Convergence of the taylor series of a branch of logarithm. Consider the branch of log defined on $\mathbb{C}$ with the negative real axis and origin removed. I was told that its Taylor series about the point $z_0 = -2 + i$ converges in a radius $\sqrt5$, which means the Taylor series actually converges for points on the negative real axis (not in its domain).
I can understand why this could be possible: the real function $f(x) = x^2$ defined on the real interval $|x| < 1$ has a Taylor series that converges everywhere to $g(x) = x^2$, where $g$ is defined on the entire real line.
But how do I actually prove that there is a branch of log that the Taylor series about $z_0 = -2 + i$ converges to? Can the Taylor series converge to a function that is not a branch of log? Why can't the radius of convergence $\sqrt5$ be increased to include the origin?
 A: 
Why can't the radius of convergence $\sqrt{5}$ be increased to include the origin?

The real part of logarithm (no matter what branch you take) is $\log |z|$. This is unbounded as $z\to 0$. But a power series is bounded on compact subsets of its disk of convergence. Therefore, no power series representing the logarithm (whatever branch) can have $0$ inside of its disk of convergence. 

Can the Taylor series converge to a function that is not a branch of $\log$? 

By definition, anything you get from Taylor series of $f$ is some branch of $f$, because it's related to $f$ by analytic continuation.  

how do I actually prove that there is a branch of log that the Taylor series about $z_0=−2+i$ converges to? 

There is nothing magical about negative real axis: you can cut the plane along any other half-line from $0$ to $\infty$ and define a branch of $\log$ in the slit plane (by interpreting $\arg z$ there). For example, the half-line $\{(2-i)t:t\ge 0\}$ would do.  Or simply define $$\operatorname{Log}z =\log(   z/(-2+i))+\log (-2+i)$$ where $\log$ is the principal branch.
