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Why is $ \vert e^{iz} \vert = e^{-Im(z)} $

I tried rewriting z as x +iy, but I can't derive the equality

4 Answers 4

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If $z=x+iy$ then $e^{iz}=e^{ix}e^{-y}$, and since $|e^{ix}|=1$ it follows that $|e^{iz}|=e^{-y}$.

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$\newcommand{\Re}{\operatorname{Re}}$You only need to prove that for all $z \in \mathbb{C}$, (Why?) $$ |e^z| = e^{\Re z} $$ This follows directly from Euler's formula, if $z = x + \mathrm iy$, $$ |e^z| = |e^x e^{\mathrm iy}| = e^x $$

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    Quick question. Why does $\left|e^x\right|=e^x$? – 2018-07-26
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    @Bell Its been a long time since I answered this question, but it seems like $x$ is real, so $e^x$ is positive. – 2018-07-26
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If $$z=x+iy\\iz=xi+i^2y=-y+ix\\\to e^{-y+ix}=x^{-y}.e^{ix}=\\e^{-y}(cosx+isinx)$$ so $$|e^{iz}|=|e^{-y}(cosx+isinx)=|e^{-y}||(cosx+isinx)|=\\e^{-y}.\sqrt{sin^2x+cos ^2x}=\\e^{-y}.1\\=e^{-y}$$

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$$|e^{iz}|=|e^{i(x+iy)}|=|e^{-y+ix}|=|e^{-y}||e^{ix}|=e^{-y}$$

Because $|e^{ix}|=1$ and $e^{-y}\gt 0$