Show that $f(a)=\frac{1} {2\pi}{\int_C {\frac{(R^2-a \overline a)f(z)}{(z-a)(R^2-z \overline a)}dz}}$

Question

The function $f(z)$ is regular when $|z|<R'$

Show that if $|a|<R<R'$ then

$$f(a)=\frac{1} {2\pi}{\int_C {\frac{(R^2-a \overline a)f(z)}{(z-a)(R^2-z \overline a)}dz}}$$

Where $C$ is the circle $|z|=R$.

Answer 1

Observe that

$$z\neq \frac{R^2}{\overline a}\;\;\text{within}\;\;|z|\le R$$

so the function

$$\frac{f(z)}{R^2-z\overline a}$$

is analytic within the integration domain, and thus

$$\int\limits_C\frac{\frac{f(z)}{R^2-z\overline a}}{z-a}dz=2\pi i\frac{f(a)}{R^2-|a|^2}$$

The equality follows (by the way, in the question there's missing a $\;i\;$ in the denominator of the constant multiplying the integral)