# How prove this inequality $e^{|\Im(z)|}\le B|\sin{z}|$

$\def\Re{\mathop{\mathrm{Re}}} \def\Im{\mathop{\mathrm{Im}}}$Let $z\in \mathbb{C}$ with $|z-n\pi|\ge\dfrac{\pi}{4}$ for all $n\in \mathbb{Z}$. If $$e^{|\Im(z)|}\le B|\sin{z}|, \quad \forall z \in \mathbb{C}$$ find the minimun $B$.

I have prove $B\ge 4$, but I think it is very ugly, do you have nice methods? And I think this $B$ is the smallest. Can you find this optimal minimun of $B$? Thank you.

Following is my solution for $B=4$:

$$e^{|\Im(z)|}\le 4|\sin{z}|.$$

Let $z=z_{1}+iz_{2},\ z_{1}\in \mathbb{R},\ z_{2}\in \mathbb{R}$, then $$e^{|\Im(z)|}=e^{|z_{2}|},$$ $$|\sin{z}|=\left|\frac{e^{iz}-e^{-iz}}{2i}\right|=\frac{1}{2} |e^{iz_{1}-z_{2}}-e^{-iz_{1}+z_{2}}|.$$ (1): If $|z_{2}|>\dfrac{\ln{2}}{2}$, then $$\dfrac{e^{|\Im(z)|}}{|\sin{z}|}=\dfrac{2e^{|z_{2}|}}{|e^{iz_{1}-z_{2}}-e^{-iz_{1}+z_{2}}|}\le\dfrac{2e^{|z_{2}|}}{e^{|z_{2}|}-e^{-|z_{2}|}}=\dfrac{2}{1-e^{-2|z_{2}|}}<4.$$ (2): If $|z_{2}|\le\dfrac{\ln{2}}{2}$, since $|z-n\pi|\ge\dfrac{\pi}{4}$, we have $$|z_{1}-n\pi+iz_{2}|\le|z_{1}-n\pi|+|iz_{2}|=|z_{1}-n\pi|-|z_{2}|$$ $$\Longrightarrow |z_{1}-n\pi|^2\ge\left(|z_{1}-n\pi+iz_{2}|+|z_{2}|\right)^2\ge\dfrac{\pi^2}{16},$$ so$$|\sin{z_{1}}|\ge\dfrac{\sqrt{2}}{2},$$ \begin{align} \sin{z}&=\dfrac{e^{iz}-e^{-iz}}{2i}=\dfrac{e^{iz_{1}-z_{2}}-e^{-iz_{1}+z_{2}}}{2i}\\ &=-\dfrac{i}{2}(e^{-z_{2}}(\cos{z_{1}}+i\sin{z_{1}})-e^{z_{2}}(\cos{z_{1}}-i\sin{z_{1}}))\\ &=-\dfrac{i}{2}(e^{-z_{2}}\cos{z_{1}}-e^{z_{2}}\cos{z_{1}})+\dfrac{1}{2}(e^{-z_{2}}\sin{z_{1}}+e^{z_{2}}\sin{z_{1}})\\ &=-\dfrac{i}{2}\cos{z_{1}}\cdot(e^{-z_{2}}-e^{z_{2}})+\dfrac{1}{2}\sin{z_{1}}\cdot(e^{-z_{2}}+e^{z_{2}})\\ &=i\sinh{z_{2}}\cdot\cos{z_{1}}+\cosh{z_{2}}\cdot\sin{z_{1}}, \end{align} so $$|\Re(\sin{z_{1}})|=|\sin{z_{1}}|\cdot|\cosh{z_{2}}|\ge|\sin{z_{1}}|,$$ so $$\dfrac{e^{|\Im(z)|}}{|\sin{z}|}\le\dfrac{e^{|z_{2}|}}{|\Re(\sin{z})|}\le\dfrac{\sqrt{2}}{|\sin{z_{1}}|}\le2.$$

Let $z=x+yi$, suppose $y\ge 0$. We shall find $\min \frac{|\sin z|}{e^y}$ when $\{x\}^2+y^2\ge (\frac{\pi}{4})^2.$
When $y\ge \frac{\pi}{4}$, $$|\sin z|^2/e^{2y}=e^{-2y}(e^{2y}+e^{-2y}-2\cos (2x))\ge (e^{-2y})(e^{2y}+e^{2y}-2):=f(y)\ge f(\pi/4)$$ When $0\le y\le \frac{\pi}{4}$ $$|\sin z|^2/e^{2y}=e^{-2y}(e^{2y}+e^{-2y}-2\cos (2x))$$ is obtained at minimal at $x^2+y^2=(\frac{\pi}{4})^2$, let $x=\frac{\pi}{4} \cos t, \ y=\frac{\pi}{4} \sin t$. $$|\sin z|^2/e^{2y}\ge 1+e^{-\pi \sin t}-2 e^{-\frac{\pi}{2}\sin t}\cos (\frac{\pi}{2}\cos t)$$ By cheating http://www.wolframalpha.com/input/?i=minimize+%7B1%2Be%5E%28%28-pi%29sin%28x%29%29+-2e%5E%28-%5Cpi%2F2+sin+x%29cos%28%28pi%2F2%29%28cos+x%29%29%7D $$\min 1+e^{-\pi \sin t}-2 e^{-\frac{\pi}{2}\sin t}\cos (\frac{\pi}{2}\cos t)=f(\pi/4),$$ it remains to prove this by hand. It seems no good way.
• Thank you ,why your only consider when $x^2+y^2\ge\dfrac{\pi^2}{4}$,you mean $|x-n\pi|\ge\dfrac{\pi}{4}$ equivalent? can you explain? May 6, 2013 at 13:06
• @math110 Not equivalent. Replace $x$ by $x-k\pi$ do not change the problem. May 6, 2013 at 13:13
• Replace $x$ by $x-k\pi$ does not change anything, so you may consider only $|x|\le \pi/2$ with $x^2+y^2\ge (\frac{\pi}{4})^2$. May 6, 2013 at 13:23