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## 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

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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]##