Find the laurent series of
$$f(z)=\frac{i}{z^2-iz+2}$$ with the values of the region $a,b$ in which it is valid $$ a<|z-1|<b $$
My attempt at a solution:
I did a partial fraction expansion and found
$$ f(z)=\frac{1}{3(z-2i)} - \frac{1}{3(z+i)} $$
Expanding about the point $|z-1|$, for the upper and lower limits $$ f(z)=\frac{1}{3(1-2i)(1-(-\frac{(z-1)}{1-2i})} - \frac{1}{3(1+i)(1-(-\frac{z-1}{1+i}))} $$
$$ f(z)=\frac{1}{3(z-1)(1-(-\frac{(1-2i)}{z-1})} - \frac{1}{3(z-1)(1-(-\frac{1+i}{z-1}))} $$
The next step through geometric expansion would be (for the first line above
$$\frac{1}{3(1-2i)} \sum (-\frac{(1-2i)}{z-1})^n -\frac{1}{3(1+1)} \sum (-\frac{(1+i)}{z-1})^n $$ And so on... But my final answer gives an $$a=\sqrt{5}, b=\sqrt{2}$$ thus $$ \sqrt{5} < |z-1| < \sqrt{2} $$
This does not seem correct, but analysing a diagram of the system shows an annulus where the laurent series cannot have an analytical expansion (its either within a disc of radius less than $\sqrt{2}$ or everywhere in the complex plane excluding a disc of radius $\sqrt{5}$ about the point. Is anyone able to give me any insight? Have I made a mistake in my reasoning? Especially assuming I can take a geometric series with imaginary terms?
Much appreciated.
You have; $$f(z)=\frac{1}{3(1-2i)}\frac{1}{(1-(-\frac{(z-1)}{1-2i})}-\frac{1}{3(1+i)}\frac{1}{(1-(-\frac{z-1}{1+i}))}$$ Expanding in a geometric series gives; $$f(z)=\frac{1}{3(1-2i)}\sum_{n=0}^{\infty}\left(\frac{z-1}{2i-1}\right)^{n} -\frac{1}{3(1+i)}\sum_{n=0}^{\infty}\left(-\frac{z-1}{1+i}\right)^{n}$$
A geometric series, $\sum_{n=0}^{\infty}r^{n}$ converges for $|r|<1$, so the the first term converges for $|z-1|<|2i-1|=\sqrt{5}$, and the second converges for $|z-1|<|1+i|=\sqrt{2}$, so as $\sqrt{2}<\sqrt{5}$, the expansion $f(z)$ converges for $|z-1|<\sqrt{2}$. Then as the modulus of a complex number is non negative, we have that this expansion converges for;
$$0\leq|z-1|<\sqrt{2}$$
Which you can confirm by drawing a diagram and looking at the positions of the poles. Looking at the poles also tells you that there should also be another expansion valid in the annulus $\sqrt{2}<|z-1|<\sqrt{5}$, and one in the region $\sqrt{5}<|z-1|$.
For the annular expansion write;
$$\frac{1}{1-(-\frac{z-1}{1+i})}=\frac{1+i}{z-1}\frac{1}{1+\frac{1+i}{z-1}}=\frac{1+i}{z-1}\sum_{n=0}^{\infty}\left(\frac{1+i}{z-1}\right)^{n}$$
Which converges for $|1+i|=\sqrt{2}<|z-1|$, so that the expansion;
$$f(z)=\frac{1}{3(1-2i)}\sum_{n=0}^{\infty}\left(\frac{z-1}{2i-1}\right)^{n} -\frac{1}{3(z-1)}\sum_{n=0}^{\infty}\left(\frac{1+i}{z-1}\right)^{n}$$
Is valid in the annulus $\sqrt{2}<|z-1|<\sqrt{5}$.
Finally, for the region $|z|>\sqrt{5}$, we expand the first term as;
$$ \frac{1}{3(1-2i)}\frac{1}{1-(-\frac{z-1}{1-2i})} = \frac{1}{3(z-1)}\frac{1}{1+\frac{1-2i}{z-1}} = \frac{1}{3(z-1)} \sum_{n=0}^{\infty}\left(\frac{1-2i}{z-1}\right)^{n} $$
Which converges for $|1-2i|=\sqrt{5}<|z-1|.$
So, as $\sqrt{2}<\sqrt{5}$, we have that the expansion;
$$ f(z)= \frac{1}{3(z-1)}\sum_{n=0}^{\infty}\left(\frac{1+i}{z-1}\right)^{n} - \frac{1}{3(z-1)} \sum_{n=0}^{\infty}\left(\frac{1-2i}{z-1}\right)^{n} $$ converges for $\sqrt{5}<|z-1|<\infty$.