Exercise 7.26: Finding some series

Find power series for each of the following functions:

a) $\displaystyle \frac{1}{1+4z}$

b) $\displaystyle \frac{1}{3-z/2}$

c) $\displaystyle \frac{z^2}{(4-z)^2}$

Describe their domains of convergence.

For a) I have $$\frac{1}{1-z} = \sum_{k=0}^{\infty} z^k \implies \frac{1}{1-(-4z)} = \sum_{k=0}^{\infty} (-4z)^k = \sum_{k= 0}^{\infty}(-4)^k z^k$$

The domain of convergence can be found by evaluating all values for which $|-4z| < 1$ which is $|z| < \frac{1}{4}$, thus the domain of convergence is ${z:|z| < \frac{1}{4} }$, which is an open disc of radius $\frac{1}{4}$ centered at the origin.

For b.) $$\frac{1}{3-\frac{z}{2}}=\frac{-1}{3}*\frac{1}{\frac{z}{6}-1}=$$
$$\frac{-1}{3{\frac{z}{6}}}*\frac{1}{1-\frac{6}{z}}$$
$$=\frac{-2}{z}*\frac{1}{1-\frac{6}{z}}=\frac{-2}{z}\sum_{0}^{\infty}\frac{6^n}{z^n}=-2\sum_{0}^{\infty}\frac{6^n}{z^{n+1}}$$
I believe the domain of convergence would be $|z|>6.$

@opernie Thus is interesting and I liked it. Somehow, you've come up with a Laurent series, though. I think it would be nice to give some thought to the domain of convergence.