# Calculate $\int_\Gamma \frac{f'(z)z}{f(z)}\, \operatorname dz$

Calculate usign the formula for zeros and poles, for a meromorphic function $f$ the following: $$\int_\Gamma \frac{f'(z)z}{f(z)}\, \operatorname dz$$

Where $\Gamma$ is simple and closed. I tried writting $\dfrac{f'(z)z}{f(z)}=\dfrac{g'(z)}{g(z)}$ for some meromorphic function (and then use the formula for zeros and poles for $g$) $g$ but I can't find $g$. I don't know if this idea is good for reducing this integral.

• Why do not you imitate the proof of the main theorem and see what you get. – Mhenni Benghorbal Dec 9 '13 at 2:36
• Yes you are right – Gaston Burrull Dec 9 '13 at 4:02
• if you got an answer you can post it, so other people will benefit from it. – Mhenni Benghorbal Dec 9 '13 at 4:07
• My answer was the same as Bruno – Gaston Burrull Dec 9 '13 at 15:55

The residue of $f'(z)/f(z)$ at a point $z_0$ is the valuation $v_{z_0}(f)$. Moreover, $f'(z)/f(z)$ has only simple poles. Therefore, the residue of $zf'(z)/f(z)$ at a point $z_0$ is $v_{z_0}(f)z_0$. By the residue theorem, the integral equals the weighed sum of the zeroes and poles of $f$ in the interior of $\Gamma$:
$$\sum_{z \in \text{ Int }\Gamma} v_{z_0}(f)z_0.$$
Hint: use the residue theorem. You can simplify $f'(z)/f(z)$ near a zero or a pole of $f(z)$.