# Abstract Nonsense

## Cute Complex Analysis Problem

Problem: Let $f:\mathbb{C}\to\mathbb{C}$ be entire with

$f(z)=f(x+iy)=\mu(x,y)+\nu(x,y)i$

Then, if either $|\mu(x,y)|\leqslant M$ or $|\nu(x,y)|\leqslant N$ for some $M,N\in\mathbb{R}$ we have that $f$ is constant.

Proof: Evidently we must find some way to employ Liouville’s theorem, but how? It may seem hard at first to do this, but in fact the reason the problem is “cute” is because it has a “cute” solution (surprisingly simple). To do this we note that if $f(z)$ is entire then so is $e^{f(z)}$. But, note that

$\left|e^f(z)\right|=\left|e^{\mu(x,y)+\nu(x,y)i}\right|=\left|e^{\mu(x,y)}\right|\left|e^{\nu(x,y)i}\right|=\left|e^{\mu(x,y)}\right|$

But, remembering that

$\displaystyle e^{z}=\sum_{n=0}^{\infty}\frac{z^n}{n!}$

and for finite $N$

$\displaystyle \left|\sum_{n=0}^{\infty}\frac{z^n}{n!}\right|\leqslant\sum_{n=0}^{\infty}\left|\frac{z^n}{n!}\right|=\sum_{n=0}^{\infty}\frac{|z|^n}{n!}$

And thus, by limiting process it follows that

$\displaystyle e^z=\lim_{N\to\infty}\left|\sum_{n=0}^{N}\frac{z^n}{n!}\right|\leqslant\lim_{N\to\infty}\sum_{n=0}^{N}\frac{|z|^n}{n!}=e^{|z|}$

And thus,

$\left|e^{f(z)}\right|=\left|e^{\mu(x,y)}\right|\leqslant e^{|\mu(x,y)|}\leqslant e^M$

And thus it follows by Liouville’s theorem that $e^{f(z)}$ is constant, but this is only possible if $f(z)$ is constant.

Now, suppose that $\left|\nu(x,y)\right|\leqslant N$, then we merely note that

$\displaystyle \frac{f(z)}{i}=\nu(x,y)-\mu(x,y)i$

and applying the exact same logic as in the previous part we may conclude that $\displaystyle \frac{f(z)}{i}=z_0$ for some $z_0\in\mathbb{C}$. Thus, $f(z)=iz_0$. $\blacksquare$