...when the loop is a positively oriented circle $C$ and $a$ lies outside of $C$?
My work so far has been to show $\frac{1}{z-a}$ is contimuous when $z\neq a$. As such, every loop integral of $\frac{1}{z-a}$ (where the loop does not contain $a$) $=0$, i.e. vanishes. Since $C$ does not contain $a$, this applies to $\int_C\frac{1}{z-a}$.
Is this correct?
On
$\frac1{z-a}$ is actually holomorphic on the simply connected region enclosed by $C$. Cauchy's theorem tells us the integral is zero. Notice this would be the winding number.
No, this is not correct. You just mentioned that $C$ does not contain $a$, but you did not actually use that hypothesis. So, the argument should also be valid when $C$ contains $a$, but then the integral is not $0$.
If $C$ doesn't contain $a$, you can find an open half-plane $H$ containing $C$ such that $a\notin H$. SInce $H$ is simply connected, the restraction of $\frac1{z-a}$ to $H$ has a primitive $F$. And so, since $C\colon[b,c]\longrightarrow H$ is a loop,$$\int_C\frac{\mathrm dz}{z-a}=F\bigl(C(c)\bigr)-F\bigl(C(b)\bigr)=0.$$