Parameterizing the curve

  • Thread starter Bashyboy
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  • #1
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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


[/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]##
 

Answers and Replies

  • #2
haruspex
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You dropped something in the final step.
 
  • #3
vela
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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
FeDeX_LaTeX
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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)?
 

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