Suppose that $f$ is analytic on and inside the circle $|z-z_0|=R$.
Show carefully that $f(z_0)$ is equal to the average of $f$ on
${|z-z_0|=R}$, i.e. show that
$$f(z_0)={1 \over {2 \pi}} \int_0^{2 \pi} f(z_0+Re^{i\theta}) \
d\theta$$
We are trying to use the cauchy integral formula to solve for it. We have it set up as $f(z_0)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z-z_0}dz = \frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{Re^{i\theta}}dz$
This is where we get suck. Need some direction for where to go!!!
Best Answer
Hint: If we let $z=z_0+Re^{i\theta}$, then $$ \frac1{2\pi}\int_0^{2\pi}f(z_0+Re^{i\theta})\,\mathrm{d}\theta =\frac1{2\pi i}\oint f(z)\frac{\mathrm{d}z}{z-z_0} $$ where the path is the circle of radius $R$ counterclockwise around $z_0$.
Next, look for the singularities of the integrand then use Cauchy's Integral Formula.