Prove that $|\sin z|\geq1$ on the boundary of a square with corners $\pi(m + 1/2)(\pm 1\pm i)$ for $m=0,1,2,3,\ldots$

Prove that $|\sin z|\geq1$ on the boundary of a square with corners $\pi(m + \frac{1}{2})(\pm 1\pm i)$ for $m=0,1,2,3,\ldots$

My approach was to write

$$|\sin z| = |\sin x\cos iy +\cos x\sin iy|$$

then I'm clueless how should I proceed there. Am I in the right direction?
Thanks!

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This was from a homework I assigned. Not happy that somebody gave a detailed answer. Please people- the student asked "am I in the right direction" not "please give a full detail solution for me to copy verbatim". (students who copy such things verbatim cheat themselves, but still...) – Daniel Moskovich Sep 5 '12 at 16:21

$\sin(x+iy)=\sin x\cos(iy)+\cos x \sin(iy)=\sin x\cosh y + i \cos x\ sinh y$

$|\sin(z)|=|\sin(x+iy)|=\sqrt{\sin^2x\cosh^2y+\cos^2x\sinh^2y}=\sqrt{\cosh^2y-cos^2x}$ as $sin^2A+cos^2A=1$ and $cosh^2B-sinh^2B=1$

Now if we consider points on the two lines $x=±(2m+1)\frac{\pi}{2}$,

$\cos x=0$ so, $|\sin(z)|=|\sin(x+iy)|=|\cosh y|=|\frac{e^y+e^{-y}}{2}|$ at $x=±(2m+1)\frac{\pi}{2}$.

Now for the positive real value of $d$, $d+\frac{1}{d}=(\sqrt d-\frac{1}{\sqrt d})^2+2≥2$

$|\sin(z)|=|\sin(x+iy)|≥1$ for any real $y$ at $x=±(2m+1)\frac{\pi}{2}$

Now if we consider points on the two lines $y=±(2m+1)\frac{\pi}{2}$,

$|\sin(z)|=|\sin(x+iy)|=\sqrt{\cosh^2y-cos^2x}=\sqrt{\sinh^2y+sin^2x}$

Now, $e^p-e^{-p} > e^q-e^{-q}$ iff $(e^p-e^q)(e^{p+q}-1)>0$ which is true if $p>q>0$

So, $(e^r-e^{-r})^2 > (e-e^{-1})^2$ if $r^2>1$

$sinh^2y=(\frac{e^y-e^{-y}}{2})^2>(\frac{e-e^{-1}}{2})^2$ if $y=±(2m+1)\frac{\pi}{2}$,

But $e-e^{-1}>2$ as $e^2-2e-1>0$ as $(e-1)>\sqrt2$

So, $sinh^2y>1$ if $y=±(2m+1)\frac{\pi}{2}$,

and $|\sin(z)|=|\sin(x+iy)|=\sqrt{\cosh^2y-cos^2x}$ $=\sqrt{\sinh^2y+sin^2x}>\sqrt{1+sin^2x}>1$ if $y=±(2m+1)\frac{\pi}{2}$,

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Hint: Use the definition $$\sin[z]=\frac{e^{iz}-e^{-iz}}{2i}$$

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