How to compute $\int_C f'(z)/f(z) dz$, when $C$ is positively oriented disk $\{ z: |z|=4\}$, when $f(z)=\frac{(z-8)^2z^3}{(z-5)^4(z+2)^2(z-1)^5}$? Hint. $ \frac{f'(z)}{f(z)} = \frac{d}{dz} log f(z) = \frac{d}{dz}(Log|f(z)|+i arg f(z))$ I'm interested especially about the fact that $\int_C \frac{f'(z)}{f(z)} dz = (Log |f(z_2)| + i arg f(z_2)) - (Log |f(z_1)| + i arg f(z_1)) = \Delta_C Log |f| + i \Delta_C arg f = i \Delta_C arg f =(*) 2 \pi i(2 \cdot 0 + 3 \cdot 1 - 4 \cdot 0 - 2 \cdot 1 - 5 \cdot 1) = -8\pi$
But now question is how do you get (*) because if $i arg f(z)=2arg(z-8)+3arg z - 4 arg(z-5) - 2 arg(z+2) - 5 arg(z-1)$ then $i\Delta_C arg f=2 \pi i(2 \cdot 0 + 3 \cdot 1 - 4 \cdot 0 - 2 \cdot 1 - 5 \cdot 1)$?
