<s>[INST] 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?
 [/INST] 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.
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