Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 08

Year : MAY 2015

By partial fractions

$$F(z)=\dfrac {7z-2}{z(z-2)(z+1)}$$

$=\dfrac 1z-\dfrac 3{z+1}+\dfrac 2{z-2}$

Put $U=z+1\\ \therefore z=U-1\\ \therefore f(z)=\dfrac 1{U-1}-\dfrac 3{(U-1)+1}+\dfrac 2{(U-1)-2}\\ \dfrac 1{U-1}-\dfrac 3U+\dfrac 2{U-3}$

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

Obviously $|U| \lt 3$

$\therefore |U| \lt 1$ and $|\dfrac U3| \lt 1$

$$\therefore f(z)=\dfrac 1{1(1-U)}-\dfrac 3U+\dfrac 2{3(\frac U3)-1}$$

$=-(1-U)^{-1}-\dfrac 3U-\dfrac 23[1-\dfrac U3]^{-1}\\ =-(1+U+U^2+...) -\dfrac 3U-\dfrac 23[1+\dfrac U3+\dfrac {U^2}{3^2}+...]\\ =-[1+(z+1)+(z+1)^2+....] -\dfrac 3{(z+1)}-\dfrac 23[1+\dfrac {(z+1)}3+\dfrac {(z+1)^2}{3^2}+...]$

Case 2 : $1\lt |U|\lt 3$

$\therefore 1 \lt|U|$ and $|U|\lt 3$

$\therefore |\dfrac 1U| \lt1$ and $|\dfrac U3| \lt1$

$$\therefore f(z)=\dfrac 1{U(1-\dfrac 1U)}-\dfrac 3U+\dfrac 2{3(\dfrac U3-1)}$$

$=\dfrac 1U[1-\dfrac 1U]^{-1}-\dfrac 3U-\dfrac 23[1-\dfrac U3]^{-1}\\ =\dfrac 1U[1+\dfrac 1U+\dfrac 1{U^2}+...] -\dfrac 3U-\dfrac 23[1+\dfrac U3+\dfrac {U^2}{3^2}+....]\\ =[\dfrac 1U+\dfrac 1{U^2}+\dfrac 1{U^3}+...]-\dfrac 3U-\dfrac 23[1+\dfrac U3+\dfrac {U^2}{3^2}+...]\\ =\dfrac 1{(z+1)}+\dfrac 1{(z+1)^2}+\dfrac 1{(z+1)^3}+.... -\dfrac 3{(z+1)}-\dfrac 23(1+\dfrac {(z+1)}3+....)$

Case 3 : for $|U| \gt 3$

Obviously $|U|\gt1$

$\therefore 1\lt | U |$ and $3 \lt |U|$

$\therefore |\dfrac 1U| \lt 1$ and $|\dfrac 3U| \lt 1$

$$\therefore f(z)=\dfrac 1{U(1-\dfrac 1U)}-\dfrac 3U+\dfrac 2{U(1-\dfrac 3U)}$$

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