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- Thread starter dwsmith
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- Thread starter
- #1

- Feb 5, 2012

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Hi dwsmith,$$

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

$$

How would this be integrated?

What is the path of integration?

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- #3

- Feb 5, 2012

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Then the parametric equation of the curve would be,Unit circle counterclockwise

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

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

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

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- #5

I wasn't thinking.Then the parametric equation of the curve would be,

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

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

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

---------- 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?

- Feb 5, 2012

- 1,621

Should be. Because,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?

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