Hi, I could use some help on part (b) of the following question from my Complex Variables, book. I didn't know if this belonged in the Calculus forum or not.

Question 3.2.5

(a) Use the identity

$\displaystyle \frac1z = \frac1{(z+1)-1}$

to establish

$\displaystyle \frac1z = - \sum_{n=0}^\infty (z+1)^n \hspace{15 mm} \left|z+1 \right| < 1$

(b) Use the above identity to also establish

$\displaystyle \frac1{z^2} = \sum_{n=0}^\infty (n+1)(z+1)^n \hspace{15 mm} \left|z+1 \right| < 1$

Check the answer by differentiating the series from (a)

Part (a) was easy:

$\displaystyle \frac1z = \frac{1}{(z+1)-1} = \frac{-1}{1-(z-1)} = -\sum_{n=0}^\infty(z+1)^n$

The check in part (b) is easy, too. But I can't do part (b) using the identity given. I tried squaring the identity, but this is as far as I've gotten (essentially, I just eliminated the negative sign):

$\displaystyle \frac1{z^2} = \frac1{z*z} = \frac1{(z+1)-1}* \frac1{(z+1)-1} = \left(- \sum_{n=0}^\infty (z+1)^n \right)* \left(- \sum_{n=0}^\infty (z+1)^n \right) = \left( \sum_{n=0}^\infty (z+1)^n \right)^2$

I've also got this (but I don't know if it's worth anything, or if it is even correct):

$\displaystyle \ \sum_{n=0}^\infty (z+1)^n \right) * \left( \sum_{n=0}^\infty (z+1)^n$

$\displaystyle = \left[1 + (z+1) + (z+1)^2 + ... \right] * \sum_{n=0}^\infty (z+1)^n$

$\displaystyle = 1 * \sum_{n=0}^\infty (z+1)^n + (z+1) * \sum_{n=0}^\infty (z+1)^n + (z+1)^2 * \sum_{n=0}^\infty (z+1)^n + ...$

$\displaystyle = \sum_{n=0}^\infty (z+1)^n + \sum_{n=0}^\infty (z+1)^{n+1} + \sum_{n=0}^\infty (z+1)^{n+2} + ...$

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

I have answered part (a) and the check in part (b) of the following question. But I cannot figure out part (b)