gtfitzpatrick
- 372
- 0
Homework Statement
Prove that if f(z) is analytic over a simply connected domain containing a simple closed curve C abd Z_{0} is a point inside C then f'(z_{0}) = \frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0)^2} dz
Homework Equations
The Attempt at a Solution
from the definition of a derivative we have f'(z_{0}) =\stackrel{lim}{h\rightarrow0} \frac{f(z_0 +h) - f(z_0)}{h}
the cauchy integral formula states f(z_{0}) = \frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0)} dz and so f(z_{0}+h) = \frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0 +h)} dz
then subing the last 2 into the first one i get f'(z_{0}) = \stackrel{lim}{h\rightarrow0} \frac{1}{2i\pi} \oint_{c} \frac{1}{h} (\frac{f(z)}{(z- z_0 - h)} - \frac{f(z)}{(z-z_0)} )dz
=f'(z_{0}) = \stackrel{lim}{h\rightarrow0} \frac{1}{2i\pi} \oint_{c} (\frac{f(z)}{(z- z_0 - h)(z-z_0)} )dz
on[/b]
= \frac{1}{2i\pi} \oint_{c} \frac{f(z)}{(z-z_0)^2} dz
is this enough of a proof or would i also have to prove cauchys integral formula?