Simple conceptual Fundamental Thm of Calculus question When applying the Fundamental Thm of Calculus in complex analysis, what does it mean for an open connected set to contain a loop? For example, does my red-color open annulus contain the black colour loop? I think so, but I'm struggling with understanding this:

When asked to integrate $\frac 1{(z-5)}$ around a circle centred at 5, why can't I use the Fundamental Thm of Calculus? After all, the function is continuous in the red annulus , and furthermore, its primitive $\ln z$ is analytic throughout the annulus which does not touch the negative real axis.
 A: The fundamental theorem of calculus is applied to a special contour: intervals of the real line.  And it applies to functions that have an antiderivative, or primitive.  In the complex plane, we have Cauchy's Theorem which states that such nicely behaved functions ("analytic") may be integrated by evaluating its antiderivative at the endpoints of a contour, no matter the shape of the contour.  
In your case, the function $1/(z-5)$ has a pole in the interior of the black contour. By Cauchy's theorem, the value of the integral is $i 2 \pi$.  You can see this from a parametrization $z=5+r e^{i \phi}$.  Now, suppose you evaluate the integral via a Fundamental Theorem of calculus mindset: let's say we want to evaluate $\log{(z-5)}$ at the beginning and end of the contour.   Parametrize by $z=5 + r e^{i \phi}$, and so we will be subtracting the value of the antiderivative at $\phi=0$ from the value at $\phi=2 \pi$:
$$\log{(r e^{i 2 \pi})} - \log{(r e^{i 0} )} = \log{r} + i 2 \pi - \log{r} - i 0 = i 2 \pi$$
It is the multivaluedness of the log function that causes the nonzero value.  But you see that it all agrees and provides for a consistent basis for computation.
A word about your annulus: yes, $1/(z-5)$ is analytic throughout your annulus.  The integral about the boundary of the annulus is indeed zero: you also have to integrate about the inner circle in the opposite direction, so the integral is $i 2 \pi - i 2 \pi = 0$.
