Find the series expansion of $f(z)=\frac{4}{(z-1)(z+3)}$ around $z_0=-1$. I have a homework problem that states: Let $$f(z)=\dfrac{4}{(z-1)(z+3)}.$$ Then $f$ has a power series expansion at each point $z_0 \in \Bbb{C}\backslash\{ 1,-3\}.$ Give a formula for the radius of convergence of the series at $z_0 \neq 1, -3.$ Find the series expansion for $z_0=-1.$
Here's what I've done so far. I know that
$$\dfrac{4}{(z-1)(z+3)} = \dfrac{1}{z-1}-\dfrac{1}{z+3}=\dfrac{-1}{1-z}-\dfrac{1}{3(1-[\frac{z}{3}])}=-\sum_{n=0}^\infty z^n -\dfrac{1}{3}\sum_{n=0}^\infty \left(-\frac{z}{3}\right)^n.$$
Now, I know the radius of convergence will be the smaller of the two radii. The series on the left has a R.O.C. of $|z|<1$ and the series on the right has a R.O.C. of $|z|<3.$ So the R.O.C. must be $|z|<1.$
However, I am stuck on how to rewrite this series around $z_0 = -1.$ I'm assuming that I can just use the formula for Taylor series expansion and just take a bunch of derivatives. But this seems tedious. I'm certain there's a way to just rewrite the current summations... but what is that method? I certainly can't just replace $z$ with $z+1.$ ...can I?
Any help would be appreciated. Thanks.
 A: No, you definitely can't just replace $z$ with $z+1$.  However, there is something else that you can do here.
Start from what you already did:
$$
\frac{4}{(z-1)(z+3)}=\frac{1}{z-1}-\frac{1}{z+3}.
$$
Now, try rewriting these as
$$
\frac{1}{z-1}-\frac{1}{z+3}=\frac{1}{(z+1)-2}-\frac{1}{(z+1)+2}=-\frac{1}{2-(z+1)}-\frac{1}{2+(z+1)}.
$$
Why is this helpful?  Because now, you can note that
$$
-\frac{1}{2-(z+1)}-\frac{1}{2+(z+1)}=-\frac{1}{2}\cdot\frac{1}{1-\frac{z+1}{2}}-\frac{1}{2}\cdot\frac{1}{1-(-\frac{z+1}{2})},
$$
and use $\sum_{n=0}^{\infty}w^n=\frac{1}{1-w}$, with $w=\frac{z+1}{2}$ for the first sum and $w=-\frac{z+1}{2}$ for the second.
Radius of convergence also follows from this immediately: remember that $\sum_{w=0}^{\infty}w^n$ converges if $\lvert w\rvert<1$ and diverges if $\lvert w\rvert>1$, and plug in your definitions for $w$ for each series.
A: sometimes it is useful to take account of some kindness on the part of the person who has set the question allowing a short-cut. this does have educational value, because we should always be alert to possible symmetries. in this case you want a Taylor series in $(z-z_0) =  z+1$. if you plug in this value you will see that the function is (as you can see from Nicholas' argument):
$$
f(z+1) = \frac4{[z+1]^2-4} = -\frac1{1-[\frac{z+1}{2}]^2} \\
= -\sum_{n=0}^{\infty} \left(\frac{z+1}{2}\right)^{2n}
$$
