Proving the cotangent function is uniformly bounded on the complex plane

Question

I'm trying to prove that the function $\cot\left(z\right)=i\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}$ is uniformly bounded in the complex plane outside $\varepsilon$ neighborhoods of the poles (with the bound depending on $\varepsilon$). The suggested method in my text is to first show that if $z=x+iy$ and $y>0$ then: $$\frac{e^{-2y}}{1+e^{-2y}}<\left|\cot\left(x+iy\right)+i\right|<\frac{e^{-2y}}{1-e^{-2y}}$$ And if $y<0$ then:$$\frac{e^{2y}}{1+e^{-2y}}<\left|\cot\left(x+iy\right)+i\right|<\frac{e^{2y}}{1-e^{-2y}}$$ The calculations are a bit tedious and they aren't coming out right for me for some reason: \begin{align*} \left|\cot\left(x+i\cdot y\right)+i\right| &=\left|i\cdot\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}+i\right| \\ &=\left|i\cdot\frac{e^{2iz}+1}{e^{2iz}-1}+i\right| \\ &=\left|\frac{i\cdot\left(e^{2iz}+1\right)+i\left(e^{2iz}-1\right)}{e^{2iz}-1}\right| \\ &=\left|\frac{2ie^{2i\left(x+i\cdot y\right)}}{e^{2i\left(x+i\cdot y\right)}-1}\right| \\ &=\left|2i\right|\cdot\left|e^{2iz}\right|\left|\frac{1}{e^{2iz}-1}\right|=2\cdot e^{-2y}\cdot\left|\frac{1}{e^{2iz}-1}\right| \\ &=2\cdot e^{-2y}\cdot\left|\frac{e^{2y}}{e^{2ix}-e^{2y}}\right| \\ &=2\left|\frac{1}{e^{2ix}-e^{2y}}\right| =2\frac{1}{\left|e^{2ix}-e^{2y}\right|} \\ &=2\cdot\frac{1}{\left|e^{ix}-e^{y}\right|\left|e^{ix}+e^{y}\right|} \\ &=\frac{2}{\sqrt{\left(e^{y}+\cos\left(x\right)\right)^{2}+\sin^{2}\left(x\right)}\cdot\sqrt{\left(e^{y}-\cos\left(x\right)\right)^{2}+\sin^{2}\left(x\right)}} \\ &=\frac{2}{\sqrt{e^{2y}+2e^{y}\cos\left(x\right)+1}\cdot\sqrt{e^{2y}-2e^{y}\cos\left(x\right)+1}} \\ &=\frac{2}{\sqrt{e^{4y}+2e^{2y}-4e^{2y}\cos^{2}\left(x\right)+1}} \\ &=\frac{2}{\sqrt{\left(1+e^{2y}\right)^{2}-4e^{2y}\cos^{2}\left(x\right)}} \end{align*} The denominator is maximal when $\cos^{2}\left(x\right)=1$ and minimal when $\cos^{2}\left(x\right)=0$ and thus: $$\frac{2e^{-2y}}{1+e^{-2y}}=\frac{2}{1+e^{2y}}\leq\left|\cot\left(x+i\cdot y\right)+i\right|\leq\frac{2}{\sqrt{\left(e^{2y}-1\right)^{2}}}=\frac{2}{\left|1-e^{2y}\right|}=\frac{2e^{-2y}}{1-e^{-2y}}.$$ I can't figure out whether I made an error in the calculations or whether there was an error in the suggested bound. I'm also not sure how to use these bounds in order to reach the required conclusion.

Regardless of this method I'm also curious whether someone has an alternative and perhaps less technical method of proving the claim.

Thanks in advance!

Answer 1

$\cot(z) = \cos(z)/\sin(z)$. The poles are at multiples of $\pi$, so you want to prove that for any $\epsilon > 0$ there is $M$ such that $|\cos(z)| \le M |\sin(z)|$ whenever $|z - n \pi| \ge \epsilon$ for all integers $n$.

By periodicity, it suffices to look at the strip $S = \{z: 0 \le \text{Re}(z) \le \pi\}$. After dealing with a compact set (on which a continuous function is bounded), it suffices to look at $\{z \in S: |\text{Im}(z)| \ge 1\}$. But if $y \ge 1$, $|e^{i(x+iy)}| = e^{-y} \le e^{y} e^{-2} = e^{-2} |e^{-i(x+iy)}|$, and then...

Answer 2

To shorten the work you have significantly you can use reverse triangle inequality to get a lower bound on the modulus of the denominator which can shortcut you to the last step. I don't know about the factor of 2 but it is unimportant for finishing the proof. To finish just optimize over $y$, most likely when y is epsilon. you can show that the ctg is bounded outside of $-\epsilon \leq y \leq \epsilon$.

To finish I think you would need to address when $z$ is between the poles which would require a separate bound. I think in this case you can Taylor expand the denominator about a given pole and get an estimate that is roughly proportional to the reciprocal of the distance to the pole.