Cauchy Integral Formula

Problem

If $f(z)$ is analytic inside and on the boundary $C$ of a simply-connected region, prove the Cauchy integral formula

$f(z_{0})={\frac {1}{2\pi i}}\oint _{C}{\frac {f(z)}{z-z_{0}}}dz$.

Solution

The function ${\frac {f(z_{0})}{z-z_{0}}}$ is analytic inside and on the boundary $C$ except at the point $z=z_{0}$. Consider a circle $C_{0}$ of radius $r_{0}$ that is centered at $z_{0}$. The equation of the circle $C_{0}$ is $|z-z_{0}|=r_{0}$ or in polar form $z-z_{0}=r_{0}e^{i\theta }$ where $0\leq \theta <2\pi$. Thus $z=z_{0}+e^{i\theta }$ and $dz=ir_{0}e^{i\theta }$. Consequently,

$\oint _{C}{\frac {f(z)}{z-z_{0}}}dz=\oint _{C_{0}}{\frac {f(z)}{z-z_{0}}}dz$

$\oint _{C}{\frac {f(z)}{z-z_{0}}}dz=\int _{0}^{2\pi }{\frac {f(z_{0}+r_{0}e^{i\theta })ir_{0}e^{i\theta }}{r_{0}e^{i\theta }}}d\theta$

$\oint _{C}{\frac {f(z)}{z-z_{0}}}dz=i\int _{0}^{2\pi }f(z_{0}+r_{0}e^{i\theta })d\theta$.

Taking the limit shows that

$\lim _{r_{0}\to 0}i\int _{0}^{2\pi }f(z_{0}+r_{0}e^{i\theta })d\theta =i(2\pi )f(z_{0})$.

Therefore,

$\oint _{C}{\frac {f(z)}{z-z_{0}}}dz=2\pi if(z_{0})$

and

$f(z_{0})={\frac {1}{2\pi i}}\oint _{C}{\frac {f(z)}{z-z_{0}}}dz$.