# Parameterizing the curve

1. Dec 13, 2014

### Bashyboy

1. The problem statement, all variables and given/known data
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 $z^{1/3}$.

2. Relevant equations

3. The attempt at a solution

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,

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

2. Dec 13, 2014

### haruspex

You dropped something in the final step.

3. Dec 14, 2014

### vela

Staff Emeritus
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.

4. Dec 20, 2014

### FeDeX_LaTeX

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