<s>[INST] Apply the Cauchy–Goursat theorem to compute the integral $\oint_C\frac{\sinh(z)}{z^{4}(1-z^{2})} dz$ where $C$ is the circle $|z|=2$. Apply the Cauchy–Goursat theorem to compute the integral
$$
\oint_C\frac{\sinh(z)}{z^{4}(1-z^{2})}dz,
$$
where $C$ is the circle $|z|=2$.
Solution:
$z^{4}(1-z^{2})=0 \Rightarrow$
$z=0,z=±1$
points where the funtion is not analytic.
 [/INST] You need to find the residue at $0$ ,$1$ and $-1$.
For the residue at $0$ you need to calculate the coefficient of $\frac{1}{z}$ in the Laurent Series expansion around $0$.
To do that use :-
$\sinh(z)=\frac{e^{z}-e^{-z}}{2}=z+\frac{z^{3}}{3!}+\frac{z^{5}}{5!}+....$.
And $\frac{1}{1-z^{2}}=1+z^{2}+z^{4}+z^{6}+....$
Combining them you need to find coefficient of $\frac{1}{z}$ in:-
$$\frac{1}{z^{4}}(z+\frac{z^{3}}{3!}+\frac{z^{5}}{5!}+....)(1+z^{2}+z^{4}+z^{6}+....)=(z+\frac{z^{3}}{3!}+\frac{z^{5}}{5!}+....)(\frac{1}{z^{4}}+\frac{1}{z^{2}}+1+z^{2}+...)$$.
Clearly the coefficient is $\frac{1}{3!}+1=\frac{7}{6}$.
Now for the residues at $1$ and $-1$. You see that
$\displaystyle\frac{\frac{\sinh(z)}{z^{4}(1-z)}}{1+z}$ has a simple pole at $z=-1$ and $\displaystyle\frac{\frac{-\sinh(z)}{z^{4}(1+z)}}{z-1}$ has a simple pole at $z=1$. So their residies are respectively $\frac{\sinh(-1)}{2}$ and $\frac{-\sinh(1)}{2}$ .
So the value of the integral by Cauchy Residue Theorem is :-
$$2i\pi\left(\frac{7}{6}-\sinh(1)\right)$$
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