Complex integration no Residue Theory everything else is ok

[Dustinsfl]

$$
\int_0^{2\pi}\frac{\bar{z}}{z^2}dz
$$

How would this be integrated?

 

[Sudharaka]

$$
\int_0^{2\pi}\frac{\bar{z}}{z^2}dz
$$

How would this be integrated?

Hi dwsmith,

What is the path of integration?

 

[Dustinsfl]

Unit circle counterclockwise

 

[Sudharaka]

Unit circle counterclockwise

Then the parametric equation of the curve would be,

\[C:~ z(\theta)=e^{i\theta};0\leq x\leq2\pi\]

\[dz=ie^{i\theta}d\theta\]

\[\therefore\int_{C}\frac{\bar{z}}{z^2}dz=\int^{2\pi}_{0}\frac{e^{-i\theta}}{e^{2i\theta}}ie^{i\theta}d\theta=i\int_{0}^{2\pi}e^{-2i\theta}d\theta=0\]

 

[Dustinsfl]

Then the parametric equation of the curve would be,

\[C:~ z(\theta)=e^{i\theta};0\leq x\leq2\pi\]

\[dz=ie^{i\theta}d\theta\]

\[\therefore\int_{C}\frac{\bar{z}}{z^2}dz=\int^{2\pi}_{0}\frac{e^{-i\theta}}{e^{2i\theta}}ie^{i\theta}d\theta=i\int_{0}^{2\pi}e^{-2i\theta}d\theta=0\]

I wasn't thinking.

---------- Post added at 10:06 PM ---------- Previous post was at 09:53 PM ----------

So to expand on this problem,

$$
f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}iz^n\int_0^{2\pi}\frac{1}{e^{(n + 1)i\theta}}d\theta = 0, \ \forall n\geq 0.
$$
Therefore, $f(z) = 0$

This would be correct then?

 

[Sudharaka]

I wasn't thinking.

---------- Post added at 10:06 PM ---------- Previous post was at 09:53 PM ----------

So to expand on this problem,

$$
f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}iz^n\int_0^{2\pi}\frac{1}{e^{(n + 1)i\theta}}d\theta = 0, \ \forall n\geq 0.
$$
Therefore, $f(z) = 0$

This would be correct then?

Should be. Because,

\[\int^{2\pi}_{0}\frac{1}{e^{(n+1)i\theta}}d\theta= \int^{2\pi}_{0}e^{-(n+1)i\theta}d\theta=\left[\frac{e^{-(n+1)i\theta}}{-i(n+1)}\right]_{0}^{2\pi}=0\]