= Solution
Set $w=z-1$. <Partial fraction decomposition> gives
$$
\frac{2z}{z^2+1}=\frac1{z-i}+\frac1{z+i}=\frac1{1-i+w}+\frac1{1+i+w}.
$$
Expand both terms as <geometric series>:
$$
f(z)=\sum_{n=0}^{\infty}(-1)^n\left[(1-i)^{-n-1}+(1+i)^{-n-1}\right]w^n.
$$
The two bracketed terms are complex conjugates. Using $1\pm i=\sqrt2e^{\pm i\pi/4}$ produces the required real coefficients:
$$
\boxed{f(z)=\sum_{n=0}^{\infty}(-1)^n2^{(1-n)/2}\cos\left(\frac{(n+1)\pi}{4}\right)(z-1)^n.}
$$
The initial terms are $1-\tfrac12w^2+\tfrac12w^3-\tfrac14w^4+\cdots$. The nearest singularities are the <simple poles> $z=\pm i$, both at distance $\sqrt2$ from the center. Hence the <Taylor series> has
$$
\boxed{R=\sqrt2,\qquad|z-1|<\sqrt2.}
$$
Back to article page
