Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 08

Year : DEC 2015

$$Let\space \dfrac 2{(z-1)(z-2)}=\dfrac a{(z-1)}+\dfrac b{(z+2)}\ $$

$Z = a (z -2) + b (z - 1)\\ When \space \space z = 1 \space \space 2 = -a \space \space a = -2 \\ When \space \space z = 2\space \space 2 = b\space \space b = 2\\ \therefore \dfrac 2{(z-1)(z-2)}=\dfrac {-2}{z-1}+\dfrac 2{z-2}$

Case 1: when $|z| \lt 1$ clearly $|z| \lt 2$

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

$=2(1-z)^{-1}-[1-(\dfrac z2)]^{-1}\\ =2[1+z+z^2+z^3...] -[1+\dfrac z2+\dfrac {z^2}4.....]$

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

$$f(z)=\dfrac {-2}{z(1-\dfrac 12)}-\dfrac 2{2(1-\dfrac z2)}$$

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

Case 3 : when $|z| \gt 2$ i.e. $\dfrac{|z|} 2 \gt 1$ i.e. $\dfrac 2{|z|} \lt 1$

$$f(z)=\dfrac {-2}z\dfrac1{(1-\dfrac 12)} +\dfrac 2z\dfrac 1{(1-\dfrac 2z)}$$

$=\dfrac {-2}z[1-\dfrac 12]^{-1}+\dfrac 2z[1-\dfrac 2z]^{-1}\\ =\dfrac {-2}z[1+\dfrac 12+\dfrac 1{3^2}....]+\dfrac 2z(1+\dfrac 2z+\dfrac {2^2}{z^2}...)$