Parameterizing the curve

[Bashyboy]

Homework Statement

Calculate $\int_C z^{1/3} dz$, where $C$ is the circle of radius $1$ centered at the origin oriented in the clockwise direction. Use the branch $0 \le \arg z \le 2 \pi$ to define $z^{1/3}$.

Homework Equations

The Attempt at a Solution

[/B]
I was hoping that someone could verify my work.

Parameterizing the curve: $z(t) = e^{-it}$, $0 \le t \le 2 \pi$.

$\int_C z^{1/3} dz = \lim_{a \rightarrow 0}_{b \rightarrow 2 \pi} \int_a^b (e^{-it})^{1/3}(-ie^{-it}) dt \iff$

$\int_C z^{1/3} dz = \lim_{a \rightarrow 0}_{b \rightarrow 2 \pi} -i \int_a^b e^{-i \frac{4}{3}t} dt \iff$

$\int_C z^{1/3} dz =\lim_{a \rightarrow 0}_{b \rightarrow 2 \pi} (-i)(i \frac{3}{4}) [e^{-i \frac{4}{3} t}|_a^b \iff$

$\int_C z^{1/3} dz = \frac{3}{4} \lim_{a \rightarrow 0}_{b \rightarrow 2 \pi} [e^{- \frac{4}{3} (b)} - e^{-i \frac{4}{3} (a)}]$

Taking the limits,

$\int_C z^{1/3} dz = \frac{3}{4} [e^{-i \frac{8}{3}} - 1]$

 

[haruspex]

You dropped something in the final step.

 

[vela]

On the contour, you should have $z^{1/3} = e^{i\theta/3}$ where $0 \le \theta \le 2\pi$. The parametrization you used doesn't give that.

 

[FeDeX_LaTeX]

On the contour, you should have $z^{1/3} = e^{i\theta/3}$ where $0 \le \theta \le 2\pi$. The parametrization you used doesn't give that.

That was my thought too, but isn't C oriented in the clockwise direction (as opposed to the usual anti-clockwise convention)?