Complex number, series Show that 
$$\frac{1}{z^2}=1+\sum_{n=1}^\infty (n+1)(z+1)^n$$ when $|z+1|<1$
I'm having problems to resolve this type of exercise since my book has virtually no exercises of this type, these expressions are based on Maclaurin series?
 A: make the substitution $w=z+1$ so $|z+1|  \lt 1 $ means $|w| \lt 1$.
now note that:
$$
\frac1{(1-w)^2} = \sum_{k=0}^\infty \binom{-2}{k} (-w)^k = 1 +2w+3w^2+...
$$
A: Write 
$$\frac{1}{z^2}=(1-(z+1))^{-2}$$
and use the binomial theorem.
$$=\sum_{n=0}^{\infty}\binom{-2}{n}(-1)^n(z+1)^n$$
Now $$\binom{-2}{n}(-1)^n=\frac{(-2)(-3)(-4)...(-(n+1))}{n!}(-1)^n=(n+1).$$
A: This more like the Taylor Expansion about $z=-1$ of the function
$$
f(z)=\frac{1}{z^2}-1
$$
Computing the first derivatives you will see that 
$$
f^{(0)}(-1)=0; \ \ f^{(1)}(-1)=2; \ \ f^{(2)}(-1)=6=3\cdot2;  \ \ f^{(2)}(-1)=14=4\cdot6
$$
So you deduce that 
$$
f^{(n)}(-1)=(n+1)n!
$$
By Taylor
$$
f(z)=\sum_{n=0}^{\infty}\frac{f^{(n)}(-1)}{n!}(z-(-1))^n=\sum_{n=1}^{\infty}(n+1)(z+1)^n
$$
FInally by the ratio test, since $\lim_{n\to \infty}\frac{n+1}{n+2}=1$, you conclude that the radius of convergence of the last power series is $1$, that is the convergence is for all $z$ such that $|z+1|<1$
A: If $$f(z)=\frac{1}{z^2}$$
then
$$f^{(n)}(z)=(-1)^{n}(n+1)!z^{-{n+2}}$$
so that $$f^{(n)}(z=-1)=(n+1)!$$
Thus, 
$$\begin{align}
f(z)&=\sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!}(z+1)^n\\\\
&=\sum_{n=0}^{\infty} (n+1)\,(z+1)^n
\end{align}$$
