The idea is that after subtracting off the pole at $z=z_0$, you are left with something holomorphic. After expanding in a Laurent series around $z=z_0$ as you suggest, and subtracting off $P(z):=\sum_{k>0} A_k (z-z_0)^{-k}$ from $f(z)$, you are left with something that is holomorphic at $z=z_0$ (since you removed the singular part) and also holomorphic in the remainder of the region $\{z\mid |z|<R\}$ (since both $f(z)$ and $P(z)$ are holomorphic in $\{z\mid z\ne z_0,\ |z|<R\}$). Therefore, $g(z)=f(z)-P(z)$ is holomorphic on $\{z\mid |z|<R\}$, so it has a power series around $0$ with radius of convergence at least $R$.
One also get at the radius of convergence like this:
Set
$\alpha = \dfrac{1 + i}{\sqrt{2}}; \tag{1}$
then
$\vert \alpha \vert = 1 \tag{2}$
and so
$\vert \alpha^{-1} \vert = 1 \tag{3}$
as well; we write
$f(z) = \dfrac{1}{1 + i - \sqrt{2}z} = \dfrac{1}{\sqrt{2}} \dfrac{1}{\alpha - z} = \dfrac{1}{\sqrt{2} \alpha} \dfrac{1}{1 - \alpha^{-1} z}; \tag{4}$
for $\vert z \vert < 1$ we have
$\vert \alpha^{-1} z \vert = \vert \alpha^{-1} \vert \vert z \vert < 1; \tag{5}$
thus the series
$\dfrac{1}{1 - \alpha^{-1} z} = 1 + \alpha^{-1}z + \alpha^{-2}z^2 + \ldots = \sum_0^\infty (\alpha^{-1} z)^n \tag{6}$
converges. Allowing $\vert z \vert = 1$, we may take $z = \alpha$ and not only is
$\dfrac{1}{(1 - \alpha^{-1} z)} = \dfrac{1}{(1 - \alpha^{-1} \alpha)} = \dfrac{1}{1 - 1} \tag{7}$
undefined, but
$\sum_0^\infty (\alpha^{-1} z)^n = \sum_0^\infty (\alpha^{-1} \alpha)^n = \sum_0^\infty 1^n = \infty; \tag{8}$
the series clearly explodes for $z = \alpha$. Since this series converges for $\vert z \vert < 1$ but not for $z = \alpha$, the radius of convergence is clearly $1$. The same holds for the series of $f(z)$, by (4).
Best Answer
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 $D$.