http://en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions
I wrote the initial draft of the article linked above in February 2004, and mostly it's still as I wrote it, although others have contributed.
Postscript:
Let $C$ be a positively oriented circle centered at $a$ that encloses a point $z$ that is closer to $a$ then is any place where $f$ blows up, and that does not enclose, nor pass through, any point where $f$ blows up.
\begin{align}f(z) &{}= {1 \over 2\pi i}\int_C {f(w) \over w-z}\,dw \tag1 \\[10pt]
&{}= {1 \over 2\pi i}\int_C {f(w) \over (w-a)-(z-a)} \,dw \tag2 \\[10pt]
&{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{1 \over 1-{z-a \over w-a}}f(w)\,dw\tag3 \\[10pt]
&{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{\sum_{n=0}^\infty\left({z-a \over w-a}\right)^n} f(w)\,dw\tag4 \\[10pt]
&{}=\sum_{n=0}^\infty{1 \over 2\pi i}\int_C {(z-a)^n \over (w-a)^{n+1}} f(w)\,dw\tag5 \\[10pt]
& = \sum_{n=0}^\infty (z-a)^n \underbrace{{1 \over 2\pi i}\int_C {f(w) \over (w-a)^{n+1}} \,dw}_{\text{No $z$ appears here!}}.\tag6
\end{align}
Step $(1)$ above is Cauchy's formula.
Step $(4)$ is summing a geometric series.
Step $(6)$ can be done because "$(z-a)^n$" has no $w$ in it; thus does not change as $w$ goes around the circle $C$.
The fact that no "$z$" appears in the expression in $(6)$ where that is noted, means that the last expression is a power series in $z-a$.
The radius of convergence will be $|z_0|$.
It is clear that it can't be greater that $|z_0|$, since $\log z$ can't be defined (holomorphically) on any neighbourhood of the origin.
On the other hand, the disc $D = \{ z : |z-z_0| < |z_0| \}$ is simply connected and doesn't contain $0$. Hence we can find a branch of the complex logarithm (that agrees with your choice of $\log$ near $z_0$. Taylor's theorem shows that the radius of convergence must be at least (and thus exactly) $|z_0|$.
Note, however, that the power series you get does not necessarily coincide with the principle branch of $\log z$ on all of $D$.
Best Answer
More precisely, the radius of convergence of the series about $z=a$ is the largest $r$ such that there is an analytic function in the open disk of radius $r$ about $a$ which has the given power series as its Taylor series around $z=a$. In particular, removable singularities don't count, and branch cuts don't count if they can be moved out of the way.
EDIT: To prove this, you can proceed as follows: