Evaluate $\oint_{\gamma}\frac{f(z)}{z^{4}} dz$, where $\gamma:[0,2\pi ]\rightarrow\mathbb{C}$, $\gamma(t) :=\cos(t) +i\sin(t)$ and $f(z)=\frac{e^{z}+e^{-z}}{2}$.
For a Theorem, I know that $f^{(n)}(z_{0})=\frac{n!}{2\pi i}\oint_{\gamma}\frac{f(z)}{(z-z_{0})^{n+1}} dz$ so, because $z_{0}=0$ we have that
$\oint_{\gamma}\frac{f(z)}{z^{4}} =2\pi i \frac{f^{'''}(z_0)}{3!}=2\pi i\frac{e^{0}-e^{0}}{3!2}=0$ Is this right?
Notice that : \begin{aligned}\int_{\gamma}{\frac{f\left(z\right)}{z^{4}}\,\mathrm{d}z}=\oint_{\left|z\right|=1}{\frac{\cosh{z}}{z^{4}}\,\mathrm{d}z}=2\pi\mathrm{i}\,\mathrm{Res}\left(z\mapsto\frac{\cosh{z}}{z^{4}},0\right)\end{aligned}
Since $ \left(\forall z\in\mathbb{C}\right),\ \frac{\cosh{z}}{z^{4}}=\sum\limits_{n=0}^{+\infty}{\frac{z^{2n-4}}{\left(2n\right)!}} $, then $ \mathrm{Res}\left(z\mapsto\frac{\cosh{z}}{z^{4}},0\right) = 0$.
Thus : $$ \int_{\gamma}{\frac{f\left(z\right)}{z^{4}}\,\mathrm{d}z}=0 $$