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In Schaum's complex variable book, there is an exercise in contour integration: $$ \int \overline{z}^{2} dz $$ over $|z|=1$.

The answer seems to be $0$, but when I integrate like this using contour integration formula, $$ \int_{0}^{2\pi}e^{-2i\theta} i e^{i\theta} d\theta $$ $$ -e^{-2i\pi}. $$ Then I get $-1$ as a result. :(

What am I doing wrong?

KCd
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simple
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1 Answers1

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$$ \int_{0}^{2\pi}e^{-2i\theta}\,i\,e^{i\theta}\,d\theta=i\int_{0}^{2\pi}e^{-i\theta}\,d\theta=i\,\frac{e^{-i\theta}}{-i}\,\Bigr|_0^{2\pi}=0. $$

Julián Aguirre
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  • I didn't understood the part before 0, shouldn't i and -i become -1 and after applying the limit and doing Euler's, we get 1. Finally 1 * -1 = -1? Sorry, if my question seems naive, I am just learning it. – simple Mar 29 '15 at 18:10
  • $e^{-2\pi i}-e^0=1-1=0$. – Julián Aguirre Mar 29 '15 at 18:12