Question #216746

Expand f(z)=๐‘ง+3/๐‘ง(๐‘ง2โˆ’๐‘งโˆ’2) in power of z where

a) |๐‘ง|<1

b) 1<|๐‘ง|<2

c) |๐‘ง|>2


1
Expert's answer
2021-07-23T08:41:38-0400

f=z+3z(z2โˆ’zโˆ’2)=z+3z(z+1)(zโˆ’2)=z+1z(1zโˆ’2โˆ’1z+1)โ€…โ€ŠโŸนโ€…โ€Šz+1z(zโˆ’2)โˆ’1z(z+1)=z+12(zโˆ’2)โˆ’12zโˆ’1z+1z+1=zโˆ’32zโˆ’1+14(z2โˆ’1)+1z+1f=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+...+arn=a(1โˆ’rn+11โˆ’r)a+ar+...+ar^n=a\left(\frac{1-r^{n+1}}{1-r}\right)โˆฃrโˆฃ<1.|r|<1. 


We take the limit and get: a+ar+...+arn+...=a1โˆ’ra+ar+...+ar^n+...=\frac{a}{1-r}.


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

โ€…โ€ŠโŸนโ€…โ€Šzโˆ’32zโˆ’1+14(z2โˆ’1)+1z+1=zโˆ’32zโˆ’1โˆ’14(1+z2+z222+z323...)+(1โˆ’z+z2โˆ’z3+z4+...)\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 12<1โˆฃzโˆฃ<1\frac{1}{2}<\frac{1}{|z|}<1. We rewrite the equality as: 


โ€…โ€ŠโŸนโ€…โ€Šzโˆ’32zโˆ’1โˆ’14(1โˆ’z2)+1z(1+1z)=zโˆ’32zโˆ’1โˆ’14(1โˆ’z2+z24โˆ’...)+1z(1โˆ’1z+1z2+...)\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: 12>1โˆฃzโˆฃ\frac{1}{2}>\frac{1}{|z|}. Thus, we get: 


โ€…โ€ŠโŸนโ€…โ€Šzโˆ’32zโˆ’1+12(1โˆ’2z)+1z(1+1z)=zโˆ’32zโˆ’1+12(1โˆ’2z+4z2+...)+1z(1โˆ’1z+1z2+...)\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)

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS