I have a following function:
$$G(z)=\sqrt{F(z)^2-1}$$
Where:
$$F=\frac{\left(1+\alpha^{*2}\right)-\left(1+\alpha^{2}\right)z}{2\left(\alpha^{*}-\alpha z\right)}$$
And $\alpha$ is a complex number.
I want to find Taylor series for $G(z)$ around $z=0$ explicitly. This is a part of a z-transform which I want to invert (of course, the original variable was $1/z$, I just changed it for convenience).
$$G(z)=\sum_{n=0}^\infty C_n z^n$$
Is there some closed form expression for the coefficients $C_n$ in this case? Or even better, a recurrence relation so I can program them?
We can kind of simplify:
$$F-1=\frac{(\alpha^{*}-1)^{2}-(\alpha-1)^{2}z}{2\left(\alpha^{*}-\alpha z\right)}$$
$$F+1=\frac{(\alpha^{*}+1)^{2}-(\alpha+1)^{2}z}{2\left(\alpha^{*}-\alpha z\right)}$$
So we get:
$$G=\frac{\sqrt{\left[(\alpha^{*}-1)^{2}-(\alpha-1)^{2} z\right]\left[(\alpha^{*}+1)^{2}-(\alpha+1)^{2} z\right]}}{2\left(\alpha^{*}-\alpha z\right)}$$
Or, even better:
$$G=\frac12 \left(\alpha^{*}-\alpha z\right)^{-1} \sqrt{a^2-2bz+c^2 z^2}$$
Where:
$$a^2=(\alpha^{*2}-1)^{2}$$
$$b=(\alpha^{*}\alpha -i\alpha+i\alpha^*-1)(\alpha^{*}\alpha+i\alpha-i\alpha^* -1)$$
$$c^2=(\alpha^{2}-1)^{2}$$
So I suppose we have a product of two known series. The square root can be expressed in terms of Gegenbauer polynomials.
Still, I would like to know the explicit terms for the whole function. Or a recurrence. Something easy to program.
Update: WA gave me a 3 term recurrence for the coefficients, so I suppose the problem is solved. I just need to implement and check that it works. If so, I will post an answer myself.