Power Series Expansion of Cauchy Integral Formula on the Unit Circle

[fauboca]

For all z inside of C (C the unit circle oriented counterclockwise),
<br /> f(z) = \frac{1}{2\pi i}\int_C \frac{g(u)}{u-z} du<br />
where g(u) = \bar{u} is a continuous function and f is analytic in C. Describe fin C in terms of a power series.

\displaystyle f(z) = \frac{1}{2\pi i}\int_C \frac{\bar{u}}{u-z} du I am confused with what I am supposed to do. I know it says describe f in terms of a power series.

 

[alanlu]

If f is analytic in C, then its Taylor series should converge to f everywhere in C, right?

 

[fauboca]

If f is analytic in C, then its Taylor series should converge to f everywhere in C, right?

Yes.

 

[fauboca]

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

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

\frac{z-z_0}{u-z_0} is uniformly bounded since the max z-z_0 is r and the min u-z_0 is s so \frac{r}{s}&lt;1.

<br /> 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<br />
The series converges uniformly for all r<s and pointwise for all z with z\in D(z_0,s). So we can integrate term by term.

<br /> 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]<br />

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

Now how can I explain why f does or does not equal g? Is f described correctly as a power series here as well?

 

[fauboca]

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

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

\frac{z-z_0}{u-z_0} is uniformly bounded since the max z-z_0 is r and the min u-z_0 is s so \frac{r}{s}&lt;1.

<br /> 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<br />
The series converges uniformly for all r<s and pointwise for all z with z\in D(z_0,s). So we can integrate term by term.

<br /> 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]<br />

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

Now how can I explain why f does or does not equal g? Is f described correctly as a power series here as well?

From here, I can expand at z_0 = 0 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?

 

[fauboca]

From here, I can expand at z_0 = 0 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 \bar{u} is not holomorphic in the disk, f\neq g.
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.