# Cauchy Integral Formula

• Nov 1st 2009, 11:41 AM
Haven
Cauchy Integral Formula
Show that if $f$ is analytic within an on a simple closed contour C and $z_{0}$ is not on C, then

$\int_{C} \frac{f'(z)}{z-z_{0}} = \int_{C} \frac{f(z)}{(z-z_{0})^2}$

I've broken this down into two cases: $z_{0}$ is interior to C and $z_{0}$ is exterior to C

Case 1:

Since $z_{0}$ is interior to C we may apply the Cauchy Integral Formula so

$f'(z_{0})*2\pi\\i = \int_{C} \frac{f'(z)}{z-z_{0}}$

and

$f'(z_{0})*2\pi\\i = \int_{C} \frac{f(z)}{(z-z_{0})^2}$

and obviously:

$\int_{C} \frac{f'(z)}{z-z_{0}} = \int_{C} \frac{f(z)}{(z-z_{0})^2}$

However I have no idea what to do when $z_{0}$ is exterior to C. We're only given that f is analytic in the closure of C. I was hoping to use contour deformation or the Cauchy-Goursat Theorem but i don't know how.
• Nov 1st 2009, 12:34 PM
shawsend
Quote:

Originally Posted by Haven

$\int_{C} \frac{f'(z)}{z-z_{0}} = \int_{C} \frac{f(z)}{(z-z_{0})^2}$

However I have no idea what to do when $z_{0}$ is exterior to C. We're only given that f is analytic in the closure of C. I was hoping to use contour deformation or the Cauchy-Goursat Theorem but i don't know how.

Arent' the integrands both analytic on $C$ (and inside) when $z_0$ is outside the contour? Then what?
• Nov 1st 2009, 12:38 PM
Haven
So if I'm getting this right. Since both integrands are analytic when $z_{0}$ is outside the countour we can apply the Cauchy-Goursat Theorem and so

$\int_{C} \frac{f'(z)}{z-z_{0}} = \int_{C} \frac{f(z)}{(z-z_{0})^2}
= 0$

Right?
• Nov 1st 2009, 12:55 PM
shawsend
That's what I'd say.