# Why $|f|=1$ on the boundary?

The function $f(z)=\exp(\exp(z))$ on the domain $U=\{z=x+iy:-\pi/2<y<\pi/2\}$. Why $|f|=1$ on the boundary of $U$?

$$|f(x\pm i\pi/2)|=\left|\exp(\exp(x\pm i\pi/2))\right|$$ I don't know why it is 1?

Based on Harald's hint, I want to prove $e^{x\pm i\pi/2}$ is imaginary. $$e^{x\pm i\pi/2}=e^x[\cos(\pi/2)\pm i\sin(\pi/2)]=\pm ie^x.$$

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If $x$ is real then $x+\pi/2$ is not on the boundary. –  Did Nov 26 '12 at 7:28
There was an obvious misprint in the question (missing $i$ in some places). I fixed that, and also replaced $+$ by $\pm$ to accomodate both components of the boundary. –  Harald Hanche-Olsen Nov 26 '12 at 7:30
Thank you, Harald. –  Sam Nov 26 '12 at 7:34

First, $\lvert e^w\rvert=1$ if and only if $w$ is imaginary. It follows that you need to show that $\exp(x\pm i\pi/2)$ is imaginary for any real $x$. As this smells a bit like homework, I am going to let you finish it on your own.