For all possible values of $x,y$ find radius of convergence of following function as $$f(z)=\frac{-x+(x-y)z}{z^2+z-1}$$
Till I am only encountered example for which I had been provided power series and I have to find its radius of convergence
I know that rational function can be written as power series, also if denominator tends to zero implies that series must diverge at that point.
My attempt:
If $z$ take value root of denominator then it diverges .
$$\alpha_1=\frac{-1+\sqrt5}2, \quad\alpha_2=\frac{-1-\sqrt5}2$$
But How to decide radius of convergece for all possible values of $x$ and $y$?
Any Help will be appreciated
Partial fraction decomposition gives
$$\frac{-x+(x-y)z}{(z - \alpha_1)(z - \alpha_2)} = -\frac{(1-\alpha_1)x + \alpha_1 y}{\sqrt{5}}\frac1{z - \alpha_1} + \frac{(1-\alpha_2)x + \alpha_2 y}{\sqrt{5}}\frac1{z - \alpha_2}$$
The radius of convergence of $$\frac1{z-\alpha_1} = -\frac1{\alpha_1} \frac1{1-\frac{z}{\alpha_1}} = -\frac1{\alpha_1} \sum_{n=0}^\infty \frac{z^n}{\alpha_1^n}$$
is $\alpha_1$ and similarly $$\frac1{z-\alpha_2} = -\frac1{\alpha_2} \sum_{n=0}^\infty \frac{z^n}{\alpha_2^n}$$ has radius of convergence equal to $-\alpha_2$.
The radius convergence of the sum is $\min\{\alpha_1, -\alpha_2\} = \alpha_1$, unless the first sum disapears (if anf only if $(1-\alpha_1)x + \alpha_1 y = 0$), in which case it is $-\alpha_2$.