There is one other notation we’re going to need below. In the complex numbers, we used the notation to mean the number with the sign of its imaginary part reversed. So, if , then . With the quaternions, we are going to use to mean the number with the sing of all of its imaginary parts reversed. So, if , then .

Transforming Space

Say you have some list of points in three-dimensional space: , , , . We are going to represent those as pure quaternions: , , , .

If we want to translate them all by two units along the x-axis, three units along the y-axis and five units along the z-axis, we can set and then add to each of our points: .

If we want to scale three-space out from the origin by a factor of seven, we can set and multiply each point by : .

Let’s say now, that we want to rotate our points parallel to some plane through the origin by an angle . The first thing we need to find is some way to describe that plane. In three-dimensions, we can simply use the surface normal of the plane. It may make more sense to you to think of this as the axis of rotation. In two- and three-dimensions, that will be fine. In higher numbers of dimensions, this will just confuse things. Let’s say our normal starts at the origin and goes through the point . To make things nice, let’s also assume it is of unit length: . We can represent our rotation as . Then, we can rotate our points by multiplying on the left by and on the right by : .

By way of an example, let us suppose we are going to rotate the point by an angle , parallel to the xy-plane. We will expect that our z-coordinate remains the same, but that the other two coordinates change. First, let us let and . The normal to the xy-plane is just the z-axis. So, our is going to be . This leaves us to multiply out (and it’s going to get messy):

Notice how the and the cancel out. Also, notice how . So, we can simplify to an extent like this:

If we go back to the double angle identites, we find that while . So, our final formula is:

Denouement

I know that in the complex number case and in the Clifford algebra case, I used . In this case, I used (swapping the roles of and ). I could have used either in the complex number case since the complex numbers are commutative: for all complex numbers and . Thus, . I probably should have used for the complex case and still used for the Clifford algebra case.

There is a technical reason for the difference. This may sail right over your head if you haven’t done much hardcore linear algebra. But, when we are using the complex numbers or the quaternions to represent rotations, we are really jamming a linear form into a vector and pretending like they’re the same thing. We are glossing over the difference between a vector space and its dual space by just pretending they are the same. We can get away with that in the complex numbers and the quaternions, but we could not get away with it in the Clifford algebra case. This is because the three-dimensional bivectors make up the dual space of the vectors. So, when we’re using three-dimensional bivectors, we are explicitly using the dual space instead of taking advantage of the isomorphism between the dual space and the original vectors. When we get to four dimensions or more, the rotations aren’t even in the dual space.

For a more English, less Math, version of the above, look at it this way. Really, a rotation is something fundamentally different than a vector. We lucked out with the complex numbers and the quaternions in that we were still able to represent rotations with something like a vector (though in the quaternion case, it was a vector plus a scalar). The luck in the quaternion case isn’t perfect, it is off by a few signs here and there. In more dimensions, we won’t be so lucky. In three dimensions, there are three degrees of freedom for a rotation. In four dimensions, there are six degrees of freedom. We are never going to jam those six degrees of freedom into our puny four-dimensional vector.

Next time

Next time, we will do these transformations of three-dimensional space using Clifford algebras instead of quaternions.

Why did I do this? Well, I mentioned it would be awkward for the two-dimensional case, but that it will be important when we get to three or more dimensions. Well, work for a moment with being a quarter rotation (ninety degrees, radians). This means our total rotation is going to be a half turn (180 degrees, radians).

For that , and so . Let’s just look at what it does to our unit vectors and to multiply on the left by and on the right by .

For , we get . Similarly, for , we get .

So far, we were only working in two dimensions. As such, there wasn’t any to worry about. But, what if there were? What happens to the -coordinate of something if you rotate things parallel to the -plane? It remains unchanged.

Well, what would happen if we multiplied on the right by ? We would end up with . We’ve ended up scaling and adding in a trivector . We’ve made a mess.

Let’s try it instead with our trick. We’re going to start with . Every time we transpose elements with different subscripts, we flip the sign. Every time we get two elements next to each other with the same subscript, they cancel out. So, switching the with the second , we get . From there, we can switch the first two elements to get which is just . We can switch the with the second to get: which is just . So, our trick leaves unchanged.

In the above, there is nothing special about the subscript three. It would work for any subscript except one or two. So, the trick allows us to break the rotation up into two parts that still do what we want with and but leave our other directions unchanged (or, maybe it’s easier to think of them as changing them and then changing them right back).

What are Clifford Algebras

In our previous post, we used two different ways to represent coordinates in the plane. We used an ordered pair of real numbers like and we used the real and imaginary parts of a complex number like . Another way we could have written the coordinates in the plane is as a vector. Typically, to express a vector, we pick an axis (say, the x-axis) and then pick a second axis perpendicular to it (say, the y-axis). Most physics books would call these and . I am going to use and instead so that there is no confusion with and because it is a notation that we will continue throughout Clifford algebras. We could then express as . Here, and are called unit vectors. We assume they have length one so that when we take three of them, we get something with length three (for arbitrary values of three).

Having done that, we can easily multiply any vector by a real number using the normal distributive law: . And, we can add vectors just like we added complex numbers: summing like parts. So, . Of course, subtraction goes the same way: .

Great. We can add and subtract vectors. We can multiply vectors by real numbers. But, what about multiplying a vector by a vector? Well, let’s try it: . Again, we’re going to use the F.O.I.L. method. We end up with: . So far, I have assumed that we can multiply real numbers by our unit vectors in any order, but that maybe multiplying would be different than . Well, to get anywhere from here, we need to be able to evaluate things like and .

[Note: the F.O.I.L. method is nothing magical. It is simply the distributive law applied three times:

Actually, this leaves me a good place to mention... if you're better at hearing someone speak while doing math than you are reading along on your own, then you can review the complex arithmetic we did in the last post with these videos at KhanAcademy.org: i and Imaginary numbers, Complex Numbers (part 1), and Complex Numbers (part 2).]

So, where are we? If we want to multiply two vectors together, we need to decide how to multiply things like or . Well, there are probably lots of ways we could do this. But, we’re going to be trying to do geometric things with our vectors. What are some geometrically useful possibilities?

We could say that any vector multiplied by itself is just itself so that . This doesn’t really help us figure out anything about and how it relates to .

Another possibility is that parallel vectors annihilate each other so that while perpendicular vectors do not. If that were the case, then . And, . so that . This has some merits. This results in what is called the Exterior (or Grassmann) algebra. There is lots of interesting stuff to learn there, but Clifford algebras take a slightly different tack.

Another natural choice might be then that if we take some vector and multiply it by itself, maybe we should get a real number that is the square of the length of . So, if we have a vector , its length is . If we want the square of the vector to be , then we need to have the following:

To make this equation work out for every and , we need to have and . This is the essence of Clifford algebra. All of the rest is deducible from here. But, we’re going to be a bit more explicit on some of the things this implies.

We can reduce further, now. When we multiplied it all out, we got: . Knowing now that and , we can rewrite this: . So, we multiplied two vectors and got a real number plus a real number times this other thing: . We’re going to call this other thing a bivector.

A bivector?

Let’s explore bivectors a little more. In particular, what is ? Well, we’ve already decided that . So, we know that . And, we already decided that , so . And, we already decided that . So, . When we square our unit bivector, we get . Our bivector is exactly like our from the complex numbers!

Recall that . Now, let’s multiply out . We get which is .

So, we could stop right here if we wanted. We could represent our points in the plane as a real number plus a real number times and do everything we did with the complex numbers. But, if we did that, then we’ve totally skipped over using the vector part of our Clifford algebra. We are only using the bivector part.

Now we’ve got real numbers and vectors and bivectors. What if we multiply ? Well, since , we are just left with . How about if we multiply ? Here, we have to make use of , so we end up with . So, our vectors and bivectors are a little weird compared to what we’re used to. We’re used to having for real numbers (and even complex numbers). But, this doesn’t work out for vectors times vectors or vectors times bivectors. And, it turns out, that will be a good thing later. In two-dimensions, it isn’t quite as big a deal. [Well, if you're familiar with vector cross-products, then maybe it is a big deal to you since you already know that .]

Anyhow, an arbitrary element of the two-dimensional Clifford algebra is of the form for real numbers , , , and . It is a real number plus a two-dimensional vector with real coordinates plus a real number times our bivector .

Transforming the Plane

It is common/natural/intuitive?/easy? to think of the coordinates of a point in the plane as a two-dimensional vector from the origin to the point. As such, we’re just going to use the coordinates as the coefficients of the unit vectors of our general element . So, the point would be represented in our algebra as . This is called a pure vector since it has zero for its real part and zero for its bivector part.

Now, what if we had a whole list of points , , , ? We can represent these as elements of our Clifford algebra as , , , . If we want to translate all of them by two units along the x-axis and seven units along the y-axis, then we can let and just add to each of our points: .

Now, what if we want to scale the whole plane up by a factor of six radiating from the origin? We can just let and multiply it by each of our points: .

How about rotation by an angle around the origin? Well, drawing on what we did with complex numbers, let’s see if we can just get away with multiplying by . Recall that we want a point to go to a point when rotated.

Let’s try multiplying . Remember, we have to be careful to keep the ’s and ’s in the order they appear because every time we swap places on them, we need to swap the sign. When we multiply this out, we get:

This is really close. We just don’t have the sign right on the sine terms. Well, we could just use the fact that and . But, that would be funky weird to use when you want to rotate by . But, we’ve seen before how switching the order of multiplying things can change the sign. Let’s try this here. Instead of multiplying with the vector on the right, let’s multiply with the vector on the left:

There we have it. So, if we let , then we can rotate all of our points by angle around the origin using: .

And, of course, if we want to scale or rotate around different points, we can translate, scale or rotate, and translate back again like we did with the complex numbers.

Not so fast

It turns out that just multiplying with the vector on the left isn’t going to serve us well as we go up in dimensions. We’re going to do something that might look really awkward now, and will be somewhat awkward for this two-dimensional case, but it’s going to prepare us to do better things in three or more dimensions.

Let’s let . And, let’s let . You might recognize this as just like the complex conjugate of .

Let’s see what we get if we multiply by on the left and by on the right.

This is getting a bit large, so let’s just concentrate for a moment on the terms that are going to contribute to in the end: namely, the term in the first factor times the real part in the second factor and the term in the first factor times the bivector part in the second factor.

We can do a similar thing with the portions that contribute to :

Now, we have to remember some trigonometric identities. In particular, we need to remember the double angle identities: and . Taking advantage of those, we have our whole product:

So, if we want to rotate by an angle around the origin, we can let . Then, we can let . And, of course, if we want to rotate around a point instead, we can translate by , do this rotation, and translate back so that .

Now, we’ve used Clifford algebras to represent rotating, translating, and scaling the plane. In our next installment, we will move on to use quaternions to rotate three-dimensional space. From there, we will move to the Clifford algebra for three-dimensional vectors .

What are Complex Numbers

Say you’re playing around with polynomials. You start playing with the equation . WIth a little fiddling, you find this is equivalent to . Then, you take the square root of both sides to find that . We started with a polynomial equation in one variable in which the highest exponent was two and we found two answers.

Pounding your chest and sounding your barbaric yawp, you move on to . This should be easy, right? With the same fiddling, we find and then .

Uh-oh. What do we do now? We can’t think of any number that when multiplied by itself gives us a negative number. If we start with zero, we end with zero. If we multiply a positive number by itself, we get a positive number. If we multiply a negative number by itself, we get a positive number. Again!

So, how do we get around this? We pull an ace out of our sleeve. We just run with the idea that there is such a number and see where it takes us. We say, There is a number such that . Everything else is going to stay the same.

Where does this take us? It turns out, it takes us very, very far. For starters, our equation , a polynomial equation in one variable where the highest exponent is two, now has two answers: .

What about ? It is a polynomial equation in one variable where the highest exponent is two. It’d be pretty spiffy if there were two answers. With the same manipulation as before, we find that .

Now, we need to remember that if and are positive numbers, then . Let’s see what happens if we extend this to allow our new number . If we said that , then . What happens if we multiply ? When we multiply real number , , and , we can do it in any order. We could do or or (or three other orders). Well, let’s assume for now that when we multiply by a real number, we can do it in either order. Then . That’s exactly what we were hoping it would be.

Good. Our equation is a polynomial equation in one variable where the highest exponent is two and it has two solutions .

As it turns out, by adding in (and real number multiples of ) to our real numbers, we have the complex numbers. These complex numbers are an algebraic completion of the real numbers. That’s just a fancy way of saying that if you make a polynomial equation in one variable where all of the coefficients are real numbers and the highest exponent is , then there will be solutions to the equation all in the complex numbers.

[It turns out that the complex numbers are algebraically complete themselves. If you make a polynomial equation in one variable where all of the coefficients are complex numbers and the highest exponent is , then there will be solutions to the equation in the complex numbers.]

A Quick Review of Complex Arithmetic

Above, we decided to say that is and go from there. We also used the idea that to find square roots of all negative numbers. And, we already played around a little bit with multiplying some numbers together. Let’s take a step back for a moment though and just add.

We still want the rest of our algebra to work. Because of that, we don’t have much choice for how imaginary numbers add together. If we take any number and add it to itself, we get . We still want that to be true when or . In general, then, we will need for any numbers , , and .

Earlier, we multiplied . What if we add one to ? Well, we can definitely write this as just like we can add one to two by just writing . In the latter case, we already have a name for . We could instead write .

We don’t already have a name for . How do we know? Well, let us assume that is some real number or some real number times . If it is some real number, we should get a positive real number when we square it. If it is a real number times , we should get a negative real number when we square it. But, if we multiply (using the old Firsts, Outers, Inners, Lasts method), we get which is not a positive real number or a negative real number.

As it happens, all of our complex numbers will have the form: for some real numbers and . When we go to add two numbers together, we just add the corresponding pieces: .

For a complex number , we call the real part and the imaginary part. When we add two complex numbers, we add the real parts together and we add the imaginary parts together. Subtraction, likewise, goes by part. If we want , we subtract the real parts and subtract the imaginary parts to obtain .

When we multiply two numbers together, we do it like we did with above with the FOIL method. . We rearrange the orders of some of the bits as we did with above to get: . And since , we have:

Transforming the Plane

You’ll notice above that a complex number is made up of two real numbers (and, of course, ). Every point in the plane has two real numbers as coordinates. We can co-opt the real numbers in the complex number to use as coordinates for points in the plane. If our points are: , , , , then we can represent them as complex numbers with , , , .

We can plot these just as we would their normal coordinates if we put the real part on the x-axis and the imaginary part on the y-axis. This is called an Argand Diagram.

Suppose now we want to translate all of our points by five units along the x-axis and negative two units along the y-axis. We can simply let and then add this to each of our points so that .

Imagine instead that we want to scale the plane radially out from the origin by a factor of three. We can simply multiply each of our points by so that . If we want to scale outward from instead, we could translate by , scale by , and then translate back.

Suppose now that we’d like to rotate the plane by some angle around the origin. For that rotation, a point should get rotated to the point . That is to say: the complex number should get rotated to the complex number .

Let’s look back to our equation for multiplying two complex numbers:

We can see the similarity to the rotation. If we take , then we get precisely: .

The points are the points on the unit circle centered at the origin. So, if we want to rotate all of our points by an angle around the origin, we simply have to multiply them by the point on the unit circle that is at angle around the origin from the x-axis: so that . Again, if we want to rotate around the point , we simply translate by , rotate by , and translate back again.

So, now we can translate points in the plane with complex addition (or subtraction). We can scale points in the plane by multiplying by a real number. We can rotate the plane by multiplying by a complex number.

Conformal maps

There are some other transformations that naturally arise from complex arithmetic. A conformal transform is one that keeps angles constant. All of the transformations we’ve done above are conformal. If you translate the whole plane, the angles between lines are unchanged. If you scale the whole plane, the angles between lines are unchanged. If you rotate the whole plane, the angles between lines are unchanged.

As it happens, a transformation of the complex plane is a conformal map if and only if the transformation has a (complex) derivative everywhere and that derivative is non-zero everywhere. Without getting into complex derivatives here, suffice it to say, they’re pretty much just like real derivatives for simple polynomials. Let’s look at our cases above. The derivative of translating by is which exists everywhere and is non-zero everywhere. The derivative of multiplying by (or ) is (or ). This exists everywhere and is non-zero everywhere (if anywhere).

Another conformal transformation of the complex plane is the Möbius transformation. The Möbius transformations preserve angles where lines meet, but they generally turn lines into curves. The Möbius transformations are of the form: for complex numbers , , , and (so long as ). You can see the Möbius transform in action on YouTube.

What’s next

Next, we’re going to see how Clifford algebras can represent all we have done here. But, that is for another day.