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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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1 Answer 1

up vote 4 down vote accepted

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

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Im really sad I need this site for such a silly answer ;) –  Applied mathematician Nov 18 '12 at 21:03
@Hempo It is fine :). It happens to everyone :-). Also, it is worth noting that $$\left \vert \cos(y) + i \sin(y) \right \vert = \sqrt{\cos^2(y) + \sin^2(y)}$$ is true only when $y \in \mathbb{R}$ and not in general when $y \in \mathbb{C}$. –  user17762 Nov 18 '12 at 21:10
Thanks for your help Marvis! –  Applied mathematician Nov 18 '12 at 22:08

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