Issues with complex integration $\int_{C} \frac{1}{z-z_{0}} = 0$

Question

We have $\int_{C} \frac{1}{z-z_{0}} = 2\pi i$ for $C$ a circle centered on $z_{0}$. However, suppose that this $C$ does not cross the real line $(-\infty,0]$ such that there exists an antiderivative to $\frac{1}{z-z_{0}}$ , namely the principal log: $Log(z-z_{0})$. Then, because $C$ is closed, the integral must equal $0$. Where is the contradiction in my logic?

Answer 1

When considering the logarithm $\log(z-z_0)$ centered at $z=z_0$, the negative real line $(-\infty,0]$ is not a (potential) branch cut. This is only a branch cut for the logarithm $\log(z)$ centered at $z=0$.

For your example, we don't care if $C$ crosses $(-\infty,0]$. Rather, we care if $C$ crosses a branch cut stretching from $z_0$ to $\infty$, such as the ray described parametrically by $z_0-t$ for $t\in\mathbb R_{\ge 0}$.

Answer 2

To define the principle log, $\operatorname{Log} (z-z_0)$, you require that the circle does not pass through the half ray starting at $z_0$ (instead of the origin $0$). But all circles centered at $z_0$ would intersect such a ray.