Theorem $10.8$
Suppose $\gamma $ is a regular closed curve. If $f$ is meromorphic inside and on $\gamma$ and contains no zeroes or oples on $\gamma$ , and if
$\mathbb{Z}$= number of zeroes of $f$ inside $\gamma$ ( a zero of order k being counted $k$ times
$\mathbb{P}=$ number of poles of $f$ inside $\gamma$ (again with multiplicity)
then $$\frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f} = \mathbb{Z}-\mathbb{P}$$
My confusion :why is $\frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f} = \mathbb{Z}-\mathbb{P}$? why not $\frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f} = \mathbb{P}?$
My attempt :At a simple pole $a$ , the residue of $f$ is given by $Res(f,a)=\lim_{z\to a}f(z)$
$$\implies Res(\frac{f'(z)}{f(z)},a)=\lim_{z\to a} (z-a) \frac{k}{z-a}+ \lim_{z\to a} (z-a)\frac{g'(z)}{g(z)}$$
$$\implies Res(\frac{f'(z)}{f(z)},a)= k$$
$$\frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f}= \sum_k Res (\frac{f'(z)}{f(z)},a_k)=\underbrace{k+k+…+k}_{k- \text{times}}=k^2=\text{ number of poles of f inside} \gamma=\mathbb{P}$$
Im not getting why is $\frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f} = \mathbb{Z}-\mathbb{P}?$
Best Answer
If $f(z)=z$, then $f$ has zero poles. But $\dfrac{f'(z)}{f(z)}=\frac1z$ and $\displaystyle\frac1{2\pi i}\oint_{|z|=1}=1$. And $1$ is the number of zeros minus the number of poles ($1-0$). So, there is no way the right formula is $\displaystyle\frac1{2\pi i}\oint_{|z|=1}=\Bbb P$. Besides, the poles of $\dfrac{f'}f$ are the poles and the zeros of $f$.