# Difference between revisions of "Cauchy Integral theorem"

The Cauchy Integral theorem states that for a function $f(z)$ which is analytic inside and on a simple closed curve $C$ in some region ${\mathcal {R}}$ of the complex $z$ plane, for a complex number $a$ inside $C$ $f(a)={\frac {1}{2\pi i}}\int _{C}{\frac {f(z)}{z-a}};dz.$ Proof:

For $f(a)$ a constant, we may rewrite $dz=d(z-a)$ , and noting that $z-a=|z-a|e^{i\phi }$ and $d(z-a)=i|z-a|e^{i\phi }\;d\phi$ , and taking the contour of integration to be a circle of unit radius $C:|z-a=1$ , we may write

${\frac {1}{2\pi i}}\int _{C}{\frac {f(a)}{z-a}}\;dz={\frac {f(a)}{2\pi i}}\int _{0}^{2\pi }{\frac {i|z-a|e^{i\phi }\;d\phi }{|z-a|e^{i\phi }}}=f(a)$ .

By Cauchy's Theorem, we may deform the contour into any closed curve that contains the point $z-a$ and the result holds.

For the case when $f(z)$ is not constant we may write

${\frac {1}{2\pi i}}\int _{C}{\frac {f(z)}{z-a}}\;dz={\frac {1}{2\pi i}}\int _{C}{\frac {f(a)}{z-a}}\;dz+{\frac {1}{2\pi i}}\int _{C}{\frac {f(z)-f(a)}{z-a}}\;dz=f(a)+{\frac {1}{2\pi i}}\int _{C}{\frac {f(z)-f(a)}{z-a}}\;dz$ .

We must show that the second term on the left is identically zero.

In a vanishingly small neighborhood $C:|z-a|=\varepsilon$ ${\frac {1}{2\pi i}}\int _{C}{\frac {f(z)-f(a)}{z-a}}\;dz\rightarrow {\frac {1}{2\pi i}}\int _{C}f^{\prime }(z)\;dz=0$ .

Because the derivative of an analytic function is also analytic, the integral vanishes identically within a neighborhood of $z=a$ . By Cauchy's theorem, the contour of integration may be expanded to any closed curve within ${\mathcal {R}}$ that contains the point $z=a$ thus showing that the integral is identically zero.