Contour Integrals

  • #1
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Find [tex] \int_{C} 3(z-i)^2 dz [/tex] where C is the circle |z-i|=4 traversed once clockwise

well i know it is zero but i just want to prove it.. kind of

so we can parametrize [tex] z(t) = i + 4e^{it}, \ 0\leq t \leq 2 \pi [/tex]

so
[tex] \int_{C} 3(z-i)^2 dz = \int_{0}^{2\pi} 3(i + 4e^{it}-i)^2 (4ie^{it}) dt [/tex]

is the setup good?

Also
Compute [itex] \int_{\Gamma} \overline{z} dz [/tex] where Gamma is the circle |z|=2 tranversed once counterclockwise

[tex] z(t) = 2e^{it} [/tex]
[tex] \int_{0}^{2\pi} (-2e^{it}) (2i e^{it}) dt [/tex]

is this correct??
Thank you for the help!
 
Last edited:

Answers and Replies

  • #2
shmoe
Science Advisor
Homework Helper
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stunner5000pt said:
Find [tex] \int_{C} 3(z-i)^2 dz [/tex] where C is the circle |z-i|=4 traversed once clockwise

well i know it is zero but i just want to prove it.. kind of

so we can parametrize [tex] z(t) = i + 4e^{it}, \ 0\leq t \leq 2 \pi [/tex]

This parameterization is going clockwise, not counterclockwise. Otherwise it looks fine.

stunner5000pt said:
Also
Compute [itex] \int_{Gamma} \overline{z} dz [/tex] where Gamma is the circle |z|=2 tranversed once counterclockwise

[tex] z(t) = 2e^{it} [/tex]
[tex] \int_{0}^{2\pi} (-2e^{it}) (2i e^{it}) dt [/tex]


Check what [tex]\overline{2e^{it}}[/tex] is again.
 

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