Pointers and help - solving an equation

I want to solve a differential equation, the exercise being, find two different independent solutions. (I don't know the proper English terms for this.)

$\displaystyle z^2 w'' - (z^2 + z)w' + (z+1)w = 0$

I start out with $\displaystyle \displaystyle w = \sum_0^\infty a_n z^{n+m}$. Differentiate to match each term in the equation and shove it in.

$\displaystyle \displaystyle \sum_0^\infty (n+m)(n+m-1)a_n z^{n+m} - \sum_0^\infty (n+m)a_n z^{n+m+1}-\sum_0^\infty (n+m)a_n z^{n+m} + \sum_0^\infty a_n z^{n+m+1} + \sum_0^\infty a_n z^{n+m} = 0$

$\displaystyle \displaystyle \sum_0^\infty [(n+m)(n+m-1) -(n+m) - 1]a_n z^{n+m} = \sum_0^\infty [(n+m) -1]a_n z^{n+m+1}$

Recursive series.

$\displaystyle \displaystyle (m+n-1)^2 a_n = (n+m -2)a_{n-1} $

Now I'm starting to get wobbly legs. I set $\displaystyle n = 0, a_0 = 1$ to see if I can pinpoint $\displaystyle m$. $\displaystyle (m-1)^2 = 0 \implies m = \pm 1 $

$\displaystyle m = -1 \implies a_n = \frac{(n-3)}{(n-2)^2} a_{n-1}$ results in a series $\displaystyle a_n = \infty, n \geq 2$. Can I render a solution from that even if it diverges?

$\displaystyle \displaystyle m = 1 \implies a_n = \frac{(n-1)}{n^2} a_{n-1}$ results in a series $\displaystyle a_n = 0, n \geq 1$ that would imply that one solution is w = z. Or..?

Now I've sort of played all my cards here. What if I state a solution w = zu and solve the equation..? Or is there a simpler way to render an other solution here?