Results 1 to 3 of 3

Thread: Complex Analysis

  1. #1
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411

    Complex Analysis

    The Cauchy Integral formula states the following:

    Suppose $\displaystyle U\in \mathbb{C}$ is an open set and $\displaystyle f$ is holomorphic on $\displaystyle U$. Let $\displaystyle z_0 \in U$ and $\displaystyle r>0$ such that $\displaystyle \overline{D}(z_0,r)\subset U$. Let $\displaystyle \gamma:[0,1]\to U$ be the $\displaystyle C^1$ curve $\displaystyle \gamma(t)=z_0+r\cos(2\pi t)+ir\sin(2\pi t)$. Then $\displaystyle f(z)=\int_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta$ for $\displaystyle z\in D(z_0,r)$

    Now I have to show the Cauchy-integral formula is valid if we only assume that $\displaystyle F\in C^0(\overline{D})$, with $\displaystyle D$ some disc, and $\displaystyle F$ is holomorphic on $\displaystyle D$

    I'm not sure if I understand what must be shown here exactly. Can someone clarify this a little?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by Dinkydoe View Post
    The Cauchy Integral formula states the following:

    Suppose $\displaystyle U\in \mathbb{C}$ is an open set and $\displaystyle f$ is holomorphic on $\displaystyle U$. Let $\displaystyle z_0 \in U$ and $\displaystyle r>0$ such that $\displaystyle \overline{D}(z_0,r)\subset U$. Let $\displaystyle \gamma:[0,1]\to U$ be the $\displaystyle C^1$ curve $\displaystyle \gamma(t)=z_0+r\cos(2\pi t)+ir\sin(2\pi t)$. Then $\displaystyle f(z)=\int_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta$ for $\displaystyle z\in D(z_0,r)$

    Now I have to show the Cauchy-integral formula is valid if we only assume that $\displaystyle F\in C^0(\overline{D})$, with $\displaystyle D$ some disc, and $\displaystyle F$ is holomorphic on $\displaystyle D$

    I'm not sure if I understand what must be shown here exactly. Can someone clarify this a little?
    My approach would be something like this:

    Let $\displaystyle g_t(z)= \frac{1}{2\pi i} \int_{\gamma _t } \frac{f(\zeta )}{\zeta -z} d\zeta$ with $\displaystyle t\in (\frac{1}{2} ,1)$ and $\displaystyle \gamma _t= te^{2\pi is}$ (assuming $\displaystyle z_0=0$) argue that $\displaystyle \lim_{t\rightarrow 1} g_t(z)$ is the function you're given and since (by the usual Cauchy integral formula) $\displaystyle g_t(z)=f(z)$ for all $\displaystyle z\in \mathbb{D}_{\frac{1}{2} }$ by Riemann's theorem we would have that $\displaystyle g_1(z)=f(z)$ for all $\displaystyle z\in \mathbb{D} $
    Last edited by Jose27; Mar 28th 2010 at 04:45 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    ok, I take it with $\displaystyle \mathbb{D}_{\frac{1}{2}}$ you mean the open ball with radius $\displaystyle r=1/2$ around $\displaystyle z_0=0$.

    Thanks for your answer however what theorem do you refer to when you say : " Riemann's Theorem''
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Oct 4th 2011, 05:30 AM
  2. Replies: 6
    Last Post: Sep 13th 2011, 07:16 AM
  3. Replies: 1
    Last Post: Oct 2nd 2010, 01:54 PM
  4. Replies: 12
    Last Post: Jun 2nd 2010, 02:30 PM
  5. Replies: 1
    Last Post: Mar 3rd 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum