10 "Real" Applications of Complex Numbers (Part 1/2)

by greenturtle3141, Aug 23, 2024, 3:33 AM

Reading Difficulty: Anywhere between 1/5 and 7/5
Prerequisites: Varies

Despite not being "real", complex numbers somehow have fascinating applications to "real" settings. That is, complex numbers can be introduced to solve problems that do not at all reference complex numbers. Despite having seen many such instances, I continue to be amused by these applications. Let's talk about 10 of them:

Geometry (Reading Difficulty: 1/5)
Trigonometric Identities (Reading Difficulty: 1-2/5)
Sums of Two Squares (Reading Difficulty: 1/5)
Roots of Unity Filtering (Reading Difficulty: 2-3/5)
Physics (Reading Difficulty: 1/5)
Solving Differential Equations (Reading Difficulty: 3-5/5)
Integrals (Reading Difficulty: 3/5)
Infinite sums (Reading Difficulty: 3-4/5)
Proof of the Central Limit Theorem (Reading Difficulty: 4-5/5)
Fractional Derivatives and Embeddings of Fractional Sobolev Spaces (Reading Difficulty: 5-7/5)

1. Geometry

Rotate points with ease!

This may be "backwards" in terms of which result depends on the other, but complex numbers are incredibly useful for rotations. This is because when multiplying complex numbers, you are really multiplying their magnitudes and adding their angles. This follows from, say, sine and cosine angle sum formulae, and it is also "easy to see" if you write complex numbers in polar form (though this is not at all a rigorous argument). But where this method really shines is when dealing with cartesian coordinates. If we have a point $(a,b) \in \mathbb{R}^2$ , and we wish to rotate this about the origin by $60$ degrees, then complex numbers makes this incredibly simple to execute: Just multiply the complex number $a+bi$ by $\cos(60) + i\sin(60)$ .

The second-best way to rotate points is via linear algebra. But even if you remember the rotation matrix, it's probably much faster to use complex numbers. Moreover, you can even use the complex number method to derive the rotation matrix! Since
$(a+bi)(\cos \theta + i\sin\theta) = a\cos\theta-b\sin\theta + (a\sin\theta + b\cos\theta)i,$ we see that the rotation of $(a,b)$ about $(0,0)$ by an angle $\theta$ is given by
$\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}.$ (Yes, we can argue this also by knowing that rotation is a linear transformation and then plugging in basis vectors. But that requires thinking. This method does not!)

Exercise: Using complex methods, find the area of an equilateral triangle whose vertices have $x$ -coordintes $0$ , $1$ , and $3$ , in some order.

Complex numbers are so intertwined with the concept of rotation that Asymptote's coordinate system is literally complex numbers, which hence allows you to multiply points together.

Other applications to Euclidean geometry

Viewing the Euclidean plane as being the same as the complex plane $\mathbb{C}$ gives some useful algebraic structure to geometry that allows one to algebraically prove certain statements. It lets you "add" points, of course, though this isn't that thrilling since this is a "standard" vector space operation. The true power (and really the only reason why you'd want to be in the complex plane versus $\mathbb{R}^2$ ) is the ability to multiply points. For example, you can prove that a triangle with vertices $a,b,c \in \mathbb{C}$ is equilateral iff
$a + \omega b + \omega^2 c = 0$ where $\omega$ is a primitive third root of unity, such as $\omega = e^{2\pi i/3} = \frac{-1}{2} + \frac{\sqrt{3}}{2}i$ .

Exercise: Prove this fact. (Avoid "bashing" it, there are nice ways to argue both directions.)

Exercise: Use this fact to prove Napoleon's Theorem.

This is just the tip of the iceberg. A more in-depth study of this topic is best deferred to some works of Evan Chen.

2. Trigonometric identities made easy

The sine and cosine angle sum identities can be quite difficult to remember on a first encounter. The way I "memorized" them was quite silly: I simply derived the on the spot whenever I needed them, and each subsequent derivation got faster and faster until I could procure them so fast that I had effectively memorized them. Indeed, "proving" the identities reduces to nothing more of the following back-of-the-envelope calculation:
$\cos(A+B)+i\sin(A+B) = e^{iA}e^{iB} = (\cos A+i\sin A)(\cos B + i\sin B) = \cos A \cos A - \sin A \sin B + (\cos A \sin B + \sin A \cos B)i.$ The angle sum identities can thus be derived in mere seconds!

Exercise: Derive the triple-angle formula $\cos(3\theta) = 4\cos^3\theta - 3cos\theta$ using complex methods.

In general, you see a good bit of "complex bashing" in Olympiad settings that concern trigonometry. Essentially, if one wishes to prove some identity that involves $\sin$ and $\cos$ , we can try writing
$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$ and
$\cos x = \frac{e^{ix}+e^{-ix}}{2}.$ While this can be quite messy, it often ends up clarifying the situation since exponentials are more "flexible" to algebraically work with than sine and cosine.

As a last remark here, let's eviscerate a particularly "fearsome" trigonometric identity: The tangent addition formula for multiple angles. That is,
$\tan(\theta_1+\theta_2+\ldots+\theta_n) = \frac{s_1-s_3+s_5-s_7 +\ldots}{1-s_2+s_4-s_6+\ldots},$ where
$s_k = \sum_{S \subseteq \{1,\cdots,n\}, |S| = k} \prod_{j \in S} \tan \theta_j$ is the $k$ th symmetric polynomial in $\tan \theta_1, \cdots, \tan\theta_n$ . For example,
$\tan(A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan A \tan C - \tan B \tan C}.$ When this was taught to me first, I recall the argument being a bit of an inductive mess. But such pains are completely sidestepped using complex numbers!

Let's start by finding a complex number $a$ whose argument is $A$ . How about $a = 1 + i\tan A$ ? Define $b$ and $c$ similarly. Then the product $abc$ has argument $A+B+C$ , and $\tan(A+B+C)$ will be given by the quotient of the imaginary and real parts of $a+b+c$ . Computing,
$abc = (1+i\tan A)(1+i\tan B)(1+i\tan C) = 1 - \tan A \tan B - \tan A \tan C - \tan B \tan C + (\tan A + \tan B + \tan C - \tan A \tan B \tan C)i.$ The tangent addition formula shows up immediately with practically no effort, and the argument is equally easy for the general case of $n$ angles.

Exercise: Evaluate the infinite sum
$\sum_{n=1}^\infty \frac{\cos n}{2^n}$ using complex numbers.

Exercise: Let $\theta \in \mathbb{R}$ be a constant. Using complex numbers, obtain a closed form for the finite sum
$a_n := \sum_{k=0}^n \sin(k\theta),$ and deduce that the sequence $\{a_n\}_{n=0}^\infty$ is bounded. (Note that if $\sin$ is replaced with $\cos$ , the same statement holds except for when $\theta$ is an integer multiple of $2\pi$ .)

3. Sums of two squares

A famous result in elementary number theory is that if there are positive integers $x$ and $y$ , each of which can be written as the sum of two perfect squares, then the product $xy$ can also be written as the sum of two squares! This miraculous statement can be proven with the help of complex numbers.

Suppose $x = a^2+b^2$ and $y = c^2+d^2$ . Then
$x = |a+bi|^2$ and
$y = |c+di|^2.$ Therefore
$xy = |a+bi|^2|c+di|^2 = |(a+bi)(c+di)|^2 = |ac-bd + (bc+ad)i|^2 = (ac-bd)^2 + (bc+ad)^2.$ Voila.

Of course, one could have jumped straight to the identity $(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (bc+ad)^2$ without ever mentioning complex numbers, but good luck finding this without the help of our imaginary friends.

Exercise: Using the help of complex numbers, show that if $x$ is a positive even integer that can be written as the sum of two squares, then $x/2$ can also be written as the sum of two squares.

4. Roots of Unity Filtering

Suppose we have a (possible infinite) polynomial $P(x) = a_0 + a_1x + a_2x^2 + \ldots.$ The sum of the coefficients of $P$ is given as $P(1)$ , clearly. How about the sum of every other coefficient? If we wish to obtain a nice expression for $a_0+a_2+a_4+\ldots$ , one trick is to observe that
$P(1) = a_0+a_1+a_2+a_3+\ldots$ $P(-1) = a_0-a_1+a_2-a_3+\ldots$ and then sum these equalities to obtain
$P(1)+P(-1) = 2a_0 + 2a_2 + 2a_4 + \ldots.$ We have hence killed all non-even terms by introducing $(-1)$ , which lets us conclude that the sum of every other coefficient, $a_0+a_2+\ldots$ , is exactly $\frac{P(1)+P(-1)}{2}$ .

What about every third coefficient? Is there a nice expression for $a_0+a_3+a_6+\ldots$ ? The answer is yes, and we can actually do the same trick to obtain it? Last time, we killed all non-even terms using the fact that $1 + (-1) = 0$ . How do we kill all non-multiple-of-3 terms? We simply go into the complex plane, using the fact that $1+\omega+\omega^2 = 0$ where $\omega$ is a (primitive) third root of unity!

If $\omega$ is, say, $-\frac12 + \frac{\sqrt{3}}{2}i$ , then $\omega^3 = 1$ and $1+\omega+\omega^2 = 0$ . So $1+\omega^n+(\omega^2)^{n}$ is $3$ exactly when $n$ is a multiple of $3$ , and otherwise it is $0$ , vanishing into thin air. A good picture to have in mind for convincing yourself that this is true, and for visualizing what exactly is going on here, is to view $1^n$ , $\omega^n$ , and $(\omega^2)^n$ as "propellors" rotating in the complex plane at different speeds. The $1^n$ propellor stays at $1$ without ever moving. The $\omega^n$ propellor starts at $1$ and rotates $120$ degrees counter-clockwise every time $n$ increments. The $(\omega^2)^n$ rotates at twice the speed, rotating $240$ degrees counter-clockwise every time $n$ goes up. In this visualization, the three propellors meet up exactly when $n$ is a multiple of $3$ , but otherwise they are evenly spaced around $0$ , which causes the cancellation.

To spell out the punchline, we plug in $1$ , $\omega$ , and $\omega^2$ into $P$ to get
$P(1) = a_0+a_1+a_2+a_3+\ldots$ $P(\omega) = a_0+a_1\omega+a_2\omega^2+a_3\omega^3+\ldots$ $P(\omega^2) = a_0+a_1\omega^2+a_2\omega^4+a_3\omega^6+\ldots$ and summing these gives
$P(1)+P(\omega)+P(\omega^2) = a_0(3) + a_1(0) + a_2(0) + a_3(3) + \ldots.$ Thus
$a_0+a_3+a_6+\ldots = \frac{P(1)+P(\omega)+P(\omega^2)}{3}.$ This neat trick works when $3$ is replaced with any prime number.

This method, known as "Roots of Unity Filtering", is especially useful when applied to certain polynomials $P$ . A common choice is $P(x) = (1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k$ , which gives a beautiful expression for sums such as
$\sum_{k=0}^{100} \binom{300}{k}.$ If you haven't seen this little sleight of hand yet, it is worth computing. (And if you're stuck, this demonstration is literally on every other site that talks about roots of unity filtering.)

Exercise: Use the methods of complex numbers to find a smooth function $f:\mathbb{R} \to \mathbb{R}$ , written in closed form, such that the sequence of derivatives of $f$ has period $3$ . (That is, $f''' = f$ but $f' \neq f$ .) Your expression for $f$ should not involve non-real numbers.

5. Acceleration of uniform circular motion

Disclaimer: I am not a physicist.

This is rather informal and, as far as I'm aware, not an actual "application" of complex numbers. It's a fun connection to make, however!

If you took some form of physics, you are probably aware that the acceleration of an object undergoing uniform circular motion about a center $O$ is a vector that points towards $O$ . If you're like me, you probably found this extremely unintuitive and/or remarkable. Why does it point exactly towards the center? Complex numbers can help us understand why. An object rotating about the complex origin $0$ with angular velocity $\omega$ will at
$e^{i\omega t}$ after $t$ seconds. This is the position of the object over time. Differentiating gives the velocity over time,
$i\omega e^{i\omega t}.$ Lastly, differentiating this gives the acceleration of the object over time,
$-\omega^2 e^{i\omega t}.$ The position vector $e^{i\omega t}$ and the acceleration vector $-\omega^2 e^{i\omega t}$ are pointing in opposite directions at all times. In other words, if we place the acceleration vector at the position of the position vector, then it will point towards $0$ , the center of rotation.

6. Differential Equations

Elementary Applications

Differential equations can be solved with the help of complex numbers. Consider, for example, the differential equation for periodic motion or whatever it's called,
$u'' = -ku$ where $k > 0$ is the spring constant. Practically speaking the "best" way to solve this in my opinion is to make the substitution $u(t) = v(\sqrt{k}t)$ to eliminate the $k$ , and then guess that $v(t) = a\sin t + b\cos t$ . But you'd only guess this if you pretty much know what the solution is (and this would be expected from you in practice), so just for fun let's try and derive the solution "from scratch".

Let $D$ be the "differentiation operator". Then the ODE may be written as $(D^2+kI)u = 0$ , where $I$ is the "identity". We can then make the factorization
$(D+\sqrt{k}iI)(D-\sqrt{k}iI)u = 0.$ (If you haven't seen this trick before, it SHOULD feel suspicious. Fortunately it turns out that this is actually perfectly rigorous if you think hard enough.)

This lets us decouple the ODE into a simple system. Let $v = (D-\sqrt{k}iI)u$ . Then we can start by solving $(D+\sqrt{k}iI)v = 0$ , i.e.
$v' + \sqrt{k}iv = 0. \qquad (*)$ Of course, it's crucial to recognize that we are now allowing our solutions to be complex-valued, i.e. of type $\mathbb{R} \to \mathbb{C}$ rather than just $\mathbb{R} \to \mathbb{R}$ . This is not a mistake because finding all solutions of the former type will also find all solutions of the latter type.

Solving $(*)$ using your favorite method (such as the rather non-rigorous "separate and integrate", or perhaps using an integrating factor, or even by guessing, like I just did), we end up with $v(t) = v(0)e^{-\sqrt{k}i t}$ . Now, from $v = (D-\sqrt{k}iI)u = u' - \sqrt{k}iu$ , we get a first-order ODE for $u$ ,
$v(0)e^{-\sqrt{k}i t} = u'(t) - \sqrt{k}iu(t).$ Using your favorite method (probably integrating factors), we get the solution
$u(t) = Ae^{-\sqrt{k}i t} + Be^{\sqrt{k}i t}$ for constants $A$ and $B$ . These are all the complex solutions. To recover the real solutions, some more work should be done, which I will omit. (Do beware that $A$ and $B$ are not necessarily real.)

Exercise: An ant is placed on each corner of a square of side length $2024$ . Each ant walks towards its counter-clockwise neighbor at a speed of $42$ units per second. Eventually the ants will converge at the center of the square. Using complex numbers, prove that each ant will have travelled exactly $2024$ units.

Advanced Applications

This is not the only way to employ complex numbers, however. We can also try using the Fourier Transform, which in this post we define to be
$\mathcal{F}(u)(xi) = \hat{u}(\xi) := \int_{\mathbb{R}} u(t)e^{-2\pi i\xi t}\,dt.$ Provided sufficient regularity and decay properties on $u$ , we have that
$\mathcal{F}(u')(\xi) = -2\pi i \xi \mathcal{F}(u)(\xi).$ This allows us to effectively turn differentiation into multiplication. Applying this to the spring equation,
$(-2\pi i \xi)^2\hat{u}(\xi) = -k\hat{u}(\xi).$ Strangely, this equation seems to suggest that the only solution is $\hat{u}(\xi) = 0$ , so that $u(x) = 0$ . What gives?

The issue is that you can only take the Fourier transform of $u$ provided that $u$ has sufficient decay properties, e.g. something like $|u(x)| \leq \frac{A}{1+|x|^2}$ . So in a way, trying to solve the spring equation using Fourier methods has given us every possible solution that decays to 0 at infinity. Indeed, $u(x) = 0$ is the only such solution.

So is this approach doomed? It is not, but fixing this will (1) require some very advanced theory, and (2) doesn't necessarily find all solutions either (the saving grace here, though, is that it does appear to find all solutions "symbolically"). The advanced theory we need is called the theory of distributions, particularly that of tempered distributions. A full discussion of this will be deferred to a different post because it really is quite advanced, but the gist is that distributions are a generalization of functions. Some properties includes:

If $f$ does not grow too fast, then it is also a tempered distribution.
If $f$ is a tempered distribution, then you can take its Fourier transform $\hat{f}$ to get another tempered distribution.
You can add tempered distributions, and this interacts well with the Fourier transform.
You cannot multiply tempered distributions. But you can multiply tempered distributions by a smooth function. For example, if $f$ is a tempered distribution, then so is $xf(x)$ .
We still have the nice identity $\mathcal{F}(u')(\xi) = -2\pi i \xi \mathcal{F}(u)(\xi)$ , which holds "in the sense of distributions". That is, you shouldn't view each side as a function. Rather you should see each side as a tempered distribution.

Now let us solve the spring equation, and I will make exactly one (fatal!) error that I shall address at the end. Taking the Fourier transform again gives
$(-2\pi i \xi)^2\hat{u}(\xi) = -k\hat{u}(\xi).$ This rearranges to
$(2\pi\xi-\sqrt{k})(2\pi\xi+\sqrt{k})\hat{u}(\xi) = 0. \qquad (*)$ It might seem like we've wound up at the same problem. Isn't the only solution here $\hat{u}(\xi) = 0$ ? This would be true if this equation was an equality of functions. But it's an equality of distributions, so the intuitive "algebra" doesn't actually apply. It turns out that the solution to a equation such as $(x-a)f(x) = 0$ for $f$ , in the sense of distributions, is given by $f = c \cdot \delta_a$ where $\delta_a$ is the Dirac delta distribution at $a$ and $c$ is any constant. This is not that easy to prove, but it is powerful here. Applying this twice to $(*)$ , we find the distributional solution
$\hat{u}(\xi) = c_1\delta_{-\sqrt{k}/(2\pi)} + c_2\delta_{\sqrt{k}/(2\pi)}$ for any constants $c_1$ and $c_2$ . Applying the inverse Fourier transform on both sides recovers the solution to the spring equation. I'll defer the details (and the more rigorous argument) for another post.

I promised that there was a fatal error. Indeed, the error here was the assumption that the solution $u$ could be treated as a (tempered) distribution. Unfortunately, we do not know a priori that this is the case, since a solution such as $u(x) = e^x$ actually grows too fast to be interpreted as a tempered distribution. Hence, this method is unable to rule out candidates such as $u(x) = e^x$ .

One silver lining is that if you just "pretend" that you could work with $e^x$ as a distribution and do the algebra to solve a differential equation such as $u' = u$ , then you will find that the method still recovers the solution, even though the steps cease to be legal. Also, this tool can be used in conjunction with other theorems to find all solutions or to even just obtain some useful information, particularly in more complicated differential equations.

The Fourier transform (in multiple dimensions) can also be used to solve partial differential equations such as $\Delta u = f$ , but I won't pursue this matter here.

(Part 2: Here)

This post has been edited 1 time. Last edited by greenturtle3141, Aug 23, 2024, 3:42 AM

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cool post

solution to exercise 1, about equilateral triangle

by fruitmonster97, Aug 23, 2024, 11:00 AM

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This is a great post! thank you for making it!

by Yummo, Sep 5, 2024, 1:08 AM

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Although I had a decent college essay, this isn't really my specialty so I don't really have anything useful to say that isn't already available online.
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Could you also make a blog post about college essay writing :skull:
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by greenturtle3141, Nov 13, 2021, 9:21 PM
hello

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67 shouts

Turtle Math

10 "Real" Applications of Complex Numbers (Part 1/2)

by greenturtle3141, Aug 23, 2024, 3:33 AM

2 Comments