Fourier series identity

I need to prove that $\dfrac{a \sin(bx)}{1 - 2a \cos(bx) + a^2} = \sum_{n=1}^\infty a^n \sin(nbx)$ where $|a| < 1$.

It seems that this can be proved by using Euler's formula identities for $\cos(bx)$ and $\sin(bx)$ and substituting $z = e^{ibx}$. From that, I get $$ \dfrac{a \sin(bx)}{1 - 2a \cos(bx) + a^2} = \dfrac{a (z - 1/z)}{2i(1 - a(z + 1/z) + a^2)}$$ $$= \dfrac{a (z^2 -1)}{2i(z - a(z^2 +1) + za^2)}$$ $$= \dfrac{a (z^2 -1)}{-2ai(z^2 - z(1-a^2)/a + 1)}$$

I don't understand what I am supposed to do next.

$$\displaystyle\sum_{n=0}^{\infty} a^{n} \sin(nbx) = \text{Im} \sum_{n=0}^{\infty} (ae^{ibx})^{n} = \text{Im} \ \frac{1}{1-ae^{ibx}}$$