# Interesting applications of major results in complex analysis

I know of two examples of major theorems in complex analysis being used to answer problems that are easy to state but otherwise not so easy to prove:

• using the Maximum modulus principle applied to $$\frac{1}{p(z)}$$ to prove that the polynomial $$p$$ has a complex root, proving the Fundamental theorem of algebra; and
• applying the Residue theorem to a well-chosen function to show that $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$, solving the Basel problem and giving a closed form for $$\zeta(2)$$.

I'm wondering what other applications of such major theorems one could add to the list. Specifically, problems that anyone with standard 'Math 101' training can understand, but that can't be solved until you bring in the tools of complex analysis, or something at least as advanced.

(There's the easy cases of complicated real trig integrals that become much neater when you turn the trig functions into complex ones and then take real parts at the end of some contour integration, but those problems aren't as interesting or elegant as the two I listed above.)

• I don't see how you use the maximum modulus principle to prove the first point? Isn't Liouville's theorem more appropriate? Sep 21 '21 at 8:59
• If you have a look at en.wikipedia.org/wiki/…, you'll see many complex-analytic ways of proving the FTA (Maximum modulus, Liouville, Rouche...).
– SPS
Sep 21 '21 at 9:10
• @spinosarus123 The argument for FTA via MMT is very simple: If $p$ is a polynomial with no zero then $f=1/p$ is entire; also $\lim_{z\to\infty}f(z)=0$, so MMT shows that $f=0$, impossible. Sep 21 '21 at 10:26

An example would be Brouwer's fixed point theorem: every continuous map $$f\colon\overline{D_1(0)}\longrightarrow\overline{D_1(0)}$$ has a fixed point. If this was not the case for some $$f$$, you could prove that for each $$z\in\overline{D_1(0)}$$ there is one and only one number $$t(z)\in(0,\infty)$$ such that$$f(z)+t(z).\bigl(z-f(z)\bigr)\in S^1\bigl(=\{z\in\Bbb C\mid|z|=1\}\bigr).$$Then the function$$\begin{array}{rccc}F\colon&\overline{D_1(0)}&\longrightarrow&S^1\\&z&\mapsto&f(z)+t(z).\bigl(z-f(z)\bigr)\end{array}$$is continuous and $$\left(\forall z\in S^1\right):F(z)=z$$. From $$F$$, we can defined a homotopy in $$\Bbb C^*$$ between a contant loop and the loop$$\begin{array}{rccc}\gamma\colon&[0,1]&\longrightarrow&\Bbb C\\&t&\mapsto&e^{2\pi it};\end{array}$$just consider$$\begin{array}{rccc}H\colon&[0,1]\times[0,1]&\longrightarrow&\Bbb C\\&(t,u)&\mapsto&F\left(ue^{2\pi it}\right).\end{array}$$But if $$\gamma$$ was homotopic in $$\Bbb C^*$$ to a constant loop, it would follow from Cauchy's theorem that $$\oint_\gamma\frac{\mathrm dz}z=0$$. However, $$\oint_\gamma\frac{\mathrm dz}z=2\pi i$$.
Using ordinary integer variables, you have to jump through hoops to get primitive Pythagorean triples. But using the fact that $$|z^2|=|z|^2$$, you may simply render
$$(m+ni)^2=(m^2-n^2)+(2mn)i$$
$$(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$$