For the following integral with $C=|z|=1$
$\displaystyle \oint_{_{C}}\frac{1}{z^{4}}dz$
I took the approach of recognizing there is a pole of order $n=4$ at $z=0$ to use the Residue Theorem for pole of order n as below:
$\displaystyle \oint_{_{C}}f(z)dz = 2\pi i~Res(f(z),z_{0})=2\pi i\left[\frac{1}{(n-1)!}\lim_{z\to~z_{0}}\frac{d^{n-1}}{dz^{n-1}}\left((z-z_{0})^n\frac{1}{z^{n}}\right)\right]$
$\displaystyle \oint_{_{C}}\frac{1}{z^{4}}dz = 2\pi i\left[\frac{1}{(4-1)!}\lim_{z\to~z_{0}}\frac{d^{4-1}}{dz^{4-1}}\left((z-0)^{4}\frac{1}{z^{4}}\right)\right]$
$\displaystyle \oint_{_{C}}\frac{1}{z^{4}}dz = 2\pi i\left[\frac{1}{6}\lim_{z\to~0}\frac{d^{3}}{dz^{3}}\left((z-0)^{4}\frac{1}{z^{4}}\right)\right]$
$\displaystyle \oint_{_{C}}\frac{1}{z^{4}}dz = 2\pi i\left[\frac{1}{6}\lim_{z\to~0}\frac{d^{3}}{dz^{3}}\left(1\right)\right]$
$\displaystyle \oint_{_{C}}\frac{1}{z^{4}}dz = 2\pi i\left[\frac{1}{6}\lim_{z\to~0}~0\right]$
$\displaystyle \oint_{_{C}}\frac{1}{z^{4}}dz = 2\pi i\cdot 0$
$\displaystyle \oint_{_{C}}\frac{1}{z^{4}}dz = 0$
Is this the wrong approach?
Anytime you have a constant in numerator and $(z-a)^{n+1}$ for some $n \in \Bbb{N}$ in the denominator, some $a \in \Bbb{R}$, if $n\geq 1$, then the result is zero. Only if $n=0$ is it the case the integral may be non-zero.