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## Homework Statement

Prove that if f(z) is analytic over a simply connected domain containing a simple closed curve C abd Z[itex]_{0}[/itex] is a point inside C then f'(z[itex]_{0}[/itex]) = [itex]\frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0)^2} dz[/itex]

## Homework Equations

## The Attempt at a Solution

from the definition of a derivative we have f'(z[itex]_{0}[/itex]) =[itex]\stackrel{lim}{h\rightarrow0} \frac{f(z_0 +h) - f(z_0)}{h}[/itex]

the cauchy integral formula states f(z[itex]_{0}[/itex]) = [itex]\frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0)} dz[/itex] and so f(z[itex]_{0}+h[/itex]) = [itex]\frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0 +h)} dz[/itex]

then subing the last 2 into the first one i get f'(z[itex]_{0}[/itex]) = [itex] \stackrel{lim}{h\rightarrow0}[/itex] [itex] \frac{1}{2i\pi}[/itex] [itex] \oint_{c} [/itex] [itex] \frac{1}{h} [/itex] ([itex] \frac{f(z)}{(z- z_0 - h)}[/itex] [itex] - \frac{f(z)}{(z-z_0)} )dz[/itex]

=f'(z[itex]_{0}[/itex]) = [itex] \stackrel{lim}{h\rightarrow0}[/itex] [itex] \frac{1}{2i\pi}[/itex] [itex] \oint_{c} [/itex] ([itex] \frac{f(z)}{(z- z_0 - h)(z-z_0)}[/itex] )dz

on[/b]

= [itex]\frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0)^2} dz[/itex]

is this enough of a proof or would i also have to prove cauchys integral formula?