### Author Topic: 2.5Q14  (Read 1193 times)

#### Ye Jin

• Full Member
•   • Posts: 17
• Karma: 9 ##### 2.5Q14
« on: November 05, 2018, 09:16:40 PM »
I have no idea about the proof. Can anyone help me with this question？
Q: If f is analytic in $|z-z_0|<R$ and has a zero of order m at $z_0$, show that $Res(\frac{f’}{f};z_0)=m$

#### Amy Zhao

• Jr. Member
•  • Posts: 7
• Karma: 9 ##### Re: 2.5Q14
« Reply #1 on: November 05, 2018, 09:51:14 PM »
Assume f is analytic at $z_0$ and has a zero of order m at $z_0$.
There exists $p>0$ and an analytic function g which does not vanish at $z_0$ on $D(z_0 ,p)$ such that $f(z)=(z-z_0)^m g(z)$ for $z$ in $D(z_0 ,p)$.
Then we can take derivatives:
$f'(z)=m(z-z_0)^{m-1} g(z) + (z-z_0)^m g'(z)$
So $\frac{f'(z)}{f(z)}$ = $\frac{m(z-z_0)^{m-1} g(z) + (z-z_0)^m g'(z)}{(z-z_0)^m g(z)}$
$=\frac{m(z-z_0)^{m} g(z)}{(z-z_0)^m (z-z_0)g(z)}$ + $\frac{(z-z_0)^{m} g'(z)}{(z-z_0)^m g(z)}$
$=\frac{m}{z-z_0}$ + $\frac{g'(z)}{g(z)}$
Now, $\frac{g'}{g}$ is analytic on D, thus $\frac{f'}{f}$ has simple pole at $z_0$
This implies that $Res(\frac{f'}{f};z_0) = m$