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Let $f = e^{2-z}$. Find and sketch the image $f(S)$ of the strip

$$S=\{1 < \mathrm{Re}(z) \leq 2, -\pi/4 < \mathrm{Im}(z) \leq 0\}.$$

I got radius of $f$ is bound by $e^3 \leq r \leq e^4, 0 < \mathrm{Arg} z < \pi/4$ but the solution is between $0 \leq r \leq e^1$.

Can someone explain why?

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Enclose LaTeX-style math in $...$ (inline) and $$...$$ (display). Makes for much more readable text. Please edit your question. – vonbrand Feb 7 '13 at 4:34

If $\newcommand{\real}{\operatorname{Re}}1 < \real z \le 2$, then $0 \le \real(2-z) \le 1$, so

$$|e^{2-z}| = e^{\real(2-z)}$$

varies between $e^0 = 1$ and $e^1$.

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