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In the introductory complex analysis course I am taking, nearly every theorem relates to rotation and argument. Why do complex numbers love doing this so much? I can understand why these theorems work; however, aside from basic knowledge of polar coordinates, I do not intuitively understand what property of complex numbers make rotation and angle such a common/convenient idea.

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    $\begingroup$ It is straightforward to prove that $$r_1e^{i\theta_1} \times r_2e^{i\theta_2} = (r_1 \times r_2)e^{i(\theta_1 + \theta_2)}.$$ This assumes that $e^{i\theta} = \cos(\theta) + i\sin(\theta).$ $\endgroup$ Jul 7, 2023 at 2:30
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    $\begingroup$ If you look at the standard matrix representation of complex numbers, then $i = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. This is a generator for $SO(2)$. As soon as you have $i$ such that $i^2 = -1$, you have to have rotations. As to why the agebraic closure of the rationals should know about rotations, I can't really say. But in that connection it's interesting how many proofs of the FTA end up thinking about how a polynomial "rotates" its arguments. $\endgroup$ Jul 7, 2023 at 12:30
  • $\begingroup$ Related: math.stackexchange.com/questions/1364439/… $\endgroup$ Jul 7, 2023 at 21:23
  • $\begingroup$ Polar coordinates are pairs of radius and angle=argument. (Which is due to an old convention to write inverse functions as $\arg\{\cos(x)=y\}$ "the argument x that gives value y under the cosine. If you teach that long enough and recursively you get a chinese whispers effect and the angle gets the additional name "argument"). Complex units correspond to rotations. $\endgroup$ Jul 8, 2023 at 16:32
  • $\begingroup$ I think this is more or less suitable for hsm $\endgroup$
    – C.F.G
    Jul 10, 2023 at 6:13

9 Answers 9

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Another way to view this is thinking about complex multiplication as a linear map on $\mathbb{R}^2$.

Fix a complex number $c = a + ib$, then for any complex number $z = x + iy$, the product $cz$ is

$$ cz = (ax - by) + i(ay + bx). $$

It's easy to see that $z\mapsto cz$ is a linear map. What if we for a moment forgot about the complex structure and thought of $z$ as a vector in $\mathbb{R}^2$? Treat the real part as the first coordinate and the imaginary part as the second coordinate as per usual. Then, the product $cz$ is represented by the matrix vector product

$$ \begin{pmatrix} a & -b \\ b & a\end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix}. $$

(Check it yourself if you are not sure.) The matrix representing $c$ is of a very special form that may look familiar. If we remember that $c$ too represented a complex number and therefore corresponds to a vector as well, we can write $a$ and $b$ in a special way with some right triangle trigonometry: $a = |c| \cos(\theta)$ and $b = |c| \sin(\theta)$. Making these substitutions, we get that $cz$ can be represented by

$$ |c|\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix}. $$

This matrix is exactly a rotation matrix, so multiplying $z$ by $c$ rotates $z$ by the angle $c$ makes with the positive $x$-axis. Of course this answer is effectively equivalent to the others, but I like the light dose of linear algebra to help make more connections to other areas of math to further solidify understanding.

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    $\begingroup$ Another reason I like this approach is that it avoids any and all muss and fuss regarding complex exponentials which require secondary discussion. Of course they're hiding in the background, but they're not necessary for understanding the scaling and rotation geometry of complex multiplication. $\endgroup$ Jul 8, 2023 at 21:37
  • $\begingroup$ I've read a lot of answers on a lot of different questions, and this is the first one which made sense to me $\endgroup$
    – Joff
    Nov 10, 2023 at 5:48
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Rather than defining the complex numbers algebraically let's instead begin with the geometric notions we're familiar with from the real numbers and extend them to the plane. Towards this end we need to distinguish between a number as a label for a position and a number as an action or a verb. Let's call the position number $z$ and the action number $\alpha$ to keep the ideas separated even though they're both real numbers.

Now how does adding $\alpha$ act on $z$? Well if we look at $z + \alpha$ we can see that it is a translation. It moves $x$ by $\vert \alpha \vert$ units in the direction of $\alpha$. In two dimensions this abstracts nicely with a vector $z$ for position and a vector $\alpha$ for displacement by componentwise addition.

For multiplication if we consider $\alpha z$ we see that we're scaling $z$ by $\vert \alpha \vert$ and if $\alpha$ is negative it flips the line. We could interpret this as either a reflection or a rotation but reflections require us to choose a specific line of reflection which is somewhat artificial in two dimensions so we'll go with rotations instead.

Towards this end we want multiplication in two dimensions to have the property that $\alpha z$ scales $z$ by $\vert \alpha \vert$ and rotates it by the direction of $\alpha$. This means if we write $z=(r,\theta)$ and $\alpha = (s, \phi)$ in polar form that we want $\alpha z = (rs, \theta + \phi)$ where the radius is scaled and the angles are added.

The rotating and scaling can be characterized by how $\alpha$ acts on the unit circle. This should be familiar from trigonometry as we define sine and cosine functions are defined on the unit circle to make scaling arguments easier. In our scheme if $z$ is on the unit circle it's of the form $z=(1,\theta)$ so that $\alpha z = (s , \theta + \phi)$. This make it easy to recover the scaling factor $s$.

Now there are two points in cartesian coordinates on the unit circle which stand out besides the usual $\pm 1$ of the real line which are $(0,1)$ and $(0,-1)$ and are called $\pm i$ for historic reasons. This allows us to write numbers in the complex plain in the form $z= x + yi$ for real $x$ and $y$ to represent the point $(x,y)$. They should have two important properties which are that $\vert \pm i \vert =1$ and $(\pm i)^2=-1$.

But now we're done. We're reconstructed the cartesian form of our rotation numbers without having to appeal to algebra at all. Why does this work? Well if we consider the standard quarter turn rotation matrix from linear algebra it has characteristic equation $x^2+1$, so it has eigenvalues $\pm i$. So there is this subtle link between the geometric idea and the algebraic one that isn't entire obvious. Despite this the geometric ideas predate the complex numbers by at least a millenia. Most of the geometric ideas actually predate algebra entirely. So in a very natural way it's the better starting point. It's for this reason I prefer to call them the circle numbers. The $\mathbb{C}$ is just sitting there and it's much more fitting.

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    $\begingroup$ The intuitive explanation that I like the most goes like this: If you multiply a number by -1, you can identify that with rotating a point on a number line (in the plane) by a half-turn around 0. Two such rotations gets you back to where you started, which lets you conclude that -1^2=1. You can then identify sqrt(-1) with a quarter-turn, and find that both the algebra and geometry seem to match: sqrt(-1)^2 =-1, and two quarter turns make a half. That this continues to keep working is due to what you've written, but this is a simpler way to start with it. $\endgroup$
    – Roy
    Jul 7, 2023 at 21:32
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"I do not intuitively understand what property of complex numbers make rotation and angle such a common/convenient idea"

It's because of the way complex number multiplication happens to be defined, and more specifically, it's due to the effect of the number $i$.

If you consider the reals as a 1D vector space, multiplication by a scalar can be thought of in the following way: you take an input number (a 1D vector), and you multiply it by a given scalar - the result then has the same relationship to the original input as the chosen scalar has to the number one (1).

E.g., suppose you're multiplying by $3$; if I denote the "input" number by underlining it, then $\underline{2} \cdot 3 = 6$ because $6/2$ is the same as $3/1$. In other words, you are scaling the input by $3$.
In the image below, imagine taking the two bars, and uniformly stretching them, keeping the starting end at zero, so that the other end of the black bar ends up above $2$. The length of the red bar is the result.

enter image description here

OK? Now watch! (To borrow a phrase from Neil deGrasse Tyson.)

Take a complex number, say $3 + i$. It has a certain relationship towards the number one ($1 + 0i$): it is at a certain angle, and its modulus is greater than one by a certain amount (by $\sqrt{10}$). Let's represent that in the image below by simply connecting its position in the complex plane to ($1 + 0i$), so that now we have the triangle depicted below. This is a graphical representation of our "complex scaling factor".

enter image description here

Now, let's take some other complex number, $2 + i$ for example. This will be our "input" number, and we are going to scale it by the first one using the same principle I outlined in the introduction.

enter image description here

Now take our triangle; keep the start of the arrows at zero, but rotate it and scale it so that what originally was $1 + 0i$ precisely overlaps our input number $2 + i$. This will place the transformed red arrow in the same geometrical relationship with the input number.

enter image description here

The result (the scaled red arrow) lands on $5 + 5i$, which is what you get when you multiply these algebraically. Because the triangles are similar, the angle is maintained. The modulus of the result is $|5 + 5i| = 5\sqrt2$, and the $|2 + i| = \sqrt5$, so their ratio is, as expected, $\sqrt{10}$.

I think you can see how there is a sense in which this extension of the notion of multiplication as defined on the reals preserves the original semantics.


So, the overall effect is that there is a scaling, and a rotation. If the number you're multiplying by falls somewhere on the unit circle, than it's just a rotation; because the modulus is $1$, there's no scaling.

In particular, the effect of multiplying by $i$ is a $90^\circ$ CCW rotation.


In that light, there's another way to look at it: I'll borrow some images from this article by Kalid Azad, as these more closely correspond to how you get there algebraically.

enter image description here
enter image description here
enter image description here

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  • $\begingroup$ This answer is the best and easiest to understand, also I found a good website (kalid azad as mentioned) that teaches math intuitively. $\endgroup$
    – Asim
    Nov 28, 2023 at 14:23
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Take two complex numbers in their polar forms, $$ z := \alpha e^{i \theta} \qquad w := \beta e^{i \varphi} $$ (Here, $\alpha,\beta$ are their magnitudes, and $\theta,\varphi$ the angle they make with the positive real axis up to a multiple of $2\pi$.)

Then $$ zw = \alpha \beta e^{i(\theta+\varphi)} $$ Think about what this means for $z$ and $w$. Multiplying $z$ by $w$ (when one thinks of $z$ as a vector, i.e. an arrow pointing from the origin to $z$ in the plane) stretches $z$ by a factor of $\beta$, and then rotates it by an angle of $\varphi$ radians about the origin.

That is, multiplication is naturally an operation about stretching and rotating.

3Blue1Brown discusses this in more detail among other items in a past livestream here.

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First of all, complex numbers are two-dimensional, having independent x (real) and y (imaginary) components. This makes it possible to define a “rotation”, which you can't really do with one-dimensional real numbers (unless you count flipping the sign).

However, this property isn't unique to complex numbers. It also works with ordinary $\mathbb{R}^2$ vectors, using the standard rectangular-polar coordinate transformation:

$$x = r \cos\theta; y = r \sin\theta$$ $$r = \sqrt{x^2 + y^2}; \theta = \operatorname{atan2}(y, x)$$

You can rotate the vector $(x, y) = (r\cos\theta, r\sin\theta)$ by adjusting the angle $\theta$ while leaving the length $r$ alone.

$$(x' = r \cos(\theta+\phi), y' = r \sin(\theta + \phi))$$

Or you can multiply the vector by a real scalar $a$, to stretch/shrink its length while preserving the angle.

$$(x' = ar\cos\theta, y' = ar\sin\theta)$$

You can do this with $\mathbb{C}$ too, but you get a bonus operation: The ability to multiply complex numbers by each other rather than just by real scalars. This is defined as:

$$(a + ib)(c + id) = (ac - bd) + i(ad + bc) $$

But an interesting thing happens when you multiply together two numbers written in polar coordinates.

A minor point of notation: For complex numbers, the vector $(x, y)$ is conventionally written as $x + iy$. Applying the coordinate transformation, this is equivalent to $r\cos\theta + ir\sin\theta = r(\cos\theta + i\sin\theta)$. For brevity, let's define $\operatorname{cis}\theta = \cos\theta + i \sin\theta$.

Now, let's consider the two complex numbers:

$$z = r\operatorname{cis}\alpha = r\cos\alpha + ir\sin\alpha$$ $$w = s\operatorname{cis}\beta = s\cos\beta + is\sin\beta$$

Multiplying them gives:

$$zw = (r\cos\alpha + ir\sin\alpha)(s\cos\beta + is\sin\beta)$$ $$=rs\cos\alpha\cos\beta + irs\cos\alpha\sin\beta + irs\sin\alpha\cos\beta - rs\sin\alpha\sin\beta $$ $$= rs (\cos\alpha\cos\beta - \sin\alpha\sin\beta + i(\cos\alpha\sin\beta + \sin\alpha\cos\beta)) $$ $$= rs (\cos(\alpha + \beta) + i\sin(\alpha + \beta)) $$ $$= rs \operatorname{cis}(\alpha + \beta) $$

Multiplying two complex numbers adds their angles. That's interesting.

If we multiply a complex number by itself, the above formula gives us:

$$(r \operatorname{cis} \theta)^2 = r^2\operatorname{cis}(2\theta)$$

It can be shown that this result generalizes to:

$$\boxed{(r \operatorname{cis} \theta)^n = r^n\operatorname{cis}(n\theta)}$$

This is called DeMoivre's Formula, and it's useful for finding powers and roots of complex numbers. And that's why we care about complex numbers' angles.

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Notice that, doing simple arithmetic with i, we get the following: $$(\cos(a)+i\sin(a))(\cos(b)+i\sin(b))=\cos(a)\cos(b)-\sin(a)\sin(b)+i(\cos(a)\sin(b)+\sin(a)\cos(b)).$$ From trigonometry, we learn $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$ and $$\sin(a+b)=\cos(a)\sin(b)-\sin(a)\cos(b),$$ thus $$(\cos(a)+i\sin(a))(\cos(b)+i\sin(b))=\cos(a+b)+i\sin(a+b).$$ In the end, it really is just a beautiful coincidence that the square root of -1 allows us to relate multiplication of points with addition of angles.

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    $\begingroup$ I don't think this is correct, or a useful way to look at it. The fact that complex number arithmetic models so many symmetries of the plane is not a coincidence. $\endgroup$
    – MJD
    Jul 7, 2023 at 2:42
  • $\begingroup$ @MJD Yeah, I don't really know what I was getting at. I fixed it. $\endgroup$ Jul 7, 2023 at 3:05
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The most illuminating answer is to consider the geometric algebra perspective; viewing rotations in the plane as bireflections over lines, and noting that the geometric product of two lines gives you an object that is algebraically isomorphic to a complex number. This view also naturally extends to arbitrary dimensions and signatures, allowing you to generalize that perspective to quaternions and biquaternions and beyond.

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Instead of asking "why is rotation special in the complex plane", let's start by asking "how can rotations even be represented on the complex plane?"

The unique property of a unit vector rotating counter-clockwise at unit-speed is that, at all times, its velocity-vector is just itself rotated 90° to the left. In the complex plane, rotating a vector 90° to the left is conveniently accomplished by multiplying by $i$

Complex rotating

Therefore, at all times, the rotation satisfies the differential equation:

$$x'(t) = i * x(t)$$

One solution to this (with the appropriate initial-condition) is $x(t) = e^{it}$.

Thus on the complex plane, exponentiation to an imaginary power is a rotation. And since exponentiation is easy to work with for many things (especially Calculus-related things), that makes complex numbers a convenient tool when dealing with rotation.

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It’s a short contribution, but I wanted to add an aspect that the other answers didn’t touch on. One of the old names for imaginary numbers was lateral numbers, and I think it’s a better name for them. The name just lost out in the course of history. The other answers explain well why they’re well suited. My point is that the name we use is ill-suited. If you asked “why are lateral numbers associated with rotation?”, you might have been able to intuit the answer yourself. Not to mention that “imaginary” adds a confusion and fear factor to learners being introduced to the term, and a feeling that the numbers are fake, even though the so called “real” numbers are entirely human invention as well. You can count whole numbers of things, you can owe integer numbers of things, you can say that five atoms out of an apple is a rational number, but the real numbers are needed to fill in the gaps necessary for more abstract things, like the ratio of circumference to diameter of a circle, and the solutions to some polynomial equations. There is nothing you can see or tangibly measure that truly requires infinite significant figures to describe. They (the real numbers that aren’t included in the rationals) strictly show up in puzzles that we invent in our minds, so why should anyone say that they’re more real than the so-called imaginary numbers?

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