Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 08

Year : MAY 2014

$$\dfrac {(z-2)(z+2)}{(z+1)(z+4)}=\dfrac {z^2-4}{z^2+5z+4}$$ ![enter image description here][1] $$\therefore \dfrac {(z-2)(z+2)}{(z+1)(z+4)}=1+\dfrac {(-5z-8)}{z^2+5z+4}$$

$=1-\dfrac 1{z+1}-\dfrac 4{z+4}$ [By partial fractions]

Case 1: for $|z| \lt 1$

Obviously .$|z| \lt 4$

$\therefore |z| \lt 1$ and $|\dfrac z4| \lt1$

$$\therefore f(z)=1-\dfrac 1{1+z}-\dfrac 4{4(z/4+1)}$$

$=1-(1+z)^{-1}-(1+\dfrac z4)^{-1}\\ =1-[1-z+z^2-z^3]-[1-\dfrac z4+\dfrac {z^2}{4^2}-\dfrac {z^3}{4^3}]\\ =[z-z^2+z^3-z^4] - [1-\dfrac z4+\dfrac {z^2}{4^2}-\dfrac {z^3}{4^3}]$

Case 2 : for $1 \lt |z| \lt 4$

$\therefore 1 \lt |z|$ and $|z| \lt 4$

$\therefore |\dfrac 1z|\lt 1$ and $|\dfrac z4|\lt1$

$$f(z)=1-\dfrac 1{z(1+\dfrac 1z)}-\dfrac 4{4(z/4+1)}$$

$=1-\dfrac 1z[1+\dfrac 1z]^{-1}-[1+\dfrac z4]^{-1}\\ =1-\dfrac 1z(1-\dfrac 1z+\dfrac 1{z^2}-\dfrac 1{z^3})-[1-\dfrac z4+\dfrac {z^2}{4^2}-\dfrac {z^3}{4^3}]\\ =-[\dfrac 1z+\dfrac 1{z^2}-\dfrac 1{z^3}-\dfrac 1{z^4}]+ [\dfrac z4-\dfrac {z^2}{4^2}+\dfrac {z^3}{4^3}-\dfrac {z^4}{4^4}]$

Case 3: for $|z|\gt 4$

Obviously $,|z| \gt 1$

$\therefore 1 \lt |z|$ and $4 \lt |z|$

$\therefore |\dfrac 1z|\lt1$ and $|\dfrac 4z| \lt 1$

$$f(z)=1-\dfrac 1{z(1+\frac 1z)}-\dfrac 4{z(1+\frac 4z)}$$

$=1-\dfrac 1z(1+\dfrac 1z)^{-1}-\dfrac 4z(1+\dfrac 4z)^{-1}\\ =1-\dfrac 1z(1-\dfrac 1z+\dfrac 1{z^2}-\dfrac1{z^3})-\dfrac 4z(1-\dfrac 4z+\dfrac {4^2}{z^2}-\dfrac {4^3}{z^3})\\ =1- [\dfrac 1z+\dfrac 1{z^2}-\dfrac1{z^3}-\dfrac 1{z^4}]-\dfrac 4z[\dfrac 4z-\dfrac {4^2}{z^2}+\dfrac {4^3}{z^3}-\dfrac {4^4}{z^4}]$