I am solving a problem involving contour integration using the Residue theorem and I have to calculate the residue of $f(z)=\frac{z^3e^{iz}}{z^4+16}$
It is clear that a pole of $f$ is $2e^{i\pi/4}$. Since it is a pole of order $1$:
$\mathrm{Res}({2e^{i\pi/4}},f(z)) =\lim_{z\to2e^{i\pi/4}}(z-2e^{i\pi/4})\frac{z^3e^{iz}}{z^4+16}$
How do I compute that limit? The result should be $\frac{z^3 e^{iz}}{4z^3}$ for $z=2e^{i\pi/4}$
Where $z_0$ is a singularity of $f(z)$ and $f(z)=\frac{h(z)}{g(z)}$: $$\lim_{z \to z_0} \left(z-z_0\right) \cdot \frac{h(z)}{g(z)}= \lim_{z \to z_0} \frac{h(z)}{g'(z)}$$