Solving two variables with one equation

Question

I have been trying to solve the following equation, but I am still stuck after trying many different methods.

I have been given this equation to solve:

$$(1 + z^{-1})^4 (a + bz^{-1} + az^{-2}) - (1 - z^{-1})^4 (a - bz^{-1} + az^{-2}) = 2z^{-d}$$

I have simplified it down to $$a(4z^{-1} + 8z^{-3} + 4z^{-5}) + b(z^{-1} + 6z^{-3} + z^{-5}) = z^{-d}$$

The goal is to find the values for '$a$' and '$b$' for ANY value for '$d$'. I don't think I made any mistakes up to the simplifications.

EDIT: I thought of having $a = 1$ and $b = 4$, but this would give me

$$ -16z^{-3} = z^{-d} $$

This is not exactly equal since it should be $1$ in front, instead of $-16$

Answer 1

The answer would be:

a = -1/16

b = 1/4

That would give us $z^{-3}$, so d = 3

Answer 2

Presumably you are given a value for $z$ as well, but you don't have enough information. You can solve for either $a$ or $b$, but not both, with one equation.

Answer 3

If you multiply both sides by $z^6$ you get \begin{equation} (z+1)(az^2+bz+a)-(z-1)(az^2-bz+a)=2z^{6-d} \end{equation} which simplifies to \begin{equation} (a+b)z^2+a=z^{6-d} \end{equation}

You may solve this for $a$, $b$, $d$ or $z$ in terms of the other three variables.