= Solution
<Partial fraction decomposition> give $h(z)=1/(z-i)+1/(z+i)$. Let $w=z-1$. Each fraction has a geometric expansion, so
$$
\boxed{h(z)=\sum_{j=0}^\infty(-1)^j\left[\frac1{(1-i)^{j+1}}+\frac1{(1+i)^{j+1}}\right](z-1)^j.}
$$
Equivalently, its coefficient is $(-1)^j2^{(1-j)/2}\cos((j+1)\pi/4)$. The first terms are
$$
h(z)=1-\frac12w^2+\frac12w^3-\frac14w^4+O(w^6).
$$
The linear and fifth-order coefficients vanish. Both <geometric series> converge for $|w|<\sqrt2$, and $h$ has genuine simple poles at $z=\pm i$, each distance $\sqrt2$ from the centre. Thus the <Taylor series> has
$$
\boxed{R=\sqrt2.}
$$
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