Expand complex function into series (factoring not working)

Question

$\frac{z}{z^2 + 9}$, I need to turn into series, where z = complex number. I have tried factoring denominator to get $\frac{1}{z} \cdot \frac{1}{(1 + \frac{9}{z^2})}$ and use binomial expansion. But solution gives answer in powers of $(\frac{z}{9})$. What am I doing wrong?

Answer 1

What about partial fraction?

$$\frac{z}{z^2 + 9} = \frac {1}{2} (\frac {1}{z+3i} + \frac {1}{z-3i})$$

$$\frac {1}{z+3i} = \frac {1}{3i} (\frac {1}{1+(\frac {z}{3i})})$$

Use geometric series for the above and similarly for the other part.

Answer 2

$$\frac{z}{z^2 + 9} = z (\frac{1}{z^2 + 9}) = \frac {z}{9} ( \frac {1}{1+(\frac {z^2}{9})})$$

$$ \frac {1}{1+(\frac {z^2}{9})} = \sum _0 ^ \infty (-1)^n (\frac {z^2}{9})^n $$