Toronto Math Forum
MAT3342018F => MAT334Lectures & Home Assignments => Topic started by: Ye Jin 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 $zz_0<R$ and has a zero of order m at $z_0$, show that $Res(\frac{f’}{f};z_0)=m$

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 $f(z)=(zz_0)^m g(z)$ for $z$ in $D(z_0 ,p)$.
Then we can take derivatives:
$f'(z)=m(zz_0)^{m1} g(z) + (zz_0)^m g'(z)$
So $\frac{f'(z)}{f(z)}$ = $\frac{m(zz_0)^{m1} g(z) + (zz_0)^m g'(z)}{(zz_0)^m g(z)}$
$=\frac{m(zz_0)^{m} g(z)}{(zz_0)^m (zz_0)g(z)}$ + $\frac{(zz_0)^{m} g'(z)}{(zz_0)^m g(z)}$
$=\frac{m}{zz_0}$ + $\frac{g'(z)}{g(z)}$
Now, $\frac{g'}{g}$ is analytic on D, thus $\frac{f'}{f}$ has simple pole at $z_0$
This implies that $Res(\frac{f'}{f};z_0) = m$