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Cauchy Integral Formula

  1. Feb 16, 2012 #1
    For all z inside of C (C the unit circle oriented counterclockwise),
    [tex]
    f(z) = \frac{1}{2\pi i}\int_C \frac{g(u)}{u-z} du
    [/tex]
    where [itex]g(u) = \bar{u}[/itex] is a continuous function and [itex]f[/itex] is analytic in C. Describe [itex]f[/itex]in C in terms of a power series.

    [itex]\displaystyle f(z) = \frac{1}{2\pi i}\int_C \frac{\bar{u}}{u-z} du[/itex] I am confused with what I am supposed to do. I know it says describe [itex]f[/itex] in terms of a power series.
     
  2. jcsd
  3. Feb 16, 2012 #2
    If f is analytic in C, then its Taylor series should converge to f everywhere in C, right?
     
  4. Feb 17, 2012 #3
    Yes.
     
  5. Feb 17, 2012 #4
    [tex]f(z)=\frac{1}{2\pi i}\int_C\frac{\bar{u}}{u-z}du\Rightarrow 2\pi if(z)=\int_C\frac{\bar{u}}{u-z_0}\frac{1}{1-\frac{z-z_0}{u-z_0}}du[/tex]

    [itex]z_0[/itex] is not necessarily on [itex]C[/itex]. Let [itex]s=\inf\{|C(t)-z_0|:C(t) \ \text{any point on} \ C\}[/itex] Since [itex]C[/itex] is compact, [itex]s>0[/itex].
    Let [itex]r[/itex] be the radius of the open disc around [itex]z_0[/itex] such that the disc doesn't intersect [itex]C[/itex]. Take [itex]z\in D(z_0,r)[/itex] fix [itex]r[/itex] such that [itex]0<r<s[/itex].

    [itex]\frac{z-z_0}{u-z_0}[/itex] is uniformly bounded since the max [itex]z-z_0[/itex] is r and the min [itex]u-z_0[/itex] is s so [itex]\frac{r}{s}<1[/itex].

    [tex]
    2\pi i f(z) = \int_C \frac{\bar{u}}{u-z_0}\sum_{n=0}^{\infty}\frac{1}{(u-z_0)^n}(z-z_0)^n du
    [/tex]
    The series converges uniformly for all r<s and pointwise for all z with [itex]z\in D(z_0,s)[/itex]. So we can integrate term by term.

    [tex]
    2\pi i f(z) = \sum_n^{\infty}\left[\int_C\frac{\bar{u}}{(u-z_0)^{n+1}}du(z-z_0)^n\right]
    [/tex]

    Let [itex]c_n=\int_C\frac{\bar{u}}{(u-z_0)^{n+1}}du[/itex]. Then
    [tex]
    2\pi i f(z) = \sum_n^{\infty}c_n(z-z_0)^n
    [/tex]

    Now how can I explain why [itex]f[/itex] does or does not equal [itex]g[/itex]? Is f described correctly as a power series here as well?
     
  6. Feb 23, 2012 #5
    From here, I can expand at [itex]z_0 = 0[/itex] around the unit circle and evaluate the coefficient.

    $$
    f(z) = \frac{1}{2\pi i}\sum_{n=0}^{\infty}\int_0^{2\pi}\frac{\bar{u}}{u^{n+1}}du z^n
    $$

    For n = 0, we have

    $$
    \frac{1}{2\pi i}\int_0^{2\pi}\frac{\bar{u}}{u}du
    $$

    How is this integral evaluated?

    Like this:

    $$
    \int_0^{2\pi}\frac{\bar{u}}{u}du = 2\pi i 0
    $$

    By the Cauchy Integral Formula, then divide by 2\pi i but here the sol is 0 regardless?

    The second term would be

    $$
    \int_0^{2\pi}\frac{\bar{u}}{u^2}du
    $$

    How would I integrate this?
     
    Last edited: Feb 23, 2012
  7. Feb 23, 2012 #6
    $$
    f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}\left[\int_C\frac{\bar{u}}{u^{n + 1}}du z^n\right].
    $$
    Since [itex]\bar{u}[/itex] is not holomorphic in the disk, [itex]f\neq g[/itex].
    Then
    $$
    f(z) = \frac{1}{2\pi i}\left[\underbrace{z^0\int_0^{2\pi}\frac{\bar{u}}{u}du}_{0} + z\int_0^{2\pi}\frac{\bar{u}}{u^2}du + z^2\int_0^{2\pi}\frac{\bar{u}}{u^3}du + \cdots\right].
    $$

    Is this how f should be described, because I have no idea how to integrate the functions when the denominator has a power of 2 or greater.
     
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