Question : Given $\text{Re}(z) \le 0$ prove that $|e^z| \le 1$.
Try: $z=x+yi$, it's given that $x \le 0$.
$|e^{z}| = |e^{x+yi}|=|e^xe^{yi}|=e^x|e^{yi}|,$ with $e^x \le e^0$ because $f(x)=e^x $ is a increasing function everywhere.
What's next? What can I say about $|e^{yi}$| ?