Upper bound for $\exp(\exp(iz))$ function

Question

Let $|z| > 1, \arg z \in [0, \pi]$.

Show that $e^{e^{iz}} \leq 1$

I tried few approaches but they led me nowhere.

Answer 1

I think that you have to show that $|e^{e^{iz}}| \leq 1.$ But this is not true (see the comment of Jean Claude Arbaut).

We have that for all $w \in \mathbb C:$

$$ (*) \quad |e^w| \le e^{|w|}.$$

Now let $z=x+iy \in \mathbb C$ with $\arg z \in [0, \pi]$, then $y \ge 0.$ Thus $iz=ix-y.$ Therefore $w:=e^{iz}= e^{ix}e^{-y}$. This gives $|w|=e^{-y} \le 1.$

From $(*)$ we now derive

$$|e^{e^{iz}}|= |e^w| \le e^{|w|} \le e^1=e.$$