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An exercise asks to find all $z\in\mathbb C$ such that $|{\sin z}|\leq 1$ and then an $n\in\mathbb N$ such that $|\sin(in)|>10000$. Here are some results we can use. For all $z=x+iy\in\mathbb C$ one has:

  1. $\overline{\exp(z)}=\exp(\overline z),\quad \overline{\sin(z)}=\sin(\overline z),\quad \overline{\cos(z)}=\cos(\overline z)$
  2. $\cos z=\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y$
  3. $\sin z=\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y$

My problem is that I end up with expressions that are as hard to handle as the starting expression.

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The latter should be simple, as you need only find an $n$ such that $\sinh n>10000.$ (Use result 3.) For the former, note that $|\sin z|\le1$ if and only if $|\sin z|^2\le1$, and apply pythagorean identities. That should make it a bit easier. –  Cameron Buie Jan 4 '13 at 15:03
    
@cameron: I have already tried that. I end up with $\sinh^2y\leq\cos^2x$. –  Kiuhnm Jan 4 '13 at 17:28
    
Excellent! That allows you to completely describe the set of points you're looking for. That is: $$\{z\in\Bbb C:|\sin z|\le1\}=\{x+iy:x,y\in\Bbb R,\sinh^2y\le\cos^2x\}.$$ If you like, you (or I, if you'd prefer) can write it up in an answer, and you can accept it (after 2 days, if you answer your own question). –  Cameron Buie Jan 4 '13 at 17:49
    
I think I'll write it up anyway. If you wish to post your own work, just comment on mine after you've posted yours, and I'll delete mine. –  Cameron Buie Jan 4 '13 at 17:56
    
@cameron: I thought we were supposed to find something more explicit. Anyway I'll accept your answer, since you took the time to write it. –  Kiuhnm Jan 6 '13 at 12:30

1 Answer 1

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As you've determined, for $z=x+iy$ with $x,y\in\Bbb R$,

$$\begin{align}|\sin z|^2 &= \sin^2x\cosh^2y+\cos^2x\sinh^2y\\ &= (1-\cos^2x)(1+\sinh^2y)+\cos^2x\sinh^2y\\ &= 1+\sinh^2y-\cos^2x.\end{align}$$ Hence, $|\sin z|\le 1$ if and only if $|\sin z|^2\le 1$ if and only if $z=x+iy$ with $x,y\in\Bbb R$ such that $\sinh ^2y\le\cos^2x$. Thus, $$\{z\in\Bbb C:|\sin z|\le1\}=\{x+iy:x,y\in\Bbb R\text{ and }\sinh^2y\le\cos^2x\}.$$

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