Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 08

Year : DEC 2014

$$f(z)=\dfrac 2{(z-1)(z-2)}$$

$\therefore f(z)=\dfrac 2{z-2}-\dfrac 2{z-1}$ (partial fraction)

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

Obviously $|z| \lt 2$

$\therefore |z| \lt 1$ and $|\dfrac z2| \lt 1$

$$\therefore f(z)=\dfrac 2{2(\frac z2-1)}-\dfrac 2{(z-1)}$$

$=-[1-\dfrac z2]^{-1}+2(1-z)^{-1}\\ =-[1+\dfrac z2+\dfrac {z^2}{2^2}+\dfrac {z^3}{2^3}+....] +2[1+z+z^2+z^3+...]$

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

$\therefore 1\lt|z|$ and $|z| \lt 2$

$\therefore |\dfrac 1z| \lt1$ and $|\dfrac z2| \lt1$

$$\therefore f(z)=\dfrac 2{2(\frac z2 -1)}-\dfrac 2{z(1-\frac 1z)}$$

$=-[1-\dfrac z2]^{-1} -\dfrac 2z[1-\dfrac 1z]^{-1}\\ =-[-1+\dfrac z2+\dfrac {z^2}{2^2} + \dfrac {z^3}{2^3}+....] -\dfrac 2z[1+\dfrac 1z+\dfrac 1{z^2}+\dfrac 1{z^3} +... ]\\ =-[1+\dfrac z2+\dfrac {z^2}{2^2} +.... ]-2[\dfrac 1z+\dfrac 1{z^2}+\dfrac 1{z^3}+...] $

Case 3 : for $|z| \gt 2$

Obviously $|z| \gt 1$

$\therefore 1\lt |z|$ and $2\lt|z|$

$\therefore |\dfrac 1z| \lt1$ and $|\dfrac 2z| \lt 1$

$$\therefore f(z)=\dfrac 2{z(1-\frac 2z)}-\dfrac 2{z(1-\frac 1z)}$$

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