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For $z_0\in{\Bbb C}$ such that $z_0\neq0$ and $R<|z_0|$, how can I compute $$ \int_{0}^{2\pi}\frac{Re^{it}}{z_0+Re^{it}}dt? $$

With help of Cauchy Theorem, one can conclude that $$ \int_{0}^{2\pi}\frac{iRe^{it}}{z_0+Re^{it}}dt=\int_{\gamma}\frac{1}{z}dz=0 $$ where $\gamma=\{z\in{\Bbb C}:|z-z_0|=R\}$. How can I do this without Cauchy Theorem?

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2 Answers 2

up vote 1 down vote accepted

Here is another way: Since $R<|z_0|$, we can write

\begin{eqnarray} \frac{Re^{it}}{z_0+Re^{it}} &=& \frac{R}{z_0}\frac{e^{it}}{1+\frac{R}{z_0}e^{it}} \\ &=& (\frac{R}{z_0}e^{it}) \sum_{n=0}^\infty (-1)^n (\frac{R}{z_0}e^{it})^n \\ &=& \sum_{n=1}^\infty (-1)^{n+1} (\frac{R}{z_0}e^{it})^n \\ &=& \sum_{n=1}^\infty (-1)^{n+1} (\frac{R}{z_0})^ne^{int} \end{eqnarray} Since the convergence is absolute, we have $\int_0^{2 \pi} \frac{Re^{it}}{z_0+Re^{it}} dt = \sum_{n=1}^\infty (-1)^{n+1} (\frac{R}{z_0})^n \int_0^{2 \pi} e^{int} dt$, and since $\int_0^{2 \pi} e^{int} dt = 0$ as long as $n\neq 0$, we have the desired result.

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$-i(\ln(z_0 + Re^{it}))' = \dfrac{Re^{it}}{z_0 + Re^{it}}$

So $-i\ln(z_0 + Re^{it})\Bigl|_0^{2\pi} = 0$ because $e^{i0} = e^{i2\pi}$.

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