(adsbygoogle = window.adsbygoogle || []).push({}); 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!

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# Contour Integrals

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