Question

find the residue of the function f(z)=1/z(z-2) where c is the circle |z|=1

Answer 1

$f(z)=\frac{1}{z(z-2)}=\frac{1}{2}(\frac{1}{z-2}-\frac{1}{z})$

The function $\frac{1}{2}\frac{1}{z-2}$ has a unique pole at $z=2$ and has no singularities inside the contour $C=\{z\in\mathbb{C}:|z|=1\}$. Therefore, the residue of this function equal to 0.

The function $-\frac{1}{2z}$ has a unique pole at $z=0$ and it is inside the contour C. The coefficient with $z^{-1}$ is equal to $-1/2$, therefore, the residue of this function is equal to $-1/2$.

Therefore, the residue of the function $f(z)=\frac{1}{z(z-2)}$ inside the contour $C$ is equal to $0-1/2=-1/2$.

Answer. -1/2