### Author Topic: Q6 TUT 0203  (Read 1172 times)

#### Victor Ivrii ##### Q6 TUT 0203
« 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
$$\Res \bigl(\frac{f'}{f}; z_0\bigr)=-l.$$

#### Jeffery Mcbride

• Full Member
•   • Posts: 24
• Karma: 19 ##### Re: Q6 TUT 0203
« Reply #1 on: November 17, 2018, 04:32:23 PM »
$\displaystyle f( z) \ =\ \frac{g( z)}{( z-z_{0})^{l}}$

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

$\displaystyle Res\left(\frac{f'( z)}{f( z)} ;l\right) \ =-l$

#### Muyao Chen

• Jr. Member
•  • Posts: 10
• Karma: 7 ##### Re: Q6 TUT 0203
« Reply #2 on: November 17, 2018, 08:50:04 PM »
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

$\frac{f'}{f}$ = $\frac{g'(z) \frac{1}{(z- z_{0})^{l}} - g(z)l\frac{1}{(z- z_{0})^{l + 1}}}{\frac{g(z)}{(z- z_{0})^{l}}}$

= $\frac{g'}{g}$ - $\frac{l}{z-z_{0}}$

Then

$Res(\frac{f'}{f}, z_{0})$ = $Res(\frac{g'}{g}$ - $\frac{l}{z-z_{0}}, z_{0})$ =  $Res(\frac{- l}{z-z_{0}}, z_{0})$  = $\frac{- l}{(z-z_{0})'}$ = $\frac{-l}{1}$ = $-l$

#### oighea ##### Re: Q6 TUT 0203
« Reply #3 on: November 18, 2018, 02:17:49 AM »
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, $\displaystyle f(z) = \frac{H(z)}{(z-z_0)^l}$, where $H$ is analytic and nonzero on all of the disk $\{z | 0 \leq |z - z_0| < R\}$.

Therefore, we express $\displaystyle H(z) = a_0 + a_1(z-z_0) + a_2(z-z_0)^2 + ... = \sum_{k=0}a_k(z-z_0)^k$

We then consider the function $f'(z)$: Using the quotient rule for differentiation, $\displaystyle f'(z) = \frac{H'(z)(z-z_0)^l - l(z-z_0)^{l-1}H(z)}{(z-z_0)^{2l}} = \frac{H'(z)(z-z_0)^l}{(z-z_0)^{2l}} - \frac{l(z-z_0)^{l-1}H(z)}{(z-z_0)^{2l}} = l\frac{H(z)}{(z-z_0)^{2l - l + 1}} - \frac{H'(z)}{(z-z_0)^{2l - l}}$
We arrive at $\displaystyle f'(z) = \frac{H'(z)}{(z-z_0)^l} - l\frac{H(z)}{(z-z_0)^{l+1}}$

It follows that $\displaystyle \frac{f'(z)}{f(z)} = \left(\frac{(z-z_0)^l}{H(z)}\right)\left[\frac{H'(z)}{(z-z_0)^l} - l\frac{H(z)}{(z-z_0)^{l+1}}\right] = \frac{H'(z)}{1H(z)} - l\frac{1}{(z-z_0)^1}$

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 $-l$.