Difference between revisions of "Cauchy Integral theorem"

The Cauchy Integral theorem states that for a function ${\displaystyle f(z)}$ which is analytic inside and on a simple closed curve ${\displaystyle C}$ in some region ${\displaystyle {\mathcal {R}}}$ of the complex ${\displaystyle z}$ plane, for a complex number ${\displaystyle a}$ inside ${\displaystyle C}$

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

Proof:

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

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

.

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

For the case when ${\displaystyle f(z)}$ is not constant we may write

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

.

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

In a vanishingly small neighborhood ${\displaystyle C:|z-a|=\varepsilon }$

${\displaystyle {\frac {1}{2\pi i}}\int _{C}{\frac {f(z)-f(a)}{z-a}}\;dz\rightarrow {\frac {1}{2\pi i}}\int _{C}f^{\prime }(z)\;dz=0}$

.

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