= Solution
The <partial fraction decomposition> is $f(z)=1/(z-i)+1/(z+i)$. Put $w=z-1$ and expand each reciprocal by a <geometric series>:
$$
\frac1{1\mp i+w}=\sum_{n=0}^\infty\frac{(-1)^nw^n}{(1\mp i)^{n+1}},\qquad |w|<\sqrt2.
$$
Adding gives \b[the <Taylor series>]
$$
\boxed{f(z)=\sum_{n=0}^\infty(-1)^n\left[(1-i)^{-n-1}+(1+i)^{-n-1}\right](z-1)^n.}
$$
Equivalently, since $1\pm i=\sqrt2e^{\pm i\pi/4}$, its coefficient is $(-1)^n2^{(1-n)/2}\cos((n+1)\pi/4)$. The first terms are $1-\frac12w^2+\frac12w^3-\frac14w^4+\cdots$.
The nearest nonremovable <poles> are $i$ and $-i$, both at distance $\sqrt2$ from the center. The function is <holomorphic> throughout the open disk of that radius and cannot be <holomorphic> across either pole. Therefore \b[the largest <radius of convergence> is $r=\sqrt2$].
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