# Cauchy's Integral formula for Derivatives

• Feb 17th 2011, 01:22 PM
Noodle
Cauchy's Integral formula for Derivatives
Higuys, I'm having a bit of trouble showing this.
Any help is much appreciated

$\displaystyle(\frac{1}{2i\pi})\oint_{C} (\frac{e^{tz}}{z^{n+1}}) dz = \frac{t^n}{n!}$

where C is the unit circle |z|=1
• Feb 17th 2011, 03:38 PM
zzzoak
$
\displaystyle
e^{tz}=\sum \ \frac{t^nz^n}{n!}
$

n term is only non zero:

$\displaystyle
(\frac{1}{2i\pi})\oint_{C} ( \ \frac{t^nz^n}{n!} \frac{1}{z^{n+1}}) dz =(\frac{1}{2i\pi}) \ \ \frac{t^n}{n!} \oint_{C} \frac{dz}{z}}= \frac{t^n}{n!}
$
• Feb 17th 2011, 04:18 PM
Noodle
Thank youu! That makes so much more sense, I had been using the wrong summation for $e^{tz}$, but got it now!