Proof inequality with complex analysis $\csc(z)$

Question

I'm studying complex analysis and i found this problem

$$ |\csc(z)|\le \frac{2e}{e^2 - 1} $$

Where $z = x + iy$ and $|y|=1$.

I started defining $\sin(x + iy)$ and put it on Euler's function but it doesn't work. Any tip? Everything will be helpful for me.

Thanks.

Answer 1

$$|\csc z|$$ $$=|\frac{1}{\sin z}|$$ $$=\frac2{|e^{iz}-e^{-iz}|} \text{(as,$sinz=\frac{e^{iz}-e^{-iz}}{2}$)} $$ $$\le\frac2{||e^{iz}|-|e^{-iz}||} \text{(as, $||x|-|y||\le |x-y| $)}$$ $$=\frac2{||e^{-y}|-|e^{y}||} \text{(as,$|e^{ix}|=1=|e^{-ix}|$)}$$ $$=\frac2{e-e^{-1}} \text{(as,$|y|=1$)}$$ $$=\frac{2e}{e^2-1}$$

Answer 2

By definition you have $$ \csc z=\frac1{\sin z} $$ and $$ \sin z=\frac{e^{iz}-e^{-iz}}{2i}\;. $$ Observe then that $z=x\pm i$ then $iz=\mp1+ix$ and $|e^w|=e^{\Re w}$ for every $w\in\Bbb C$, so you have to distinguish the two cases according with the sign. Il will write the case $z=x+i$: \begin{align*} |\csc z| &=\frac2{|e^z-e^{-z}|}\\ &\le\frac2{|e^{-iz}|-|e^{iz}|}\\ &=\frac2{e-e^{-1}}\\ &=\frac{2e}{e^2-1}\;. \end{align*} The other is similar, you just have to "swap" the summand in the reverse triangular inequality.