Primitive of $ z \longmapsto (z-z_0)^k $ in $ \mathbb{C} \backslash \{z_0 \}.$

Question

During my lecture on complexe analysis I came across the folliging statement, which I can intuitively understand but I can find a proper explanation.

Statement: $ \forall k \in \mathbb{Z} \backslash \{-1\}$ the function z $\longmapsto (z-z_0)^k $ have a primitive in $ \mathbb{C} \backslash \{z_0 \}.$

Any explanation or hint would be really appreciated.

Answer 1

It means that the function has anti-derivative. Since $k \not = -1$, $\frac{(z-z_{0})^{k+1}}{k+1}$ is well defined on $\mathbb{C}-\{z_{0}\}$ and is an antiderivative for $(z-z_{0})^{k}$. If $k = -1$, there is no primitive, since $\int_{|z-z_{0}| = 1}\frac{dz}{z-z_{0}} = 2\pi i$, where the circle is oriented in the anticlockwise sense.