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

S= { 1 < Rez =< 2, -pi/4 < lmz =< 0}

I got radius of f is bound by e^3 =< r =< e^4 , 0 < Argz < pi/4 but the solution is between 0 =< r =< 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

1 Answer 1

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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