| course |
course_year |
question_number |
tags |
title |
year |
Complex Analysis or Complex Methods |
IB |
11 |
IB |
2012 |
Complex Analysis or Complex Methods |
|
Paper 1, Section II, 13A |
2012 |
Using Cauchy's integral theorem, write down the value of a holomorphic function $f(z)$ where $|z|<1$ in terms of a contour integral around the unit circle, $\zeta=e^{i \theta} .$
By considering the point $1 / \bar{z}$, or otherwise, show that
$$f(z)=\frac{1}{2 \pi} \int_{0}^{2 \pi} f(\zeta) \frac{1-|z|^{2}}{|\zeta-z|^{2}} \mathrm{~d} \theta$$
By setting $z=r e^{i \alpha}$, show that for any harmonic function $u(r, \alpha)$,
$$u(r, \alpha)=\frac{1}{2 \pi} \int_{0}^{2 \pi} u(1, \theta) \frac{1-r^{2}}{1-2 r \cos (\alpha-\theta)+r^{2}} \mathrm{~d} \theta$$
if $r<1$.
Assuming that the function $v(r, \alpha)$, which is the conjugate harmonic function to $u(r, \alpha)$, can be written as
$$v(r, \alpha)=v(0)+\frac{1}{\pi} \int_{0}^{2 \pi} u(1, \theta) \frac{r \sin (\alpha-\theta)}{1-2 r \cos (\alpha-\theta)+r^{2}} \mathrm{~d} \theta$$
deduce that
$$f(z)=i v(0)+\frac{1}{2 \pi} \int_{0}^{2 \pi} u(1, \theta) \frac{\zeta+z}{\zeta-z} \mathrm{~d} \theta$$
[You may use the fact that on the unit circle, $\zeta=1 / \bar{\zeta}$, and hence
$$\left.\frac{\zeta}{\zeta-1 / \bar{z}}=-\frac{\bar{z}}{\bar{\zeta}-\bar{z}} \cdot\right]$$