Applications of complex numbers to solve non-complex problems

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:

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

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 $c=a+bi$).

While studying complex numbers from the book Complex Numbers from A to Z By Titu Andreescu, I found an interesting problem which was as follow:

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.

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:

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.)

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):

Cubic formula:

Quartic formula:

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.

$\newcommand{\SLp}[1]{\mathrm L^{#1}}\newcommand{\norm}[1]{\lVert#1\rVert}$ The famous Reisz-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 $[a,b]$.

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.

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 $n,k$:

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

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.

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 $\pm i$.