<s>[INST] Laurent Series Expansion of $f(z) = \frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz}$ at $z=0$ For my function $$f(z) = \frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz}$$ I have determined that it contains $3$ poles at $z=0,z=-a,z=-\frac{1}{b}$
I believe these to be the only singularities.
I found them by factoring the denominator
$$f(z) = \frac{bz^2+2z+b}{z(bz^2+(1+b^2)z+b)}$$
Then clearly there is a pole at $z=0$ and the remaining singularities are found by solving the quadratic
$$bz^2+(1+b^2)z+b = 0$$
$z = \frac{-(1+b^2) + \sqrt{(1+b^2)^2-4(b)(b)}}{2b}$ and $z =\frac{ -(1+b^2) - \sqrt{(1+b^2)^2-4(b)(b)}}{2b}$
This simplifies to
$z = \frac{-1-b^2 -b^2 +1}{2b}$ and $z=\frac{-1-b^2+b^2-1}{2b}$
$z = \frac{-2b^2}{2b}$ and $z= \frac{-2}{2b}$
So,
$z=-b$ and $z=-\frac{1}{b}$
I want to Laurent Expand these singularities and determine the radius of convergence for each singular point.
All help is appreciated. I want to focus on the $z=0$, once I have clarified the process I will be able to expand the remaining myself.
 [/INST] Before factoring take notice of the similarity of the coefficients in the numerator and denominator. This suggests that there is a short cut that can be taken before attempting to partial fraction that would make our lives easier.
$$\frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz} = \frac{bz^2+(1+b^2)z+b+(1-b^2)z}{bz^3+(1+b^2)z^2+bz} = \frac{1}{z}+\frac{1-b^2}{bz^2+(1+b^2)z+b}$$
Now partial fractions is automatic
$$\frac{1}{z}+\frac{1-b^2}{(z+b)(bz+1)} = \frac{1}{z}+\frac{1}{z+b}-\frac{1}{bz+1}$$
$z=0$ won't actually provide illumination on what to do for the other singularities, so instead I will show the procedure for $z=-b$. Every term needs to be centered at $z=-b$ like so
$$\frac{1}{(z+b)-b}+ \frac{1}{z+b} + \frac{1}{b(z+b)+1-b^2} = -\frac{1}{b}\frac{1}{1-\frac{(z+b)}{b}}+ \frac{1}{z+b} + \frac{1}{1-b^2}\frac{1}{\frac{b(z+b)}{1-b^2}+1}$$
The scaling is done because it is important to have the $+1$ to apply geometric series. Now we assume the quantity $z+b$ is small so we can use the Taylor expansion
$$-\frac{1}{b}\sum_{n=0}^\infty \frac{(z+b)^n}{b^n} + \frac{1}{z+b} + \frac{1}{1-b^2}\sum_{n=0}^\infty (-1)^n\left(\frac{b}{1-b^2}\right)^n(z+b)^n$$
The case $b=1$ needs to be handled separately since in that case $b=\frac{1}{b}$
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