Radius of convergence of power series of complex $\log$ Let $f(z) = \log(z)$ for $z\in \Bbb{C}\setminus (-\infty,0]$.
Since $f$ is holomorphic on its domain, we know it has a power series development about each point $z_0\in \Bbb{C}\setminus (-\infty,0]$. We also that its radius of convergence will be at least the distance of $z_0$ to the nearest singularity.
How can we determine the exact radius of convergence?
 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 $D$.
A: On the open ball $B(z_0,|z_0|)$, we can define the logarithm as
$$
f(z)=\log(z_0)+\int_{z_0}^z\frac{\mathrm{d}z}z\tag{1}
$$
where the contour from $z_0$ to $z$ is contained completely in $B(z_0,|z_0|)$. Cauchy's Integral Theorem guarantees we get the same result for any such contour. That is because the difference of any two such contours is a closed contour which does not contain the only singularity of $\frac1z$ at $z=0$.
The radius of convergence is no more than $|z_0|$ since the $\log(z)$ is not analytic in any neighborhood of $z=0$. Furthermore, by Cauchy's Differentiation Formula
$$
\frac{\mathrm{d}^n}{\mathrm{d}z^n}\log(z_0)=\frac{n!}{2\pi i}\int_{\gamma_{\large z_{\small0},r}}\frac{\log(w)\,\mathrm{d}w}{(w-z_0)^{n+1}}\tag{2}
$$
where
$$
\gamma_{z_0,r}(t)=z_0+re^{it}\qquad\text{for }t\in[0,2\pi]\tag{3}
$$
is a counter-clockwise circle of radius $r$ centered at $z_0$ where $r\lt|z_0|$.
For $w\in\gamma_{z_0,r}$, $\frac{|z_0|-r}{|z_0|}\le\frac{|w|}{|z_0|}\le\frac{|z_0|+r}{|z_0|}$. Therefore,
$$
|\log(w)|\le\log\bigg(\frac{|z_0|^2}{|z_0|-r}\bigg)\tag{4}
$$
Combining $(2)$ and $(4)$, we get that for any $r\lt|z_0|$,
$$
\begin{align}
\limsup_{n\to\infty}\left|\frac1{n!}\frac{\mathrm{d}^n}{\mathrm{d}z^n}\log(z_0)\right|^{1/n}
&\le\limsup_{n\to\infty}\left|\frac1{r^{n+1}}\log\bigg(\frac{|z_0|^2}{|z_0|-r}\bigg)\right|^{1/n}\\
&=\frac1r\tag{5}
\end{align}
$$
Using the Root Test, we get that the radius of convergence of the Taylor series is at least $r$ for any $r\lt|z_0|$. Thus, the radius of convergence is at least $|z_0|$.
Therefore, the radius of convergence is precisely $|z_0|$.
A: We first choose a point a to expand round it and measure its modulus and expand the function using Taylor's Theorem.  We know that this power series converge with radius of convergence equal to modulus a and no more except possibly on the boundary but not all points of the boundary the function is analytic at.  
