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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 $|e^{yi}$| ?

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$\vert e^{iy} \vert = \underbrace{\vert \cos(y) + i \sin(y) \vert = \sqrt{\cos^2(y) + \sin^2(y)}}_{\because y \in \mathbb{R}} = 1$

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    Thanks for your help Marvis!2012-11-18