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- #1
Hi dwsmith,$$
\int_0^{2\pi}\frac{\bar{z}}{z^2}dz
$$
How would this be integrated?
Then the parametric equation of the curve would be,Unit circle counterclockwise
I wasn't thinking.Then the parametric equation of the curve would be,
\[C:~ z(\theta)=e^{i\theta};0\leq x\leq2\pi\]
\[dz=ie^{i\theta}d\theta\]
\[\therefore\int_{C}\frac{\bar{z}}{z^2}dz=\int^{2\pi}_{0}\frac{e^{-i\theta}}{e^{2i\theta}}ie^{i\theta}d\theta=i\int_{0}^{2\pi}e^{-2i\theta}d\theta=0\]
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?