Answer to Question #216746 in Complex Analysis for subodh gupta

Question

Expand f(z)=𝑧+3/𝑧(𝑧2−𝑧−2) in power of z where

a) |𝑧|<1

b) 1<|𝑧|<2

c) |𝑧|>2

Answer 1

$f=z+\dfrac{3}{z(z^2-z-2)}=z+\dfrac{3}{z(z+1)(z-2)}=z+\dfrac{1}{z}\left(\dfrac{1}{z-2}-\dfrac{1}{z+1}\right)\\\implies z+\dfrac{1}{z(z-2)}-\dfrac{1}{z(z+1)}=z+\dfrac{1}{2(z-2)}-\dfrac{1}{2z}-\dfrac{1}{z}+\dfrac{1}{z+1}\\=z-\dfrac{3}{2}z^{-1}+\dfrac{1}{4(\frac{z}{2}-1)}+\dfrac{1}{z+1}$

We remind the formula: $a+ar+...+ar^n=a\left(\frac{1-r^{n+1}}{1-r}\right)$, $|r|<1.$ 

We take the limit and get: $a+ar+...+ar^n+...=\frac{a}{1-r}$.

a). For $|z|<1$ we use the formula and get:

$\implies z-\dfrac{3}{2}z^{-1}+\dfrac{1}{4(\frac{z}{2}-1)}+\dfrac{1}{z+1}\\=z-\dfrac{3}{2}z^{-1}-\dfrac14\left(1+\dfrac{z}{2}+\dfrac{z^2}{2^2}+\dfrac{z^3}{2^3}...\right)+\left(1-z+z^2-z^3+z^4+...\right)$

b). We point out that $\frac{1}{2}<\frac{1}{|z|}<1$. We rewrite the equality as: 

$\implies z-\dfrac{3}{2}z^{-1}-\dfrac{1}{4(1-\frac{z}2)}+\dfrac{1}{z(1+\frac{1}{z})}\\=z-\dfrac{3}{2}z^{-1}-\dfrac{1}{4}\left(1-\dfrac{z}{2}+\dfrac{z^2}4-...\right)+\dfrac{1}{z}\left(1-\dfrac{1}{z}+\dfrac{1}{z^2}+...\right)$

c). We have: $\frac{1}{2}>\frac{1}{|z|}$. Thus, we get: 

$\implies z-\dfrac{3}{2}z^{-1}+\dfrac{1}{2(1-\frac{2}{z})}+\dfrac{1}{z(1+\frac{1}{z})}\\=z-\dfrac{3}{2}z^{-1}+\dfrac12\left(1-\dfrac{2}{z}+\dfrac{4}{z^2}+...\right)+\dfrac{1}{z}\left(1-\dfrac{1}{z}+\dfrac{1}{z^2}+...\right)$