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Contour integral

Let f be analytic in the disk |z|<2. Prove that:

    \[\frac{1}{2\pi i} \oint \limits_{\left | z \right |=1} \frac{\overline{f(z)}}{z-\alpha} \, \mathrm{d}z = \left\{\begin{matrix} \overline{f(0)} & , & \left | \alpha \right |<1 \\\\ \overline{f(0)} - \overline{f\left ( \frac{1}{\bar{\alpha}} \right )} & , & \left | \alpha \right |>1 \end{matrix}\right.\]

Solution

It follows from Taylor that f(z)=\sum \limits_{n=0}^{\infty} c_n z^n and the convergence is uniform. Hence,

    \begin{align*} \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\overline{f(z)}}{z-\alpha} \,\mathrm{d}z &=\frac{1}{2\pi i }\oint \limits_{|z|=1} \sum_{n=0}^{\infty} \frac{\overline{c_n} \bar{z}^n}{z-\alpha} \,\mathrm{d}z \\ &= \sum_{n=0}^{\infty} \frac{\overline{c_n}}{2\pi i }\oint \limits_{|z|=1}\frac{ \bar{z}^n}{z-\alpha} \,\mathrm{d}z \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{\left | z \right |=1\Rightarrow \bar{z}^n = \frac{1}{z^n}}{=\! =\! =\! =\! =\! =\! =\!=\! =\!=\!}\sum_{n=0}^{\infty} \frac{\overline{c_n}}{2\pi i }\oint \limits_{|z|=1}\frac{1}{z^n(z-\alpha)} \,\mathrm{d}z \end{align*}

We have that \displaystyle \mathrm{Res}\left( \frac{1}{z^n(z-\alpha)};\alpha\right) = \frac{1}{\alpha^n} and for n \geq 1 we also have that \displaystyle \displaystyle \mathrm{Res}\left( \frac{1}{z^n(z-\alpha)};0\right) = -\frac{1}{\alpha^n} due to

    \[\frac{1}{z^n(z-\alpha)} = -\frac{1}{\alpha z^n} \frac{1}{1-z/\alpha} = -\frac{1}{\alpha z^n}\left(1 + \frac{z}{\alpha} + \frac{z^2}{\alpha^2} + \cdots \right)\]

So if |\alpha|<1 then \alpha lies within the disk |z|=1; hence the integral equals \overline{c_0} = \overline{f(0)} whereas if |\alpha|>1 then \alpha lies outside the disk |z|=1; hence

    \begin{align*} \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\overline{f(z)}}{z-\alpha} \,\mathrm{d}z &= -\sum_{n=1}^{\infty} \frac{\overline{c_n}}{\alpha^n} \\ &= \overline{c_n} - \sum_{n=0}^{\infty} \frac{\overline{c_n}}{\alpha^n} \\ &= \overline{f(0)} - \overline{\left ( \sum_{n=0}^{\infty} \frac{c_n}{\bar{\alpha}^n} \right )}\\ &= \overline{f(0)} - \overline{f\left ( \frac{1}{\bar{\alpha}} \right )} \end{align*}

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