Vtyjkyuk

We just need to show that $f(z)$ can be written as a power series.

Indeed beginalign*
f(z)&= frac12pi iint_Cfracf(w)w-z,dw \
&= frac12pi iint_Cfracf(w)(w-z_0)-(z-z_0),dw \
& = frac12pi iint_Cfracf(w)w-z_0frac11-fracz-z_0w-w_0,dw\
&= frac12pi iint_Cfracf(w)w-z_0sum_n=0^inftyleft(fracz-z_0w-w_0right)^n,dw \
& = sum_n=0^inftyleft(frac12pi iint_Cfracf(w)(w-z_0)^n+1dwright)(z-z_0)^n\
endalign*

Do note that $1-fracz-z_0w-w_0 = sum_n=0^inftyleft(fracz-z_0w-w_0right)^n$ is valid, and see that $left|dfracz-z_0w-w_0right|< 1$ (note that $|w-z_0| = r$ where $w$ lies on the boundary of circle wrt $z_0$.

However the main issue is how do I know if this series converges uniformly? The answer did not explain that.

To prove analyticity you only have to prove the existence of a power series expansion in some ball around any given point. Take a disk of radius $frac r 2$. In that disk you get $|frac z-z_0 w-_0| <frac 1 2$ and apply M-test to get uniform convergence in the smaller disk.

Theorem: Let $sum_n=0^inftyb_n(z-z_0)^n$ be a power series with radius of convergence $R>0$ and let $C$ be a compact subset of $z in mathbb C: $.

Then $sum_n=0^inftyb_n(z-z_0)^n$ converges uniformly on $C$.