A First Course in Complex Analysis · Problem 31

Consequences of Cauchy's Theorem — Beck, Marchesi, Pixton & Sabalka

Mathematics — Complex Analysis · cauchys-integral-formula

Problem

Prove that for

0 < r < 1

\frac{1}{2\pi} \int_0^{2\pi} \frac{1 - r^2}{1 - 2r\cos(\phi) + r^2}\, d\phi = 1.

(The function

P_r(\phi) := \dfrac{1-r^2}{1-2r\cos(\phi)+r^2}

is the Poisson kernel and plays an important role in the world of harmonic functions, as we will see in Exercise 6.3.13.)