isolated point, Cauchy integral formula

**zelda2139** · November 11th 2009, 06:03 PM

Let $z_0$ be an isolated point of a function $f$ and suppose that $f(z)=\frac{\phi(z)}{(z-z_0)^m}$ , where $m$ is a positive integer and $\phi(z)$ is analytic and nonzero at $z_0$ . By applying the extended form of the Cauchy integral formula to the function $\phi(z)$ , show that $\text{Res}_{z=z_0}=\frac{\phi^{(m-1)}(z_0)}{(m-1)!}$ .

I do not see how to do this. Our book says:
Since there is a neighborhood $|z-z_0|< \epsilon$ throughout which $\phi(z)$ is analytic, the contour used in the extended Cauchy integral formula can be the positively oriented circle $|z-z_0|< \frac{\epsilon}{2}$ . How do I use this suggestion? I don't see that now. Thanks in advance.

**chisigma** · November 11th 2009, 10:49 PM

The defintion of 'residue' of an $f(z)$ which has a pole of order m in $z=z_{0}$ and is analytic elsewhere in a region around $z_{0}$ is ...

$Res_{z=z_{0}} f(z) = \lim_{z \rightarrow z_{0}} \frac{1}{(m-1)!} \frac{d^{m-1}}{dz^{m-1}} \{(z-z_{0})^{m}\cdot f(z)\}$ (1)

Now if is...

$f(z)= \frac{\phi(z)}{(z-z_{0})^{m}}$ (2)

... where $\phi(*)$ is analytic in $z=z_{0}$ the (1) gives to You...

$Res_{z=z_{0}} f(z) = \frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}} \phi(z)_{z=z_{0}}$ (3)

Kind regards

$\chi$ $\sigma$

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