Proving Zero Contour Integral |z-i|=4 traversed once clockwise

[stunner5000pt]

Find \int_{C} 3(z-i)^2 dz 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 z(t) = i + 4e^{it}, \ 0\leq t \leq 2 \pi

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

is the setup good?

Also
Compute \int_{\Gamma} \overline{z} dz [/tex] where Gamma is the circle |z|=2 tranversed once counterclockwise<br /> <br /> z(t) = 2e^{it}<br /> \int_{0}^{2\pi} (-2e^{it}) (2i e^{it}) dt<br /> <br /> is this correct??<br /> Thank you for the help!

 

[shmoe]

Find \int_{C} 3(z-i)^2 dz 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 z(t) = i + 4e^{it}, \ 0\leq t \leq 2 \pi

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

Also
Compute \int_{Gamma} \overline{z} dz [/tex] where Gamma is the circle |z|=2 tranversed once counterclockwise<br /> <br /> z(t) = 2e^{it}<br /> \int_{0}^{2\pi} (-2e^{it}) (2i e^{it}) dt<br />

<br /> <br /> Check what \overline{2e^{it}} is again.