# A tricky contour integral.

Let $$f$$ be holomorphic on a domain $$D$$ containing $$\overline{D}(0,1)$$, and $$\gamma$$ be the positively oriented circular contour centred at 0 of radius 1. Prove that $$\frac{1}{2 \pi i} \int_{\gamma} \frac{\overline{f(z)}}{(z-z_0)} dz= \begin{cases} \overline{f(0)} &\mbox{if} \,\, |z_0|<1, \\ \overline{f(0)}-\overline{f(1/ \overline{z_0})} &\mbox{if}\,\, |z_0|>1. \end{cases}$$

In the previous part of the question I was asked to prove that for a continuous function $$\phi:A=\{z\in\mathbb{C}:|z|=1\}\rightarrow\mathbb{C}$$ and $$\gamma$$ as above,

$$\overline{\int_{\gamma}\phi(z)\,\, dz} = -\int_{\gamma} \overline{\phi(z)}\,\,\frac{dz}{z^2}$$

which I've already done. So it's reasonable to think that the above would help solve the integral (perhaps along with Cauchy's Integral Formula) for the different cases of $$z_0$$ but I don't know how to get to a point where I can use it.

I've seen some suggestions of a variable change from $$z$$ to $$1/z$$ as $$\overline{f(\bar{z})}$$ is in fact holomorphic and so continuous on A since for $$|z|=1 \,,\bar{z}=1/z$$ by the identity $$\bar{z}z=|z|^2$$ but I don't know how to implement the change of variables on a contour integral because we haven't been taught how to do so yet which leads me to think that maybe it isn't what they want us to do.

Any guidance would be appreciated. Thank you in advance.

One of the quickest way to show this is to use power series. If $$|\zeta|<1$$, we may write $$\frac{1}{z-\zeta} = \sum_{j=0}^\infty \frac{\zeta^j}{z^{j+1}}$$, and if $$|\zeta|>1$$, then we have $$\frac{1}{z-\zeta} = -\sum_{j=0}^\infty \frac{z^j}{\zeta^{j+1}}.$$ Note that $$\overline{z} = z^{-1}$$ on the unit circle and $$\frac{1}{2\pi i}\int_\gamma z^k dz = 1_{\{k=-1\}}.$$ Let $$f(z) = \sum_{i=0}^\infty a_i z^i$$. In case $$|\zeta|<1$$, $$\frac{1}{2\pi i} \int_\gamma \left(\sum_{i,j\geq 0} \overline{a_i}z^{-i-j-1}\zeta^j\right) dz = \overline{a_0} = \overline{f(0)}.$$ Otherwise, $$-\frac{1}{2\pi i} \int_\gamma \left(\sum_{i,j\geq 0} \overline{a_i}z^{-i+j}\zeta^{-j-1}\right) dz = -\sum_{j=0}^\infty \overline{a_{j+1}}\zeta^{-j-1} = -\overline{f(\frac{1}{\overline{\zeta}})} + f(0),$$as desired.

• We've only just started doing power series in complex analysis so this is a little too advanced for me at this point. But thank you for taking the time. – user337254 Nov 18 '18 at 14:05