Cauchy Integral Formula

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  • #1
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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.
 

Answers and Replies

  • #2
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If f is analytic in C, then its Taylor series should converge to f everywhere in C, right?
 
  • #3
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If f is analytic in C, then its Taylor series should converge to f everywhere in C, right?
Yes.
 
  • #4
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[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?
 
  • #5
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[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?
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:
  • #6
158
0
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?
$$
f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}\left[\int_C\frac{\bar{u}}{u^{n + 1}}du z^n\right].
$$
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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