Contour Integral of $\frac{z^{*}}{z-1}$, where $z^{*}$ is the complex conjugate of $z$. 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.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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}
A: Observe that
$$\oint \frac{z^{*}}{z-1} dz= 25i\int_{0}^{2\pi}\frac{1}{5e^{i\theta}-1}d\theta= 25 \oint \frac{1}{(z-1)z} dz.$$
Can you proceed ?
A: Your approach tries the substitution $z=5e^{i\theta}$, but then reverts to $z$ (getting the same intermediate result as my comment noting that on $|z|=5$, $\bar z=\frac{25}z$), where I would normally use residues.
It is best only to use logarithms in complex analysis when one is willing to define and use a branch cut for the logarithm. Otherwise, the hidden branch cut can cause confusion.
However, let's try to continue with the $z=5e^{i\theta}$ approach.
$$
\begin{align}
\oint\frac{\color{#C00}{\bar z}}{\color{#090}{z-1}}\,\color{#00F}{\mathrm{d}z}
&=\int_0^{2\pi}\frac{\color{#C00}{5e^{-i\theta}}\,\color{#00F}{5ie^{i\theta}}}{\color{#090}{5e^{i\theta}-1}}\,\color{#00F}{\mathrm{d}\theta}\tag1\\
&=5i\int_0^{2\pi}\frac{e^{-i\theta}}{1-\frac15e^{-i\theta}}\,\mathrm{d}\theta\tag2\\
&=5i\int_0^{2\pi}\sum_{k=0}^\infty\frac{e^{-i(k+1)\theta}}{5^k}\,\mathrm{d}\theta\tag3\\
&=5i\sum_{k=0}^\infty\frac1{5^k}\int_0^{2\pi}e^{-i(k+1)\theta}\,\mathrm{d}\theta\tag4
\end{align}
$$
Explanation:
$(1)$: $z=5e^{i\theta}$
$(2)$: cancel common factors
$(3)$: use the Taylor series for $\frac1{1-x}$
$(4)$: swap order of sum and integral
Each of the integrals in $(4)$ is pretty easy.
