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

$$z(t) = 2e^{it}$$
$$\int_{0}^{2\pi} (-2e^{it}) (2i e^{it}) dt$$

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

$$z(t) = 2e^{it}$$
$$\int_{0}^{2\pi} (-2e^{it}) (2i e^{it}) dt$$

Check what $$\overline{2e^{it}}$$ is again.