Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz6 => Topic started by: Victor Ivrii on November 17, 2018, 04:07:17 PM

$\newcommand{\Res}{\operatorname{Res}}$
If $f$ is analytic in $\{z\colon z  z_0 < R\}$ and has a zero of order $m$ at $z_0$ , show that
$$
\Res \bigl(\frac{f'}{f}; z_0\bigr)=m.
$$

check attached file

$Since\ f(z)\ has\ a\ zero\ of\ order\ m\ at \ z_{0}$,
$$\quad\therefore f(z) = (zz_{0})^mg(z),where g'(z_{0})\ne0$$
$$\quad\therefore f'(z) = (zz_{0})^mg'(z)+m(zz_{0})^{m1}g(z)$$
$$ f'(z) = (zz_{0})^m(g'(z)+m(zz_{0})^{1}g(z)) $$
$$\quad\therefore {{f'(z)}\over {f(z)}} ={{(zz_{0})^m(g'(z)+m(zz_{0})^{1}g(z))}\over {(zz_{0})^mg(z)}} $$
$$= {{g'(z)}\over {g(z)}}+m(zz_{0}^{1}) $$
$$\quad\therefore Res({{f'}\over f}, z_{0})=m$$

Thus there exists analytic $g$ s.t. $f(z) = (zz_0)^mg(z)$ where $g(z_0) \neq 0$.
Thus there exists a small ball around $z_0$ s.t. $g(z) \neq 0$ (by continuity) and analytic , which means $\frac{1}{g(z)}$ is analytic as well, thus $\frac{g'(z)}{g(z)}$ is analytic on that ball as well.
Since $m \ge 1$,
\begin{align*}
\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)^{m1}g(z) + (zz_0)^m g'(z)}{(zz_0)^m g(z)} \\
&= \frac{g'(z)}{g(z)} + m \frac{1}{zz_0}
\end{align*}
We have shown $\frac{g'}{g}$ is analytic on that ball. Thus the residue, which means the coefficient of $(zz_0)^{1}$ is only $m$ .