# Integrate $\int_C(\frac{1}{z-4}+\frac{i}{z-i})dz$

Integrate $$\int_C\left(\frac{1}{z-4}+\frac{i}{z-i}\right)dz$$ where $C$ is the straight line starting from $2 + 2i$ to $2 − 2i$

As far as I understand, branch cut is $$z\leq4\quad\text{and}\quad z\leq i$$ But $2 + 2i$ and $2 - 2i$ are both in these intervals. How would I have to change the argument then?

• This integral is integrated over the curve C which is a straight line starting from 2 + 2i to 2 − 2i – JMK Mar 9 '17 at 16:33
• Just write the line $C$ in function of $z$, after make a change of variable to make it a real line and integrate as if it was over some real interval. – Masacroso Mar 9 '17 at 17:28

Hint: Let $C: z=2-2ti$ where $t\in[-1,1]$ then set it to your integral with $dz=-2i\,dt$.