Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz6 => Topic started by: Victor Ivrii on November 17, 2018, 04:10:26 PM

$\newcommand{\Res}{\operatorname{Res}}$
If $f$ is analytic in $\{z\colon 0< z  z_0 < R\}$ and has a pole of order $l$ at $z_0$ , show that
$$
\Res \bigl(\frac{f'}{f}; z_0\bigr)=l.
$$

$\displaystyle f( z) \ =\ \frac{g( z)}{( zz_{0})^{l}}$
where $\displaystyle g( z)$ is analytic in $\displaystyle z\ \ z_{0} \ < \ R\ and\ g( z_{0}) \ \neq 0$
Then,
$\displaystyle \frac{f'( z)}{f( z)} \ =\ \frac{g'( z)}{g( z)} \ \ \frac{l}{z\ \ z_{0}}$
Since $\displaystyle g( z_{0}) \ \neq \ 0,\ then\ \frac{g'( z)}{g( z)}$ is analytic at $\displaystyle z_{0}$ , so the residue of that first term is 0. So for the residue of the whole thing, we read the coefficent of the second term which is l. So we get
$\displaystyle Res\left(\frac{f'( z)}{f( z)} ;l\right) \ =l$

Write
f(z) = $\frac{g(z)}{(z z_{0})^{l}}$
so
f '(z) = $\frac{g'(z)(z z_{0})^{l}  g(z)(z z_{0})^{l 1}}{(z z_{0})^2}$
=g'(z) $\frac{1}{(z z_{0})^{l}}$  g(z)l$\frac{1}{(z z_{0})^{l + 1}}$
so that
$\frac{f'}{f}$ = $\frac{g'(z) \frac{1}{(z z_{0})^{l}}  g(z)l\frac{1}{(z z_{0})^{l + 1}}}{\frac{g(z)}{(z z_{0})^{l}}}$
= $\frac{g'}{g}$  $\frac{l}{zz_{0}}$
Then
$Res(\frac{f'}{f}, z_{0})$ = $Res(\frac{g'}{g}$  $\frac{l}{zz_{0}}, z_{0})$ = $Res(\frac{ l}{zz_{0}}, z_{0})$ = $\frac{ l}{(zz_{0})'}$ = $\frac{l}{1}$ = $l$

This proof does not require the direct evaluation of Laurent/Power series, but instead relies on algebraic and calculus manipulation.
Let $f(z)$ be a function analytic on the punctured disk of radius $R$ centered on $z_0$: $\{z  0 < z  z_0 < R\}$, and has a pole of order $l$ @ $z_0$.
Then, $\displaystyle f(z) = \frac{H(z)}{(zz_0)^l}$, where $H$ is analytic and nonzero on all of the disk $\{z  0 \leq z  z_0 < R\}$.
Therefore, we express $\displaystyle H(z) = a_0 + a_1(zz_0) + a_2(zz_0)^2 + ... = \sum_{k=0}a_k(zz_0)^k$
We then consider the function $f'(z)$: Using the quotient rule for differentiation, $\displaystyle f'(z) = \frac{H'(z)(zz_0)^l  l(zz_0)^{l1}H(z)}{(zz_0)^{2l}} = \frac{H'(z)(zz_0)^l}{(zz_0)^{2l}}  \frac{l(zz_0)^{l1}H(z)}{(zz_0)^{2l}} = l\frac{H(z)}{(zz_0)^{2l  l + 1}}  \frac{H'(z)}{(zz_0)^{2l  l}}$
We arrive at $\displaystyle f'(z) = \frac{H'(z)}{(zz_0)^l}  l\frac{H(z)}{(zz_0)^{l+1}}$
It follows that $\displaystyle \frac{f'(z)}{f(z)} = \left(\frac{(zz_0)^l}{H(z)}\right)\left[\frac{H'(z)}{(zz_0)^l}  l\frac{H(z)}{(zz_0)^{l+1}}\right] = \frac{H'(z)}{1H(z)}  l\frac{1}{(zz_0)^1}$
From now on, we can see that the residue could be $l$, but considering $\frac{H'(z)}{1H(z)}$, we conclude that the $H$ being analytic and nonzero on the entire disk implies $H'$ being analytic as well, and finally $\frac{H'(z)}{H(z)}$, and so that fraction has no principal part. This leaves the maximum negative degree of the principal part to be 1.
We conclude that the residue has to be $l$.