Compute the number of zero of $f(z)=(z-1)(z-2)$ inside $C= \{|z|=3\}$ using the argument principal.
I am not sure what is the difference between a pole and a zero. I know the pole is where the singularity occurs and that is when the function is not holomorphic( analytic).
$z=1$, $z=2$( would be the zeros?) and they both lie within $C$ since the radius is 3.
After, when using the argument principal formula which says $ \frac{1}{2\pi i} \int \frac{f'(z)}{f(z)} dz =1+1 =2 $ I obtain $1+1$ at the end but do not understand where the $\frac{1}{2\pi i}$ vanishes
Any clarification will be helpful. Thank u