Q6 TUT 0101

[Victor Ivrii]

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

[XueGe Huang]

check attached file

[Qi Cui]

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

[Ende Jin]

Thus there exists analytic $g$ s.t. $f(z) = (z-z_0)^mg(z)$ 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(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-1}g(z) + (z-z_0)^m g'(z)}{(z-z_0)^m g(z)} \\
    &= \frac{g'(z)}{g(z)} + m \frac{1}{z-z_0}
\end{align*}
We have shown $\frac{g'}{g}$ is analytic on that ball. Thus the residue, which means the coefficient of $(z-z_0)^{-1}$ is only $m$ .