I have to find power series of $\frac{1}{(z-1)(z-2)}$ centered at $3+i$ and give its radius of convergence.
I just simply transformed it using partial fractions and geometric series expansion and radius of convergence turned out to be $\sqrt{2}$. Is that a correct answer for the radius?
My attempt:
$$\frac{1}{(z-1)(z-2)}=\frac{1}{z-2}-\frac{1}{z-1}=\frac{1}{1-z}-\frac{1}{2-z}=\frac{1}{-2-i-(z-(3+i))}-\frac{1}{-1-i-(z-(3+i))}=\frac{1}{-2-i}\frac{1}{1-\frac{z-(3+i)}{-2-i}}-\frac{1}{-1-i}\frac{1}{1-\frac{z-(3+i)}{-1-i}}$$
Now I check when $| \frac{z-(3+i)}{-2-i} | < 1$ and $|\frac{z-(3+i)}{-1-i}|<1$.
Writing $z=x+iy$ we have:
$$\sqrt{(x-3)^2+(y-1)^2}<\sqrt{5}$$
and
$$\sqrt{(x-3)^2+(y-1)^2}<\sqrt{2}$$
So $| \frac{z-(3+i)}{-2-i} | < 1$ and $|\frac{z-(3+i)}{-1-i}|<1$ whenever $z$ lies inside an open ball on complex plane centered at $3+i$ and with radius $\sqrt{2}$.
Now I use power series expansion and have:
$$\frac{1}{-2-i}\frac{1}{1-\frac{z-(3+i)}{-2-i}}-\frac{1}{-1-i}\frac{1}{1-\frac{z-(3+i)}{-1-i}}=\frac{1}{-2-i}\sum_{n=0}^{\infty}(\frac{z-(3+i)}{-2-i})^n-\frac{1}{-1-i}\sum_{n=0}^{\infty}(\frac{z-(3+i)}{-1-i})^n$$
And radius of convergence of this series is $\sqrt{2}$. Is this ok?