How to determine the residues of $z^2 \displaystyle\log \frac{z+1}{z-1}$ in $z=-1$? $z \in \mathbb{C}$ i need to evaluate $\displaystyle\int_{\gamma}z^{2}\log\frac{z+1}{z-1}dz$
where $\gamma$ is $C(0,2)$ with starts and end at $z=2$.
Well, i know that $\displaystyle\log \frac{z+1}{z-1}$ has poles in $-1$ and $1$ and by his expansion we have 
$\displaystyle\log \frac{z+1}{z-1}$  =$\displaystyle\log (1 + \frac{2}{z-1}) = \sum_{n=1}^{\infty}(-1)^{n+1}\frac{2^n}{(z-1)^n}\frac{1}{n}$ 
and $z^{2} = (z+1-1)^2 = (z-1)^2 +2(z-1) + 1$
then the terms of the wanted form are 
$\displaystyle \frac{2}{z-1} + \frac{-4}{z-1} + \frac{8/3}{z-1}$
so
$\underset{1}{Res}$ $z^2 \displaystyle\log \frac{z+1}{z-1} = -2 + 8/3$
but the problem now is how can i find the residue of $z=1$? how can i manipulate de $\log$ to attain this.
 A: Note the we can choose branch cuts that begin at the branch point and extend to $-\infty$.  The result of these coalescing branch cuts is that $f(z)=z^2\log\left(\frac{z+1}{z-1}\right)$ is analytic in $\mathbb{C}\setminus [-1,1]$.

METHODOLOGY $1$:  Residue at Infinity

Hence, $f(z)$ is analytic on the annulus for $2<|z|>R$ for all $R>2$ and the Residue at Infinity is given by 
$$\begin{align}
\text{Res}\left(f(z),z=\infty\right)&=\text{Res}\left(-\frac1{z^2}f\left(\frac1z\right),z=0\right)\\\\
&=-\text{Res}\left(\frac1{z^4}\log\left(\frac{1+z}{1-z}\right),z=0\right)\\\\
&=-\text{Res}\left(\frac1{z^4}\left(2z+\frac23z^3+O(z^5)\right),z=0\right)\\\\
&=-\text{Res}\left(\frac2{z^3}+\frac{2/3}{z}+O(z),z=0\right)\\\\
&=-\frac23
\end{align}$$
Therefore, we find that

$$\bbox[5px,border:2px solid #C0A000]{\oint_{|z|=2}z^2\log\left(\frac{z+1}{z-1}\right)\,dz=i \frac43\pi }$$

Note that the integral $\oint_{|z|=2}f(z)\,dz=-2\pi i \text{Res}\left(f(z),z=\infty\right)$, where the appearance of the additional minus sign is due to the reversed orientation from the transformation $z\to 1/z$.


METHODOLOGY $2$:  Laurent Series for $2<|z|<R$ as $R\to \infty$

Note that since $f(z)$ is analytic on the annulus $2<|z|<R$, then application of Cauchy's Integral Theorem shows that
$$\begin{align}
\oint_{|z|=2}f(z)\,dz&=\lim_{R\to \infty}\oint_{|z|=R} f(z)\,dz\\\\
&=\lim_{R\to \infty}\int_0^{2\pi }R^2e^{i2\phi}\log\left(\frac{Re^{i\phi}+1}{Re^{i\phi}-1}\right)\,(iRe^{i\phi})\,d\phi\\\\
&=\lim_{R\to \infty}i\int_0^{2\pi} R^3e^{i3\phi}\left(2\sum_{n=1}^\infty \frac{R^{-(2n-1)}e^{-i(2n-1)\phi}}{2n-1}\right)\,d\phi\\\\
&=i\frac43 \pi
\end{align}$$
as expected!  

Note that we exploited the facts that (i) the legitimacy of interchanging the integration and summation uniform is granted by the uniform convergence of the series $\sum_{n=1}^\infty \frac{R^{-(2n-1)}e^{-i(2n-1)\phi}}{2n-1}$ and (ii) the integral $\int_0^{2\pi }e^{im\phi}\,d\phi$ simplifies as
$$\int_0^{2\pi} e^{im\phi}\,d\phi=\begin{cases}0&,m\ne 0\\\\2\pi&,m=0
\end{cases}$$
