# Thread: isolated point, Cauchy integral formula

1. ## isolated point, Cauchy integral formula

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

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

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

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

Now if is...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$