# Applications of complex numbers to solve non-complex problems

Recently I asked a question regarding the diophantine equation $x^2+y^2=z^n$ for $x, y, z, n \in \mathbb{N}$, which to my surprise was answered with the help complex numbers. I find it fascinating that for a question which only concerns integers, and whose answers can only be integers, such an elegant solution comes from the seemingly unrelated complex numbers - looking only at the question and solution one would never suspect that complex numbers were lurking behind the curtain!

Can anyone give some more examples where a problem which seems to deal entirely with real numbers can be solved using complex numbers behind the scenes? One other example which springs to mind for me is solving a homogeneous second order differential equation whose coefficients form a quadratic with complex roots, which in some cases gives real solutions for real coefficients but requires complex arithmetic to calculate.

(If anyone is interested, the original question I asked can be found here: $x^2+y^2=z^n$: Find solutions without Pythagoras!)

EDIT:

I just wanted to thank everyone for all the great answers! I'm working my way through all of them, although some are beyond me for now!

• The proof of Stirling's Formula and the calculation of $\int_0^{+\infty}\frac{\sin x}{x}dx$. They become easier if you use the properties of holomorphic functions. – Li Li Dec 28 '16 at 17:34
• "the shortest path between to truths in the real domain passes through the complex domain"-Hadamard – Jorge Fernández Hidalgo Dec 28 '16 at 17:37
• The purpose for which complex numbers originally were introduced into standard mathematics was to find real-valued solutions of cubic equations with real-valued coefficients--a "non-complex" problem. – David K Dec 28 '16 at 18:50
• @DavidK Your comment should be THE answer to the question. Why don't you make it one? – Ari Brodsky Dec 28 '16 at 23:01
• Possible duplicate of Interesting results easily achieved using complex numbers – Jack Mar 13 '17 at 15:57

I suppose the most common one on this site is an application of the Residue Theorem. That is:

$$\int_\gamma f(z) dz = 2\pi i \sum_k Res(f; z_k)$$

where $f$ is an analytic function with only finitely many isolated singularities $z_k$ inside a closed curve $\gamma$ in the complex plane.

While this theorem is clearly a result of Complex Analysis, it in fact has many uses in computing integrals along the real line. Indeed, by constructing $\gamma$ to be semicircular contours, we can immediately compute the real integral $\int_{-\infty}^\infty f(x) dx$ for functions $f(z)$ that are the complex extension of real-valued $f(x)$ (as long as $f(z)$ disappears as $|z|\rightarrow \infty$).

This typical contour $\gamma$ appears as: where $j$ is an isolated singularity of $f(z)$ and we take $a\rightarrow \infty$.

Here is a straight-forward example. We attempt to compute the definite integral:

$$\int_{-\infty}^\infty \cfrac{dx}{(1+x^2)^2}$$

Defining $f(z):= \cfrac{1}{(1+z^2)^2} = \cfrac{1}{(z+i)^2(z-i)^2}$ where $z\in \mathbb{C}$, and the complex contour $\gamma_a$ to be the semicircle in the upper-half plane, we have by the Residue Theorem: $$\int_{\gamma_a} f(z) dz = 2\pi i Res(f; i) = \cfrac{2\pi i}{4i} = \cfrac{\pi}{2}$$

Now, noting that as $|z|\rightarrow \infty, |f(z)| \rightarrow 0$, so

$$\cfrac{\pi}{2} = \lim_{a\rightarrow \infty} \int_{\gamma_a} f(z) dz = \lim_{a\rightarrow \infty} \left(\int_{-a}^a f(x) dx + \int_{|z|=a,\theta \in [0,\pi]} f(z) dz \right) = \int_{-\infty}^\infty f(x) dx$$

and we have computed our real-valued integral of a real-valued function using Complex Analysis.

• Using complex residues, evaluating $\int_0^{\infty}(1+x^4)^{-1}dx$ is a simple exercise. Using only real analysis you are probably stuck. – DanielWainfleet Dec 29 '16 at 17:49
• @user254665 Not really stuck, as $1+x^4$ factors into two real quadratics. It's a little easier to do a linear change of variables to get the denominator to $4+x^4$ which factors into rational quadratics. – Erick Wong Dec 29 '16 at 20:24

I always found it pleasing how easily one can compute $\int e^{ax}\cos bx \; dx$ and $\int e^{ax}\sin bx \; dx$ by regarding them as the real and imaginary components of $\int e^{cx}\; dx$ (where $c=a+bi$).

An interesting problem is show that:

If $n$ is even then:

$${n \choose 0}\cos^{n}x-{n \choose 2}\cos^{n-2}x\cdot\sin^{2}x+{n \choose 4}\cos^{n-4}x\cdot\sin^{4}x-{n \choose 6}\cos^{n-6}x\cdot\sin^{6}x+...+(-1)^{n/2}{n \choose n}\sin^nx=\cos(nx)$$

That comes from the simple fact:

$$(\cos x+i\sin x)^n=\cos nx+i\sin nx$$

Just write $(\cos x+i\sin x)^n$ using binomial expansion and look to the real part.

Furthermore if we choose some values of $x$ we can find many real sums.

For example, take $x=\pi/4$:

$$\left(\frac{\sqrt{2}}{2}\right)^n\left[{n \choose 0}-{n \choose 2}+{n \choose 4}-{n \choose 6}+...+(-1)^{n/2}{n \choose n}\right]=\cos(n\pi/4)$$

We just found a way, using complex number, to calculate:

$${n \choose 0}-{n \choose 2}+{n \choose 4}-{n \choose 6}+...+(-1)^{n/2}{n \choose n}$$

The real-valued function \begin{align*} &f:\mathbb{R}\rightarrow\mathbb{R}\\ &f(x)=\frac{1}{1+x^2}=1-x^2+x^4-x^6+\cdots \end{align*} allows a series representation around $0$ with radius of convergence $1$.

Although the denominator never attains the value zero, the radius of convergence is restricted to $1$. The reason are the singularities at $\pm i$.

• Yes, and expanding that function around an arbitrary $x = k$ has radius of convergence $R=\sqrt{k^2+1}$, which if you don't think about the singularities at $\pm i$ is even more spooky! – Ant Dec 30 '16 at 2:31

Suppose you want to find the real solutions to the differential equation $$y'' - 2 y' + 2y = 0$$ you consider the associated algebraic equation $$z^2 - 2z + 2 = 0$$ solve it in the complex field: $$z_{1,2} = 1 \pm i$$ then a base for all the complex solutions is: $$y_{1,2}(x) = \exp((1\pm i)x) = e^x e^{\pm i x} = e^x (\cos x \pm i \sin x)$$ all the real solutions are: $$y(x) = c_1 e^x \cos x + c_2 e^x \sin x$$

There are many examples from number theory. One of the simplest is the parametrization of primitive Pythagorean triples. When viewed in the Gaussian integers $\,\Bbb Z[i] = \{ a + b\,i\,: a,b\in\Bbb Z\}$ the parametrizaton arises immediately from the fact that coprime factors of a square must themselves be squares (up to unit multiples), which is true for the Gaussian integers because they too, like natural integers, enjoy unique prime factorization.

Indeed if $\ z^2 = x^2 + y^2 = (x-y\,i) (x+ y\,i)$ and $\,x,y\,$ are coprime then one easily checks that $\,x-y\,i,\,x+y\,i\,$ are coprime, so being coprime factors of the square $\,z^2$ they must themselves be squares (up to a unit factor). Thus e.g. $\ x + y\ i\, =\, (m + n\ i)^2 =\ m^2 - n^2 + 2mn\, i,\,$ hence $\,x = m^2-n^2,\ y = 2mn\,$ (using the unit factor $1$; using the other unit factors $\, -1,\pm i\,$ merely changes signs or swaps $\,x,y\,$ values). Notice how very simple the solution is from this perspective.

Similarly one can solve low degree cases of Fermat's Last Theorem by employing analogous factorizations over certain rings of algebraic integers. For example, Gauss showed there are no solutions for exponent $3$ by working in the ring of integers of $\rm\ \mathbb Q(\sqrt{-3})\:,\:$ and Dirichlet did similarly for exponent $5$ using $\rm\ \mathbb Q(\sqrt{5})\:.$ Later Kummer generalized these techniques to handle all regular prime exponents by working over rings of cyclotomic integers. For a nice exposition see Ribenboim: 13 lectures on Fermat's last theorem. Weil nicely summarizes the essence of these techniques in his Number Theory, Ch.IV,S.VI,p.335:  A friend of mine came up with a proof of Heron's formula for the area of a triangle which uses complex numbers. I can't find the original publication, but the proof appears on pages 2-4 in Six mathematical gems from the history of Distance Geometry by Leo Liberti and Carlile Lavor. For completeness, I'll reproduce it here.

Heron's Formula: The area of a triangle with sidelengths $a$, $b$, and $c$ is given by $K = \sqrt{s(s-a)(s-b)(s-c)}$, where $s = \tfrac{1}{2}(a+b+c)$ is the semi-perimeter.

Proof: Let's start by drawing a triangle $\Delta ABC$ with sidelengths $BC = a$, $CA = b$, and $AB = c$, along with the inscribed circle centered at $O$ with radius $r$. Let the external tangents to the circle from $A$, $B$, and $C$ have lengths $x$, $y$, and $z$ respectively. Also, let $u = OA$, $v = OB$, $w = OC$, and let $\alpha$, $\beta$, and $\gamma$ be the angles centered at $O$ as shown in the diagram below. The area of $\Delta ABC$ is the sum of the areas of $\Delta OBC$, $\Delta OCA$, and $\Delta OAB$ which is $K = \dfrac{1}{2}ra + \dfrac{1}{2}rb + \dfrac{1}{2}rc = rs$, so we simply need to find an expression for $r$ in terms of $a,b,c$. This is where complex numbers becomes useful.

Since $r = u\cos\alpha$ and $x = u\sin\alpha$, we can write $r+ix = ue^{i\alpha}$. Similarly, $r+iy = ve^{i\beta}$ and $r+iz = we^{i\gamma}$. Then, since $2\alpha+2\beta+2\gamma = 2\pi$, we have $\alpha+\beta+\gamma = \pi$. Therefore, \begin{align} (r+ix)(r+iy)(r+iz) &= (ue^{i\alpha})(ve^{i\beta})(we^{i\gamma}) \\ r^3+ir^2(x+y+z)+i^2r(xy+yz+zx)+i^3xyz &= uvwe^{i(\alpha+\beta+\gamma)} \\ r^3+ir^2(x+y+z)-r(xy+yz+zx)-ixyz &= uvwe^{i\pi} \\ \underbrace{[r^3-r(xy+yz+zx)]}_{\text{real}}+\underbrace{i[r^2(x+y+z)-xyz]}_{\text{imaginary}} &= \underbrace{-uvw}_{\text{real}}. \end{align}

By equating imaginary parts, we get that $r^2(x+y+z)-xyz = 0$, i.e. $r = \sqrt{\dfrac{xyz}{x+y+z}}$.

Then, since $a = y+z$, $b = z+x$, and $c = x+y$, we have $2s = a+b+c = 2x+2y+2z$. Hence, $s = x+y+z$, and so, $x = s-a$, $y = s-b$, and $z = s-c$.

Therefore, $r = \sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}}$, and thus, $K = rs = \sqrt{s(s-a)(s-b)(s-c)}$.

One example I particularly like, because it very strongly geometric in flavor rather than algebraic, is the use of conformal transformations for studying two-dimensional fluid flow.

• I'm glad someone else brought this up. As someone who never got a solid handle on vector calculus (or, for that matter, any vector operations more complex than basic arithmetic), I had a much easier time grasping and performing fluid flow calculations when complex numbers were used instead. – JDM Dec 31 '16 at 18:09

I'm surprised no one has mentioned Dirichelet's theorem on primes. Dirichilet proved that if the base and offset of an arithmetic sequence do not have a common factor, there are infinitely many primes in the sequence. Dirichelet's proof used complex analysis, and it is very surprising that to this day there is no known proof of this simple fact about integers without using complex analysis.

To see how complex analysis could possibly come into play here, let's step back and look at the fact that there are infinitely many primes at all. Of course, this special case of Dirichelet's theorem has a trivial proof: If there are only finitely many primes, multiply them all together and add 1. That number cannot be divisible by any prime, so it itself is new prime, contradicting our assumption that we already had all primes.

Somehow, though, that proof does not capture much about the nature of primes. Euler gave a different proof which, although at first it is not obvious how, does give more insight about primes. Euler's proof is as follows: If there are only finitely many primes, let $P$ be the set of all primes, and compute the following product $Q$: $$Q=\prod_{p\in P}{1 \over {1-p^{-1}}}$$ Expanding these terms using Euler's formula $${1 \over {1-x}}=1+x+x^2+\centerdot\centerdot\centerdot$$ and multiplying out, we see that since every whole number is uniquely determined by its prime factorization we have $$Q=1+{1\over 2}+{1\over 3}+{1\over 4}+\centerdot\centerdot\centerdot$$ It is not hard to prove directly that this sum diverges, giving us our contradiction. But a more interesting observation is that this divergence comes from the fact that the Riemann zeta function $$\zeta(z)=\prod_{p\in P}{1\over{1-p^{-z}}}=1+{1\over{2^z}}+{1\over{3^z}}+\centerdot\centerdot\centerdot$$ has a pole at $z=1$.

In the proof of Dirichelet's theorem, we look at a generalization of the Riemann zeta function, the L-functions: $$L(z,\chi)=\prod_{p\in P}{1\over{1-\chi(p)p^{-z}}}$$ where $\chi$ is a character, a map from $\mathbb{Z}$ to $\mathbb{C}$ which defines a homomorphism from the multiplicative group of $\mathbb{Z}/n\mathbb{Z}$ to the multiplicative group of $\mathbb{C}$. It turns out that each of our arithmetic sequences corresponds to a character, and by techniques similar to the above, it can be shown that an arithmetic sequence has infinitely many primes if and only if its corresponding L-function has a pole, in this case at $z=0$. But the only way known to show that such poles exist in general is to use purely analytic methods from complex analysis.

One of the best constructed demonstrations using complex numbers to solve a problem which did not explicitly require them appears in that classic Math text: One Two Three... Infinity by George Gamow.

It seems there was a buried treasure on an island, on which one found where to dig by counting the paces from an old gallows pole to a pine tree, turning 90 degrees right, and walking the same number of paces again then marking that point. Then starting again from the gallows walk to an oak tree and do the same but turn left. The treasure will be found buried at the mid-point of these two marked points. Ah but, by the time our heros get to the island the gallows had rotted completely away! There are the pine and oak trees, but without a starting point how can the treasure be found?? Turns out a little application of complex analysis does the trick most efficiently.

(the starting point cancels out of the complex/vector equation)

Gamow was a great puzzle master as well as a master mathematician.

• Marvelous puzzle, I am attempting to solve it. Would you be willing to pm me your solution when I am ready? – electronpusher Dec 31 '16 at 18:26
• @electronpusher the worthy and detailed solution to this puzzle (as a gentle back-door to learning complex analysis) is given in Gamow's text. This link contains scans of the old text: jwilson.coe.uga.edu/EMT725/Treasure/Gamow/Gamow41.47.html – Theophrastus Dec 31 '16 at 19:13
• Today I saw Bottema's Theorem in cut-the-knot (cut-the-knot.org/Curriculum/Geometry/Bottema.shtml). It is exactly the treasure hunt problem given in "One Two Three ... Infinity" ! – Cyriac Antony Apr 17 '18 at 10:37

The cubic and quartic formulas involve manipulations of imaginary numbers, even when dealing with equations whose roots are all real (note the factors $\sqrt{-3}$ in the formulas below):

• Nah, just use trig identities and this all fades away ;) – Simply Beautiful Art Mar 14 '17 at 14:15

The eigenvalues of a symmetric real matrix are real.

Let $A$ be a symmetric real matrix and let $\lambda$ be a possibly complex eigenvalue. If $v\in\mathbb{C}^n$ is an eigenvector of $A$ relative to $\lambda$, we have, denoting by $^H$ the Hermitian transpose, $$\lambda(v^Hv)=v^H(\lambda v)= v^H(Av)=v^H(A^Tv)=v^HA^Hv= (Av)^Hv=(\lambda v)^Hv=\bar{\lambda}(v^Hv)$$ Since $v\ne0$, also $v^Hv\ne0$ and therefore $\lambda=\bar{\lambda}$ is real.

Since there is already a lot of discussion and the question has recieved much attention too, so I think there is something more to say.

While studying complex numbers from the book Complex Numbers from A to Z By Titu Andreescu I realised that complex numbers has got extensive application in geometry, let us see how:

# Problem $$1$$

Two regular polygons are inscribed in the same circle. The first polygon has $$1982$$ sides and the second has $$2973$$ sides. if polygons have any common vertices then how many such vertices will be there??

Believe it or not, I have asked this question from almost $$20$$ people and only one of them thought of applying complex numbers there. The answer of this tough looking questions is $$991$$ which is simply the HCF of $$2973$$ and $$1982$$.

If you want to know how? Then go to the page number $$63$$ of the link I have provided.

The most of math enthusiasts of my age usually like the contest problem most. Usually geometric questions in international level competitions include those problem where we need to prove something (Regarding relations in geometric figure). I found three problems which are entirely based on a single concept (The real product of two complex numbers), out of these three, two have been asked in International Olympiad.

# Problem$$2$$:

Let $$ABCD$$ be a convex quadrilateral. Prove that:

$$AB^2+CD^2=AD^2+BC^2$$ If and only if $$AC$$ is perpendicular to $$BD$$.

# Problem$$3$$:

Let $$M,N,P,Q,R,S$$ be the midpoints of sides $$AB,BC,CD,DE,EF,FA$$ of a hexagon. Prove that $$RN^2=MQ^2+PS^2$$ If and only if $$MQ$$ is perpendicular to $$PS$$.

# Problem$$4$$:

Let $$O$$ be the circumcentre of $$\triangle ABC$$. Let $$D$$ be the midpoint of the segment $$AB$$ and let $$E$$ is the centroid of the triangle $$ACD$$. Prove that lines $$CD$$ and $$OE$$ are perpendicular if and only if $$AB=AC$$.

These questions can be solved involving the concepts which are taught at first section of lesson $$4$$ of the given link.

Now, I also think that these things are not obvious. :) :) :) :)

• Isn't it clear that the answer will be the gcd of the two numbers, though? Complex numbers seem overkill here, although it is nice to see the connection to roots of polynomials. – Théophile Dec 28 '16 at 20:15
• Why do you think it is obvious @Théophile, Everything comes with a concept behind it, please tell me about concept you are applying while saying it obvious. – Vidyanshu Mishra Dec 28 '16 at 20:33
• It's obvious because every divisor of one polygon corresponds to a way of dividing it into rotationally symmetric sections, so every common divisor represents a subset of the intersecting vertices: if $d$ divides the number of vertices in both polygons, then, starting at one common vertex, skipping 1982/$d$ vertices of the first polygon and 2973/$d$ vertices of the other, you'll get to another common vertex. The number of shared vertices in the subset is clearly $d$, and therefore the total set is the maximum value of $d$, i.e., the GCD. – Kyle Strand Dec 29 '16 at 0:27
• .....honestly, the basic concept is so intuitively obvious to me (and, apparently, to @Théophile) that any way of trying to explain it rigorously (such as in my comment above) just makes it seem much more complicated. Just picture two regular polygons with arbitrary numbers of faces inscribed in a circle as the question describes, and think about where the vertices will line up if there's at least on shared vertex; it should quickly become clear. – Kyle Strand Dec 29 '16 at 0:31
• To all those who commented on answer I would ask whatever your argument is, what is the problem blemishes with answer?? Fundng common point is a real concept and the lone wolf has shown it using complex numbers. Everything is okay for me. – Atul Mishra Dec 29 '16 at 4:31

There is a very interesting application of modular forms to prove the identity $$\sigma_7(n) = \sigma_3(n) + 120\sum_{i=1}^{n-1}\sigma_3(i)\sigma_3(n-i)$$ for $n \geq 1$, where $\sigma_k(n) = \sum_{d | n} d^k$. In particular, let $$E_k(\tau) = 1 - \frac{2k}{B_k}\sum_{n=1}^\infty \sigma_{k-1}(n)q^n$$ where $B_k$ is the $k$-th Bernoulli number and $q = e^{2 \pi i \tau}$. It turns out that $E_k$ is a weight $k$ modular form, and the ring of modular forms of all weights is graded by weight, so in particular both $E_4(\tau)^2$ and $E_8(\tau)$ are weight 8 modular forms. It furthermore turns out that the weight 8 modular forms comprise a one-dimensional complex vector space, so $E_4(\tau)^2$ and $E_8(\tau)$ are at most different by a constant factor. Since both of their constant coefficients are 1, however, we actually have $$E_4(\tau)^2 = E_8(\tau).$$ Expanding and collecting coefficients gives us the identity I wrote at the top.

Without using complex analysis and the theory of modular forms, one might never expect such a relationship between the different $\sigma_k$ functions.

(See chapter 1 of "A First Course in Modular Forms" by Diamond and Shurman for further details of this extraordinary topic.)

(1). A power series $f(x)=\sum_{n=0}^{\infty}a_nx^n$ with non-zero radius of convergence can be differentiated term-by-term within its circle of convergence. This can be more easily shown in $\mathbb C.$

Theorem: Let $D$ be a non-empty open subset of $\mathbb C.$ Let $(f_m)_m$ be a sequence of functions, each analytic on $D,$ that converges uniformly to $f$ on any compact subset of $D.$ Then $f$ is analytic on $D$ and $(f'_m$) converges uniformly to $f'$ on any compact subset of $D.$

This can shown by taking,for any $x\in D,\;$ $r>0$ small enough that $\{y:|y-x|<r\}\subset D.$ For any $s\in (0,r)$ the set $\{y:|y-x|\leq s\}$ is a compact subset of $D .$ And $f_m(x)=\frac {1}{2\pi i}\int_{|y-x|=s}\frac {f_m(y)}{(y-x)}dy$ and $f'_m(x)=\frac {1}{2\pi i}\int_{|y-x|=s}\frac {f_m(y)}{(y-x)^2}dy.$... et cetera.

(2). Any real polynomial is the product of real polynomials $q_1,...,q_n$ where each $q_i$ is quadratic or linear. Obvious from the Fundamental Theorem of Algebra. I dk how you could do it without $\mathbb C.$

Digression: Littlewood wrote: "Could a fellowship be awarded for a dissertation of just 2 lines? Perhaps. Consider: Theorem. A non-constant complex polynomial $p$ has a zero. Proof. If not, then $1/p$ is a bounded entire function.".... Hypothetically. Littlewood would have known that the Fundamental Theorem of Algebra historically preceded Liouville's theorem that a bounded entire function is constant.

A practical example: electrical impedance and admittance in physics and engineering. From this question:

Analyzing AC circuits is easier with complex numbers. I understand you have to replace the regular representations with those containing complex numbers representations.

My answer there explains the purpose:

The imaginary part of $Y$ [admittance] is a clever substitute for differentiation and integration.

There are more physical quantities that may be represented in a similar manner, e.g. permeablility. The general term for this usage of complex numbers is "phasor".

One really essential application of complex number would be to solve the problem the number of sum of $k$ squares, ie. given a number $n$, how many distinct ways can this be written as a sum of $k$ squares, for $k=4,6,8$. I believe the use of complex number here is completely essential. Jacobi solved this using the theta function. This is getting into a bit of advanced number theory, so I will keep it simple.

Without getting into too much technical details, here is the ideas:

1. Just like any combinatorics problem, we starts out by writing down the generating function. The coefficients of this are the answer to the problem, and now the goal is to find what the coefficient is.

2. This function have a nice property, called being a modular form (of a certain level). We now called this function the theta series.

3. Now there are other functions with the same properties, some of them are called the Eisenstein series. We know the coefficient of the Eisenstein series very well, since we can explicitly construct them.

4. Now we can try to write the theta series as a linear combination of the Eisenstein series. There is no reasons to think this is possible of course, but we try to at least make the first few coefficients match.

5. Once we do that, if you do some further calculation, you will notice that other coefficients, even those you did not try to match, seemed to match. At this point you should suspect that all coefficients match. If they really match, then you have solved the problem, as you can deduce the coefficients of the theta series from the Eisenstein series. So how can we try to prove that? Well, at this point we should suspect that not just the theta series, but all modular form of that level must be able to be written as a linear combination of these Eisenstein series, or in other word, the dimension of the space of modular form of that level is small enough that these Eisenstein series are sufficient to span it.

6. So far at this point, nothing we have done truly use complex number. Now it gets more interesting. By some simple transformation, modular form, which can be considered to be complex function, can now be considered also as a function on certain Riemann surface (if you don't know what it is, think of a different kind of complex plane) with specified bound on how bad they can blow up, and where can they blow up.

7. At this point, complex analysis take place (in particular, partial differential equation problem of finding a analytic complex function with specified value on the boundary), giving you a formula to calculate the dimension.

8. It turns out at this point that, only for $k=4,6,8$ does the dimension being small enough for these Eisenstein series to span the space of modular form. This solve the problem for $k=4,6,8$.

As far as I know, there is no other methods for this problem. In fact, the failure of the method on higher $k$ did prevent us from solving it for higher $k$ completely.

SIDE NOTE: even a bigger application, of course, is the proof of Fermat's Last Theorem. Unfortunately, I do not know the proof, so I am not sure how much of it truly use complex number; and beside, at that level, complex numbers is already ingrained everywhere that it is hard to separate them.

My personal favorite example involves the solution of the cubic by radicals. When the cubic has three real roots, application Cardano's formula requires extraction of the cube root of a complex number. It can be shown that for all-real-root cubics, NO solution by radicals can avoid the use of intermediate non-real complex numbers.

All of the trigonometric identities easily follow from $e^{bi} = \cos b + i \sin b$. Sine and cosine sum from $e^{(a+b)i} = e^{ai}e^{bi}$, De Moivre's theorem, and so on.

Another way complex numbers can be used to solve a non-complex problem is in solving a system of equations. This is especially useful when there is $$x^2+y^2$$ in the denominator. Just for illustration you can see the Daniel Fischer's solution to one of my problem here. Of course the problem can be solved by arithmetic means, but this method is much shorter and more elegant. Even more we avoid the two additional "quasi-solutions" to the equation (Solving the system without complex numbers gives us 4 potential solutions).

The reason why this is useful is that when we add two equations some of the information is "lost", but when we multiply one of them by $$i$$ we avoid that when we add them. So simply said we can manipulate the equations without losing any informations.

Furthermore some geometric problem in IMO (International Math Olympiad) can be simplified by using complex numbers. For example

Prove that there exists a convex 1990-gon with the following two properties :

a.) All angles are equal.

b.) The lengths of the 1990 sides are the numbers $$1^2$$, $$2^2$$, $$3^2$$, $$\cdots$$, $$1990^2$$ in some order.

(IMO $$1990$$, question 6)

can be solved by using the 1990-th roots of unity and Euler's Identity. In addition I've seen the use of complex numbers in some contest problems using coloring and tiling.

Similarly to Arnaldo Nascimento's solution, the complex numbers can be used to calculate certain sums of binomial coefficient. In particular for given $$n,k$$:

$$\sum_{i = 0}^{ik \le n} \binom{n}{ik} = \frac 1k \sum_{j=0}^{k-1}\left(1+e^{\frac{(2 i \pi)j}{k}}\right)^n$$

Many problems in planar Euclidean geometry can be solved with complex numbers by interpreting the space as an Argand diagram. This approach is especially fruitful if there are known angles in the problem, because rotation multiplies by a known unit complex number. My favourite example is Napoleon's theorem, so named because a legend attributes it to Bonaparte. The result states:

• Erecting equilateral triangles outward on a triangle's edges, the new triangles' centres are the vertices of an equilateral triangle;
• Erecting the equilateral triangles inward instead, this result still applies (the equilateral triangle thereby constructed is just a point if the original triangle is equilateral);
• The two equilateral triangles obtained above differ in area by the area of the original triangle (in particular, if its side lengths were $a,\,b,\,c$ and its area $\Delta$ then our constructed equilateral triangles have areas $\dfrac{a^2+b^2+c^2}{8\sqrt{3}}\pm\dfrac{\Delta}{2}$, proving Weitzenböck's inequality with equality iff the original triangle is equilateral).

Cut the Knot has at least three different proofs of this using complex numbers; their first is here.

There are many parts of physics and engineering where pairs of inherently real physical quantities (say $x$ and $y$) appear together in two equations in a pattern like $\dots ax \dots by \dots$ and $\dots -bx \dots ay \dots$, for some constants $a$ and $b$. This pattern can be reformulated as a single equation for the complex numbers $\dots (x + iy)(a - ib) \dots$, where the real and imaginary parts of the complex equation correspond to the two real equations.

In numerical methods applied to these problems, the complex formulation typically halves the memory requirements ($a$ and $b$ only appear once in the complex equation, but twice in the two real ones), can reduce the amount of computation by a factor of 4 or more.

Even bigger savings can be made in some applications. For example, when modelling the vibration of a structure with cyclic symmetry (e.g. a fan with $n$ identical blades), in the general case the motion of each blade is different, but the motion of the $n$ corresponding points on the complete set of blades can be described by a discrete Fourier series, requiring only one complex number for any value of $n$ (which in real engineering applications may be 100 or more) rather than $2n$ real numbers (note, $2n$ not $n$, because the corresponding points on each blade move in different phases relative to each other as well as with different amplitudes).

This decouples one very large problem into $n$ much smaller ones, and there is the added advantage that often only a few of the small problems have "interesting" solutions from a practical point of view (and the "interesting" subset is known a priori), so most of the $n$ small problems need not be solved at all.

Indirectly, I used complex numbers to solve the distribution of returns for stocks and all asset classes. The inspiration was in the complex, but because the algebra works out the same I solved it in $R^2$. My field doesn't tend to use complex numbers except incidentally. Nonetheless, I do not think I would have found a general solution without the complex numbers.

Returns are the sell price divided by the purchase price minus one. It is a ratio distribution. In its simplest construction, it is the Cauchy distribution.

There are two main ways to derive the Cauchy distribution. The first and most obvious is that it is the ratio of two independent random normal variates centered on $\mu$. The second is a little less obvious, that is to convert it to polar coordinates and solve it using trigonometric methods. In particular, it forces the arc tangent into the equations somewhere. The CDF of the Cauchy is the arc tangent.

Traditionally, you center it on $(0,0)$. Where complex numbers came in for me was the realization that any point in $\mathbb{C}$ can be considered $(0,0)$ because it is not an ordered set. The first realization was I could slide the bivariate normal distribution around anywhere. The second was I could distort the shape simply with a covariance term. The third was that it did not have to be jointly normal, it could be any pairing of distributions. Since realized return is future value divided by present value minus one, I could handle any type of asset and any set of rules for purchases and sales by allowing it to slide around the plane freely using the distributions that are logical to the problem.

It also links, loosely, returns to planar Brownian motion by thinking of errors as having two components, an initial error and a future error. Indeed, in a curious sense, your mentally modeled error term for the sale price is "imaginary" and the error realized when the purchase went through was "real," at least until the closing transaction.

The Egorychev method: using line integrals to calculate combinatorial identities. Many examples in Egorychev method and the evaluation of binomial coefficient sums and here:

Prove an identity in a Combinatorics method

proving an invloved combinatorial identity

An identity involving binomial coefficients

...

In discrete mathematics and in statistical physics, one often has to deal with summations over variables that are not completely independent due to constraints, such as the sum of all these variables being fixed. An example from statistical mechanics is this solution of a problem posted here on StackExchange. This problem is a good discrete math example. The solution is easily obtained using generating functions, but calculating the series expansion coefficients of the generating function, particularly the asymptotic behavior of these coefficients, is most easily done using complex analysis.

$\newcommand{\SLp}{\mathrm L^{#1}}\newcommand{\norm}{\lVert#1\rVert}$ The famous Riesz-Thorin Interpolation Theorem:

Let $(X,\mathcal M,\mu),(Y,\mathcal N,\nu)$ be measure spaces, $p_0,p_1,q_0,q_1 \in [1,\infty]$. (If $q_0 = q_1 = \infty$, $\nu$ is also required to be semi-finite) Define, for $t \in \left]0,1\right[$, $$\frac{1}{p_t} = \frac{1-t}{p_0} + \frac{t}{p_1}, \qquad \frac{1}{q_t} = \frac{1-t}{q_0} + \frac{t}{q_1}$$ If $\Phi\in\operatorname{Hom}(\SLp{p_0}(\mu) + \SLp{p_1}(\mu),\SLp{q_0}(\nu) + \SLp{q_1}(\nu))$ such that $\Phi\restriction_{\SLp{p_0}},\Phi\restriction_{\SLp{p_1}}$ are bounded. Then $$\norm{\Phi}_{\SLp{p_t} \to \SLp{q_t}} \leq \norm{\Phi}_{\SLp{p_0} \to \SLp{q_0}}^{1-t}\norm{\Phi}_{\SLp{p_1} \to \SLp{q_1}}^t$$

which does not seem to be related to complex analysis, is usually proved using the Three Lines Lemma (a.k.a. Hadamard Three Line Theorem):

Let $f$ be a bounded continuous complex function on the strip $E = \{z \in \mathbb{C}\mid a \leq \Re z \leq b\}$ that is holomorphic on $E^\circ$. Define $m(x) = \sup_{y \in \mathbb{R}} |f(x + iy)|$. Then $m(x)$ is logarithmically convex on $[a,b]$.