Cauchy-Goursat theorem. If a function $f$ is analytic at all points interior to and on a simple closed contour $C$, then $$\int_C f(z) \,dz=0.$$
This is a problem from Churchill's Complex Variables.
Problem: Apply the Cauchy-Goursat theorem to show that
$$\int_C f(z) \,dz=0$$
when the contour $C$ is the unit circle $|z|=1$, in either direction, and when
$f(z)=\operatorname{Log}(z+2)$, where $\operatorname{Log}$ is the principal branch.
I don't understand how I can apply the theorem in this case, since the branch of $f(z)$ is not defined on the ray $\theta=\pi$; hence does not satisfy the condition of analytic at all points interior to and on a simple closed contour $C$, of the theorem.
How can I make sense of this problem? I would greatly appreciate any help.
If we take $\operatorname{Log}(z)$ along its principal branch, wherein $\operatorname{Log}(z) = \ln|z| + i \operatorname{Arg}(z)$, for $- \pi < \operatorname{Arg}(z) \leq \pi$, then we know that $\operatorname{Log}(z)$ is analytic everywhere except the real axis where $z \leq 0$.
It follows that $\operatorname{Log}(z+2)$ is analytic everywhere except along the real axis where $z+2 \leq 0$. Rearranging, we see that $\operatorname{Log}(z+2)$ is analytic everywhere except the set of points $\{z \in \mathbb{C} \ | \ \operatorname{Im}(z) = 0 \text{ and } \operatorname{Re}(z) \leq -2 \}$.
Now sketch $C$ and this set of points, and you'll find that the stipulations of Cauchy-Goursat are satisfied.
Footnote: The principal branch of $\operatorname{Log}(z)$ often works, but in general you can take any branch cut you'd like to fit your needs. Ultimately, $\operatorname{Log}(z)$ with any branch will always be an antiderivative of $\displaystyle \frac{1}{z}$ at all points not along the cut.