Contour integral with cauchy Calculate integral
$$\oint_{\gamma} \frac{e^{2i z}}{z^4}-\frac{z^4}{(z-i)^3}dz$$
when $\gamma$ is circles $S(0,6)$ parameterization once rotated over space $[2\pi]$.
Is there more to it than just calculate it with cauchy.
\begin{align}
\oint_{\gamma} \frac{e^{2i z}}{z^4}dz- \oint_{\gamma}\frac{z^4}{(z-i)^3}dz&=\oint_{\gamma} \frac{\overbrace{e^{2i z}}^{f_1(z)}}{(z-0)^{3+1}}dz- \oint_{\gamma}\frac{\overbrace{z^4}^{f_2(z)}}{(z-i)^{2+1}}\\
&=\frac{2\pi i}{3!}\frac{\partial^3}{\partial z^3}(e^{2i z_0})_{z_0=0}-\frac{2\pi i}{2!}\frac{\partial^2}{\partial z^2}(z_0^4)_{z_0=i} \\
&=\frac{2\pi i}{3!}\left( -8i e^0\right)-\frac{2\pi i}{2!}(12i^2) \\
&=\frac{\pi i}{3}(-8 i) - \pi i(-12) \\
&=\frac{8 \pi}{3} + 12\pi i
\end{align}
Or... Is there some pitfalls with the analyticity and you can't do it straight forwardly.
Any hints appriciated.

Your use of $\;n_i\;$ is odd. The order of the derivative should be put directly, for example
$$\oint\limits_\gamma\frac{e^{2iz}}{z^4}dz=\frac{2\pi i}{3!}\frac{d^3}{dz^3}\left(e^{2iz}\right)_{z=0}=\frac{\pi i}3(-8i)=\frac{8\pi}3$$
