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

$$\left.\frac{\zeta}{\zeta-1 / \bar{z}}=-\frac{\bar{z}}{\bar{\zeta}-\bar{z}} \cdot\right]$$