In this page, lets learn Cauchy integral theorem and formula with examples.

Cauchy’s Integral Theorem Statement

If f(z) is continuous and analytic at each point within and on a closed curve C, then

$\boxed{\displaystyle \int_C f(z)~dz=0}$.

Cauchy’s Integral Formula

If f(z) is analytic within and on a closed contour C and if $a$ is any point within C, then

$\boxed{\dfrac{1}{2\pi i}\displaystyle \int_C \dfrac{f(z)}{z-a}~dz=f(a)}$.

Cauchy’s Integral Formula for Derivatives:

$\boxed{\dfrac{n!}{2\pi i}\displaystyle \int_C \dfrac{f(z)}{(z-a)^{n+1}}~dz=f^{(n)}(a)}$.

Solved Examples

Question 1: Evaluate $\displaystyle\int_C \dfrac{z^2-z+1}{z-1}~dz$ where

1. C:= |z|=1
2. C:= |z|=1/2.

Answer:

Part 1: In Cauchy’s integral formula, f(z)=z²-z+1, a=1 lies on C: |z|=1.

So, $\dfrac{1}{2\pi i}\displaystyle\int_{|z|=1} \dfrac{z^2-z+1}{z-1}~dz$ $=f(1)=1^2-1+1=1$.

⇒ $\displaystyle\int_{|z|=1} \dfrac{z^2-z+1}{z-1}~dz=2\pi i$.

Part 2: a=1 lies outside the curve C: |z|=1/2. Therefore, by Cauchy’s integral theorem, the integral is zero.

Question 2: Evaluate $\displaystyle\int_{|z|=3} \dfrac{e^{2z}}{(z+1)^4}~dz$.

Answer:

Apply the above Cauchy’s integral formula for derivatives: $\dfrac{n!}{2\pi i}\displaystyle \int_C \dfrac{f(z)}{(z-a)^{n+1}}~dz=f^{(n)}(a)$, we have that

$n+1=4$, $a=-1$ and $f(z)=e^{2z}$.

So $\dfrac{3!}{2\pi i}\displaystyle \int_{|z|=3} \dfrac{e^{2z}}{(z+1)^4}~dz=f^{”’}(-1)$

⇒ $\displaystyle \int_{|z|=3} \dfrac{e^{2z}}{(z+1)^4}~dz=\dfrac{2\pi i}{6} \times f^{”’}(-1)$

⇒ $\displaystyle \int_{|z|=3} \dfrac{e^{2z}}{(z+1)^4}~dz=\dfrac{\pi i}{3} \times f^{”’}(-1)$.

Now, $f^{”’}(z)=2^3 e^{2z}$, so $f^{”’}(-1)=8 e^{-2}$

∴ $\displaystyle \int_{|z|=3} \dfrac{e^{2z}}{(z+1)^4}~dz=\dfrac{\pi i}{3} \times \dfrac{8}{e^{2}}$ $=\dfrac{8\pi i}{3e^2}$.

Also Study:

• Complex Differentiation
• Cauchy-Riemann Equation
• Cauchy-Riemann Equation in Polar Form
• Analytic Function