Radius of convergence - Complex plane A example in my textbook explain that $x^2-2x+2=0$ has the solutions $x = 1± i$. The distance between $x=0$ and $x = 1± i$ is $\sqrt{2}$ in the complex plane. So the radius of convergence of the Taylor series for $(x^2-2x+2)^{-1}$ is $\sqrt{2}$ around $x=0$. 
I looked for why this result is true, but I didn't find it. Is anyone could explain to me why this is true or just give me the theorem related?
 A: Hint: Compute the partial fraction decomposition of $\frac{1}{x^2-2x+2}$ and consider the radius of convergence of the power series for $\frac{1}{1-x}$.
More precisely, we can write
$$
\frac{1}{x^2-2x+2}=\frac{A}{x-(1+i)}+\frac{B}{x-(1-i)}.
$$
Using standard techniques from calculus, one can find the constants $A$ and $B$.
Now, the power series expansion of 
$$
\frac{A}{x-(1+i)}
$$
can be computed as follows:
$$
\frac{A}{x-(1+i)}=\frac{-A}{1+i}\cdot\frac{1}{1-\frac{x}{1+i}}.
$$
Using the substitution $y=\frac{x}{1+i}$, we have that 
$$
\frac{A}{x-(1+i)}=\frac{-A}{1+i}\cdot\frac{1}{1-y}
$$
which is a common function from calculus whose power series centered at $y=0$ converges for $|y|<1$.  By substitution, $|x|<|1+i|=\sqrt{2}$.  Approach the other fraction similarly.
A: Suppose a function $f$ is analytic on a domain $D$ with $D\supset C(r)=\{x:|x|<r\},$ where $0<r<\infty.$ Then $f(x)$ is equal to its power series expansion about $0$, for every $x\in C(r).$ 
Suppose the radius of convergence of this series is $R,$ with $R>r.$
The series then defines an analytic function $g$ with domain $C(R)=\{x:|x|<R\},$ and $g(x)=f(x)$ whenever $|x|<r.$ 
But $g(x)$ is continuous at $x$ whenever $|x|=r.$ So whenever  $|x|=r$ and $(x_n)_{n\in N}$ is any sequence in $C(r)$ converging to x,  then, since $f(x_n)=g(x_n)$ for every $n\in N,$ we have $g(x)=\lim_{n \to \infty} g(x_n)=\lim_{n\to \infty} f(x_n).$
Hence,when $f(x)=(x^2-2 x+2)^{-1}$ and $r=\sqrt 2,$ the radius $R$ of convergence about $0$ of the power series for $f$ cannot exceed $\sqrt 2.$  Otherwise there exists $g(1+i)$ such that for any sequence $(x_n)_n$ in $C(\sqrt 2)$ converging to $1+i,$ the sequence $(f(x_n))_{n\in N}$  converges to $g(1+i).$
But $(f(x_n))_{n\in N}$ never converges when $(x_n)_{n\in N}$ converges to $1+i.$  
