Proof using Cauchy's integral formula

malawi_glenn
Science Advisor
Messages
6,735
Reaction score
2,433

Homework Statement


Let f and g ve analytic functions inside and on a simple connected contour \Gamma. If f(z) = g(z) for all z on \Gamma, prove that
If f(z) = g(z) for all z inside \Gamma


Homework Equations



<br /> f(z_{0}) = \int_{\Gamma}\dfrac{f(z)}{z-z_{0}}dz

if f is analytic in a simple connected domain containing \Gamma and z_{0} is a point insinde \Gamma.

The Attempt at a Solution



I know I must (?) prove that f(z_{0}) = g(z_{0}) for all z_0, but i have no idea how to use the fact that f and g are equal on all point on \Gamma, I have no Lemma or Theoreme for this in my book (I am not allowed to use the theory of bounds for analytical functions, just Cauchy's integral formula)

Can someone give me a small hint ? =)
 
Physics news on Phys.org
Well, what happens if you write out:
f(z_0)-g(z_0)
in the format you already have?
 
NateTG said:
Well, what happens if you write out:
f(z_0)-g(z_0)
in the format you already have?

okay i try that =) thanks!
 
<br /> f(z_{0}) - g(z_{0}) = \dfrac{1}{2\pi i}\int_{\Gamma}\dfrac{f(z)}{z-z_{0}}dz - \dfrac{1}{2\pi i}\int_{\Gamma}\dfrac{g(z)}{z-z_{0}}dz

I still don't know how to use the fact that they are equal for all z ON \Gamma =(
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top