This one seems a lot harder...
Find the power series about the origin of:
z^2/(4-z)^2
You only need to bother calculating the series for 1/(4 - z)^2 (why?).
Did you bother to read the link I gave in my reply here: http://www.mathhelpforum.com/math-he...er-series.html
Now look for the pattern in $\displaystyle f^{n}(0)$ where n = 0, 1, 2, 3, ....
If you need more help, please say where you are stuck.
If you know the power series of $\displaystyle \frac{1}{(1-z)^2}$ you are essentially done. Consider:
$\displaystyle \frac{1}{(1-z)^2}=\frac{d}{dz}\frac{1}{1-z}=\frac{d}{dz}\sum_{n=0}^\infty z^n=\sum_{n=1}^\infty n z^{n-1}=\sum_{n=0}^\infty (n+1)z^n$
To know this is perhaps generally useful, but now let's solve your original problem:
$\displaystyle \frac{z^2}{(4-z)^2}=\left(\frac{z}{4}\right)^2\cdot\frac{1}{\lef t(1-\left(\frac{z}{4}\right)\right)^2}=\left(\frac{z}{ 4}\right)^2\cdot\sum_{n=0}^\infty(n+1)\left(\frac{ z}{4}\right)^n=\ldots$