# Cauchy Integral Formula

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• November 1st 2009, 10:41 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.
• November 1st 2009, 11:34 AM
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?
• November 1st 2009, 11:38 AM
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?
• November 1st 2009, 11:55 AM
shawsend
That's what I'd say.