Power Series Expansion of Cauchy Integral Formula on the Unit Circle

[fauboca]

For all z inside of C (C the unit circle oriented counterclockwise),
$$f(z) = \frac{1}{2\pi i}\int_C \frac{g(u)}{u-z} du$$
where $g(u) = \bar{u}$ is a continuous function and $f$ is analytic in C. Describe $f$in 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, $s>0$.
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 $0<r<s$.

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

$$2\pi i f(z) = \int_C \frac{\bar{u}}{u-z_0}\sum_{n=0}^{\infty}\frac{1}{(u-z_0)^n}(z-z_0)^n du$$
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.

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

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

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, $s>0$.
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 $0<r<s$.

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

$$2\pi i f(z) = \int_C \frac{\bar{u}}{u-z_0}\sum_{n=0}^{\infty}\frac{1}{(u-z_0)^n}(z-z_0)^n du$$
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.

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

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

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.