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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

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    A zero is just where the function is zero. One can think of a pole as occurring where the denominator is zero.2012-11-20

1 Answers 1