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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. –  Christopher A. Wong Nov 20 '12 at 22:33
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1 Answer

Using the method of partial fractions, we get $\frac{f'(z)}{f(z)}=\frac{2z-3}{(z-1)(z-2)}=\frac{1}{z-1}+\frac{1}{z-2}$. Now apply Cauchy's integral formula, stating that

$$\int_{\gamma} \frac{\text{d}z}{z-a} =2\pi i$$

where $\gamma$ is a simple closed curve enclosing $a$, for example $\gamma(t)=\epsilon e^{2\pi i t} + a$ with $\epsilon>0$ and $t\in [0,1]$.

This gives you your result and shows where the $2\pi i$ comes from/dissapears to.

See http://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem

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