Contour integral

Complex Analysis
Post Reply
User avatar
Posts: 178
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Contour integral


Post by Riemann »

Let $f$ be analytic in the disk $|z|<2$. Prove that:

$$\frac{1}{2\pi i} \oint \limits_{\left | z \right |=1} \frac{\overline{f(z)}}{z-\alpha} \, \mathrm{d}z = \left\{\begin{matrix} \overline{f(0)} & , & \left | \alpha \right |<1 \\\\ \overline{f(0)} - \overline{f\left ( \frac{1}{\bar{\alpha}} \right )} & , & \left | \alpha \right |>1 \end{matrix}\right.$$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

User avatar
Tolaso J Kos
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa

Re: Contour integral


Post by Tolaso J Kos »

It follows from Taylor's theorem that $f(z)=\sum \limits_{n=0}^{\infty} c_n z^n$ and that the convergence is uniform. Thus,

\begin{align*} \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\overline{f(z)}}{z-\alpha} \,\mathrm{d}z &=\frac{1}{2\pi i }\oint \limits_{|z|=1} \sum_{n=0}^{\infty} \frac{\overline{c_n} \bar{z}^n}{z-\alpha} \,\mathrm{d}z \\
&= \sum_{n=0}^{\infty} \frac{\overline{c_n}}{2\pi i }\oint \limits_{|z|=1}\frac{ \bar{z}^n}{z-\alpha} \,\mathrm{d}z \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{\left | z \right |=1\Rightarrow \bar{z}^n = \frac{1}{z^n}}{=\! =\! =\! =\! =\! =\! =\!=\! =\!=\!}\sum_{n=0}^{\infty} \frac{\overline{c_n}}{2\pi i }\oint \limits_{|z|=1}\frac{1}{z^n(z-\alpha)} \,\mathrm{d}z

On the other hand we have that $\displaystyle \mathfrak{Res} \left( \frac{1}{z^n(z-\alpha)};\alpha \right) = \frac{1}{\alpha^n}$ και για $n \geq 1$ έχουμε $\displaystyle \mathfrak{Res}\left( \frac{1}{z^n(z-\alpha)};0\right) = -\frac{1}{\alpha^n}$ since

$$\frac{1}{z^n(z-\alpha)} = -\frac{1}{\alpha z^n} \frac{1}{1-z/\alpha} = -\frac{1}{\alpha z^n}\left(1 + \frac{z}{\alpha} + \frac{z^2}{\alpha^2} + \cdots \right)$$

If $|\alpha|<1$ then $\alpha$ lies inside the disk $|z|=1$ and the integral equals $\overline{c_0} = \overline{f(0)}$ whereas if $|\alpha|>1$ then $\alpha$ lies outside the disk $|z|=1$ and hence:

\frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\overline{f(z)}}{z-\alpha} \,\mathrm{d}z &= -\sum_{n=1}^{\infty} \frac{\overline{c_n}}{\alpha^n} \\ &= \overline{c_n} - \sum_{n=0}^{\infty} \frac{\overline{c_n}}{\alpha^n} \\
&= \overline{f(0)} - \overline{\left ( \sum_{n=0}^{\infty} \frac{c_n}{\bar{\alpha}^n} \right )}\\
&= \overline{f(0)} - \overline{f\left ( \frac{1}{\bar{\alpha}} \right )}
Imagination is much more important than knowledge.
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute


Sign in

Who is online

Users browsing this forum: No registered users and 1 guest