# Math Help - complex analysis continuity

1. ## complex analysis continuity

Suppose that $f: A \rightarrow \mathbb{C}$ is continuous, where $A$ is a region in $\mathbb{C}$. Let $\gamma$ be a simple closed curve in $A$. Show that for each $n$, $f_n(z)=\int_\gamma \frac{f(\zeta)}{(\zeta-z)^n}d\zeta$ is continuous on $A-\gamma$.

Let $\epsilon>0$. Suppose that there is $\delta>0$ such that $|z-z_0|<\delta$. I need to show $|f_n(z)-f_n(z_0)|<\epsilon$.
$|f_n(z)-f_n(z_0)|=|\int_\gamma \frac{f(\zeta)}{(\zeta-z)^n}d\zeta-\int_\gamma \frac{f(\zeta)}{(\zeta-z_0)^n}d\zeta|=\int_\gamma f(\zeta)(\frac{1}{(\zeta-z)^n}-\frac{1}{(\zeta-z_0)^n})d\zeta$.

I got stuck here... can I get some help please?

2. Originally Posted by dori1123
Suppose that $f: A \rightarrow \mathbb{C}$ is continuous, where $A$ is a region in $\mathbb{C}$. Let $\gamma$ be a simple closed curve in $A$. Show that for each $n$, $f_n(z)=\int_\gamma \frac{f(\zeta)}{(\zeta-z)^n}d\zeta$ is continuous on $A-\gamma$.

Let $\epsilon>0$. Suppose that there is $\delta>0$ such that $|z-z_0|<\delta$. I need to show $|f_n(z)-f_n(z_0)|<\epsilon$.
$|f_n(z)-f_n(z_0)|=|\int_\gamma \frac{f(\zeta)}{(\zeta-z)^n}d\zeta-\int_\gamma \frac{f(\zeta)}{(\zeta-z_0)^n}d\zeta|=\int_\gamma f(\zeta)(\frac{1}{(\zeta-z)^n}-\frac{1}{(\zeta-z_0)^n})d\zeta$.

I got stuck here... can I get some help please?
There's also a more general proof using uniform continuity; but let's do it your way.

Remember $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+a b^{n-2} + b^{n-1})$. We have

$\frac{1}{(\zeta-z)^n}-\frac{1}{(\zeta-z_0)^n}=\frac{(\zeta-z_0)^n-(\zeta-z)^n}{(\zeta-z)^n(\zeta-z_0)^n}$

hence, using the above formula,

$\left|\frac{1}{(\zeta-z)^n}-\frac{1}{(\zeta-z_0)^n}\right|\leq$ $|z-z_0|\frac{\sum_{k=0}^{n-1}|\zeta-z|^k |\zeta-z_0|^{n-1-k}}{|\zeta-z|^n|\zeta-z_0|^n}$.

Since the curve $\gamma$ is a bounded closed set (thus a compact set), you can choose $\delta>0$ such that the closed ball $\bar{B}(z_0,\delta)$ does not meet $\gamma$. Then we can find $m,M>0$ such that, for any $z$ in this ball, $m<|\zeta-z| for all $\zeta\in \gamma$. Using these bounds in the previous inequality enables to bound the right-hand side by $C|z-z_0|\leq C\delta$ for some constant $C$, and to conclude quickly from there.

3. Thank you very much.