How to Express Cauchy Integral Formula in Terms of a Power Series?

[Dustinsfl]

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 $g(u) = u^7$ is a continuous function and $f$ is analytic in C. Describe $f$ in C in terms of a power series.

$\displaystyle f(z) = \frac{1}{2\pi i}\int_C \frac{u^7}{u-z} du$ I am confused with what I am supposed to do. I know it says describe $f$ in terms of a power series.

 

[chisigma]

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 $g(u) = u^7$ is a continuous function and $f$ is analytic in C. Describe $f$ in C in terms of a power series.

$\displaystyle f(z) = \frac{1}{2\pi i}\int_C \frac{u^7}{u-z} du$ I am confused with what I am supposed to do. I know it says describe $f$ in terms of a power series.

Applying the Cauchy integral formula You have...

$ \displaystyle \int_{C} \varphi(u)\ du = 2 \pi i \sum_{k} r_{k}$ (1)

... where $r_{k}$ is the residue of any singularity of $\varphi(*)$ inside C. In Your case is $\varphi(u)= \frac{u^{7}}{u-z}$ and the only singularity is in $u=z$ so that is...

$\displaystyle f(z) = \lim_{u \rightarrow z}\ \varphi(u) (u-z)= z^{7}$ (2)

... which of course is the Taylor expansion of itself...

Kind regards

$\chi$ $\sigma$

 

[Dustinsfl]

$\displaystyle f(z) = \lim_{u \rightarrow z} \varphi(u) (u-z)= z^{7}$ (2)

... which of course is the Taylor expansion of itself...

Kind regards

$\chi$ $\sigma$

z^7 or u^7?

 

[chisigma]

z^7 or u^7?

$z^{7}$

Kind regards

$\chi$ $\sigma$

 

[Dustinsfl]

$z^{7}$

Kind regards

$\chi$ $\sigma$

Why did you say $f(z) = \lim \varphi(u)(u-z)$?

 

[chisigma]

Why did you say $f(z) = \lim \varphi(u)(u-z)$?

If $\varphi(z)$ has a pole of order 1 in $z=z_{0}$ then its residue is...

$r_{z=z_{0}} \varphi(z)= \lim_{z \rightarrow z_{0}} \varphi(z)\ (z-z_{0})$ (1)

Kind regards

$\chi$ $\sigma$

 

[Dustinsfl]

Ok so to see if I understand correctly, I am going to solve $g(u) = \bar{u}$.

$f(z)=\frac{1}{2\pi i}\int_C\frac{\bar{u}}{u-z}du$ again let $\varphi(u) = \frac{\bar{u}}{u-z}$

Then $2\pi i\sum_na_n=\int_C\varphi(u)du$

$\displaystyle\lim_{u\to z}\varphi(u)(u-z) = $ since this is the conjugation of u, u would have to always be real correct?

 

[Dustinsfl]

Ok so to see if I understand correctly, I am going to solve $g(u) = \bar{u}$.

$f(z)=\frac{1}{2\pi i}\int_C\frac{\bar{u}}{u-z}du$ again let $\varphi(u) = \frac{\bar{u}}{u-z}$

Then $2\pi i\sum_na_n=\int_C\varphi(u)du$

$\displaystyle\lim_{u\to z}\varphi(u)(u-z) = $ since this is the conjugation of u, u would have to always be real correct?

So redoing this problem:

$$f(z)=\frac{1}{2\pi i}\int_C\frac{\bar{u}}{u-z}du\Rightarrow 2\pi if(z)=\int_C\frac{\bar{u}}{u-z_0}\frac{1}{1-\frac{z-z_0}{u-z_0}}du$$

$z_0$ 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?

 

[Dustinsfl]

So redoing this problem:

$$f(z)=\frac{1}{2\pi i}\int_C\frac{\bar{u}}{u-z}du\Rightarrow 2\pi if(z)=\int_C\frac{\bar{u}}{u-z_0}\frac{1}{1-\frac{z-z_0}{u-z_0}}du$$

$z_0$ 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?

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

How would this be evaluated?

 

[Dustinsfl]

So f would be described as

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 there a way to evaluate these integrals or must it be left like this?

 

[Dustinsfl]

So f would be described as

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 there a way to evaluate these integrals or must it be left like this?

This post didn't apply to this part of the problem.