Computing $\int_{\gamma} {dz \over (1-z)^3}$ 
(a) Let $\gamma$ be the circle of radius ${1 \over 2}$ centered at the
  origin, oriented counter-clockwise.  
Compute
$$ \int_{\gamma} {dz \over (1-z)^3} $$
(b) Same as above, except $\gamma$ is centered at $1$ with radius ${1 \over 2}$.

Part (a) attempt:


*

*Let $\gamma$ be the counter-clockwise circle of radius ${1 \over 2}$ centered at the origin parameterized by $\gamma(t) = {1 \over 2}e^{-it}$ on $0 \le t \le 2 \pi$.

*Then we have
$$
\int_\gamma {dz \over (1 - z)^3} = \int_{-\gamma} {(-1)^2 dz \over (z-1)^{2+1}} = {2 \pi i \over 2!} f^{(2)}(1) \text{ s.t. }f(z) = 1
$$
and since $f^{(2)}(1) = 0$ we have that our integral is equal to zero.

*Another way of seeing this integral is equal to $0$ is to observe that since $1$ is outside the circle $-\gamma$, we have that $n(-\gamma, 1) = 0$ by Cauchy's Theorem in a disk (i.e., $-\gamma$ is a closed curve inside an open disk of radius greater than $1/2$ s.t. $f(z)$ is completely analytic on this disk; hence $\int_{-\gamma} {dz \over 1 - z} = 0$ so that the integral above must also equal $0$).
Part (b) attempt:


*

*Now if we assume that $\gamma$ is centered at $1$ with radius $1/2$, then we have that $n(-\gamma,1) = 1$ so that $-n(\gamma,1) = -1$.  This makes me suspect our integral in question isn't equal to zero.

*Yet we still seem to have
$$
\int_\gamma {dz \over (1 - z)^3} = \int_{-\gamma} {(-1)^2 dz \over (z-1)^{2+1}} = {2 \pi i \over 2!} f^{(2)}(1) \text{ s.t. }f(z) = 1
$$
But it seems like this value is still $0$ since $f^{(2)}(1) = 0$ regardless of where $\gamma$ is centered.  This doesn't seem right -- so I think I'm misapplying Cauchy's Integral Formula.  Is this the case?

You are correct.  To see this, simply set $z=1+\frac12 \, e^{i \phi}$, and the integral becomes
$$-i 8 \int_0^{2 \pi} d\phi \, e^{-i 2 \phi}$$
which is clearly zero.  
Hopefully, this should help you see why the only nonzero contribution to an integral over a closed circle comes from the $1/(z-a)$ term.
