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A homework: Calculate the integral $$\int_{|z|=r}\frac{|dz|}{z-1}$$, $z$ is a complex variable, and $r\neq1$.

When $r<1$, it can be obtained by the mean value formula, I get stacked in the case $r>1$.

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Could you give a little more background on what you tried and where you get stuck? Thank you. –  vonjd Oct 22 '12 at 16:18
    
of course we can solve it by $z=r e^{i \theta}$, but it is a long calculation. –  van abel Oct 22 '12 at 16:23
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1 Answer

up vote 4 down vote accepted

Hint:

Use polar coordinates $z=r{e^{i\varphi}}$ on the circle $|z|=r \ $; then $$|dz|=r\ d\varphi, \quad 0 \leqslant \varphi< 2 \pi,\\ \int\limits_{|z|=r}\frac{|dz|}{z-1}=\int\limits_{0}^{2\pi}\frac{r\ d\varphi}{r{e^{i\varphi}}-1}.$$ \begin{gather} \int\limits_{0}^{2\pi}\frac{r\ d\varphi}{r{e^{i\varphi}}-1}=\int\limits_{0}^{2\pi}\frac{r\ d\varphi}{r{e^{i\varphi}}\left(1 - \frac{e^{-i\varphi}}{r}\right)}. \end{gather} In the case $r>1$ we can expand \begin{gather} \frac{1}{1 - \frac{e^{-i\varphi}}{r}}=\sum\limits_{k=0}^{\infty}{\frac{e^{-ik\varphi}}{r^k}}. \end{gather} Note that series is absolutely convergent, so we can integrate it: \begin{gather} \int\limits_{0}^{2\pi}\frac{r\ d\varphi}{r{e^{i\varphi}}\left(1 - \frac{e^{-i\varphi}}{r}\right)}=\sum\limits_{k=0}^{\infty}\int\limits_{0}^{2\pi}{\frac{e^{-i(k+1)\varphi}}{r^k}d\varphi}. \end{gather} Since $k \ne -1, \quad \int\limits_{0}^{2\pi}e^{-i(k+1)\varphi}\ d\varphi=0,$ therefore, $$\int\limits_{0}^{2\pi}\frac{r\ d\varphi}{r{e^{i\varphi}}\left(1 - \frac{e^{-i\varphi}}{r}\right)}={0}$$

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too tedious! Can you give another solution? –  van abel Oct 22 '12 at 16:44
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@van abel: It's not so hard — I add details to previous answer –  M. Strochyk Oct 22 '12 at 17:25
    
thanks, just note that $k\neq-1$. –  van abel Oct 23 '12 at 14:43
    
@van abel: Thanks, You are absolutely right! –  M. Strochyk Oct 23 '12 at 15:04
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