Suppose we have some complex function, say $f(z)=\log(1+z)$ with branch cut along $(-\infty,-1]$. This has Laurent (Taylor in this case) Series about $z=0$: $f(z)=z-\frac{z^2}{2!}+\frac{z^3}{3!}\,-\,...$.
How does the branch cut affect the series? I believe that due the fact that there is a branch point at $z=-1$, the radius of convergence of the Laurent Series can be affected, but is anything else?
Upon selecting the principal branch, $\log(1+z)$ is analytic in $\mathbb{C}\setminus(-\infty,-1]$. Hence, we write its Taylor series about any point that excludes the branch cut.
Inasmuch as $z=0$ is not on the branch cut, then we can write
$$\log(1+z)=\sum_{n=1}^\infty \frac{(-1)^{n-1}z^n}{n}$$
for $|z|<1$.
If we selected instead to cut the plane along $[-1,\infty)$, $z=0$ is on the branch cut and we cannot represent $\log(1+z)$ on this branch by a Taylor series about $z=0$.
We can write, however, the Taylor series of $\log(1+z)$ about any point, say $z_0$, that is not on the branch cut. In that case, we have
$$\log(1+z)=\log(1+z_0)+\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(\frac{z-z_0}{1+z_0}\right)^n$$
for $|z-z_0|<\begin{cases}|1+z_0|&,\text{Im}(z_0)\le -1\\\\ \min\left(\text{Im}(z_0),|1+z_0|\right)&,\text{Im}(z_0)>-1\end{cases}$.