So I have to evaluate $\oint \frac{z^{*}}{z-1}$ on a circle of radius 5 centered at the origin of the complex plane, where $z^{*}$ is the complex conjugate of $z$ and the orientation is anticlockwise.
I know z* isn't analytic anywhere and I can assume $z=5e^{i\theta} \implies dz = 5ie^{i\theta}d\theta.$
This transforms my integral to $\int_{0}^{2\pi}\frac{5e^{-i\theta}(5ie^{i\theta})} {5e^{i\theta}-1}d\theta = 25i\int_{0}^{2\pi}\frac{1}{5e^{i\theta}-1}d\theta$
I let $ u = 5e^{i\theta} => du = i5e^{i\theta}d\theta = iu d\theta $
$\implies \frac{25i}{i} \int \frac{1}{(u-1)u}du = 25 (\int \frac{1}{u-1}du – \int \frac{1}{u}du)$
$ = 25(ln(u-1)-ln(u)) = 25 ln (\frac{u-1}{u}) $
$\implies 25 [ln (\frac{5e^{i\theta}-1}{5e^{i\theta}})]_{0}^{2\pi}$
I'm not sure if I've made a mistake uptil now and what exactly I need to do further.
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \oint_{\verts{z}\ =\ 5}{\overline{z} \over z - 1}\,\dd z & = \oint_{\verts{z}\ =\ 5}{\verts{z}^{2} \over z\pars{z - 1}}\,\dd z = 25\oint_{\verts{z}\ =\ 5}{\dd z \over z\pars{z - 1}} \\[5mm] & = 25\lim_{R \to \infty}\oint_{\verts{z}\ =\ R\ >\ 1} {\dd z \over z\pars{z - 1}} = \bbx{\Large 0} \\[5mm] & \mbox{because}\quad 0 < \verts{\oint_{\verts{z}\ =\ R} {\dd z \over z\pars{z - 1}}} < {2\pi \over R} \end{align}