I want to find the Laurent series of the following function around the point $z=2$, $z$ is a complex number.
$f(z)= \frac {z^2-2z+5}{(z-2)(z^2+1)}$
After the fraction decomposition we received a new expression of my given function:
$f(z) = \frac {1}{z-2} + \frac {i}{z-i} - \frac {i}{z+i}$
So we need to sum up just last two terms (regular part of the LS), the first one is already a main part of the LS.
$\frac {i}{z+i} = \frac {i}{z+i -2 +2} = ... = \frac {i}{i+2} \frac {1}{1 - \frac {z-2}{-i-2}} =\frac {i}{i+2} \sum_{n} (-1)^n(z-2)^n (\frac {1}{i+2})^n = \sum_{n} i (-1)^n(z-2)^n ( \frac {1}{i+2})^{n+1}$
$\frac {i}{z-i} = \frac {i}{z-i -2 +2} = ... = \frac {i}{2-i} \frac {1}{1 - \frac {z-2}{i-2}} =\frac {i}{2-i} \sum_{n} (-1)^n(z-2)^n (\frac {1}{2-i})^n = \sum_{n} i (-1)^n(z-2)^n ( \frac {1}{2-i})^{n+1}$
Altogether:
$f(z)= \frac {z^2-2z+5}{(z-2)(z^2+1)} = \frac {1}{z-2} + \sum_n i(-1)^n(z-2)^n \frac {(2+i)^{n+1}-(2-i)^{n+1}}{5^{n+1}} $
But my textbook says:
$f(z)= \frac {z^2-2z+5}{(z-2)(z^2+1)} = \frac {1}{z-2} + \sum_n (-1)^n(z-2)^n \frac {(2+i)^{n+1}-(2-i)^{n+1}}{5^{n+1}} $ (missing $i$)
I tried to calculate it many times, so did I do some wrong assumptions while calculating because I don't think I just forgot somewhere to divide that $i$.
Is this mathematically correct, or do I have to prove something to use the Laurent series properly?