Find the expansion of $\frac {1}{z^2+3}$ valid for $|z-1|<2$

Question

Find the expansion of $$\frac {1}{z^2+3}$$ valid for $|z-1|<2$.

Should I start with considering $z-1=u$ ? It gets messy that way. I tried reorganizing the denominator of $f(z)$ but that didn't work too.

Answer 1

Because $|z-1|<1<2$, you can just consider that $u=z-1$ is a small variable, $|u|<1$.

$$\frac{1}{z^2+3} = \frac{1}{(z-1)^2 + 2(z-1) + 4}=\frac{1}{4}\frac{1}{1+u/2+u^2/4}$$

Mathematica 7.0 showed that

$$\frac{1}{u^2/4 + u/2 + 1}=\left(1-\frac{u}{2}\right)\sum_0^{\infty}\left(\frac{u}{2}\right)^{3k}$$

This leads us to write ($v=u/2$),

$$\frac{1}{1+v+v^2}=\frac{1-v}{1-v^3}=(1-v)\sum_0^{\infty}v^{3k}$$

This also converges for $|v|<1$ (i.e.,$|z-1|<2$).

mike