$\mathbf {The \ Problem \ is}:$ Question no. 9
$\mathbf {My \ approach}:$ Actually, firstly I was thinking of expanding $\frac{1}{\xi-z}$ but here $1=|\xi|>|z|$ ; so I can't approach .
Then I was trying to use the formula $\frac{f(z+h)-f(z)}{h}$ alongside the fact that $g$ is continuous on a compact domain .
But, I am getting stuck during the above calculation .
A small hint is warmly appreciated, thanks in advance .
You can use the fact that, when $|\xi|=1$ and $|z|<1$,$$\frac1{\xi-z}=\frac1\xi\frac1{1-z/\xi}=\sum_{n=0}^\infty\frac{z^n}{\xi^{n+1}}.$$Therefore, you have$$f(z)=\sum_{n=0}^\infty\int_{|\xi|=1}\left(\frac{g(\xi)}{\xi^{n+1}}\,\mathrm d\xi\right)z^n.$$