Cauchy's Integral formula for Derivatives

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February 17th 2011, 12:22 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
February 17th 2011, 02:38 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!} $
February 17th 2011, 03:18 PM
Noodle

Thank youu! That makes so much more sense, I had been using the wrong summation for $e^{tz}$ , but got it now!

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