complex integral, help.

Question

Proof that, if $f$ is holomorphic in the ring $U = {r <| z - a | <R}$, then the integral of $f$, along the circumference with center $a$ and radius $ρ$, $\int_{| z - a |=ρ}f(z)dz$, has the same value, for all $r <ρ <R$.

idk how to start :(

Answer 1

We can define a closed contour in four pieces; one around $|z-a|=R$ One around $|z-a| =p$ and two line segments between these two circles along the positive real line. Let's call this contour $\gamma$

$\int_{|z-a|= p} f(z)\ dz + \int_{\gamma} f(z)\ dz = \int_{|z-a| = R} f(z)\ dz$

As $f$ is holoporphic everywhere inside the contour of $\gamma, \int_{\gamma} f(z)\ dz = 0$

$\int_{|z-a|= p} f(z)\ dz = \int_{|z-a| = R} f(z)\ dz$ for all $p$ such thata $r<p<R$