Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : DEC 2015

Now equating denominator with zero

$$Z(z-1) (z-2)= 0$$

We get $z = 0$ or $z = 1$ or $z = 2.$

The circle $|z|=1.5$ which has centre at origin and radius $= 1.5$ the point $z = 2$ lies outside the circle and $z = 0$ and $z = 1$ lies inside circle.

Now by Cauchy integral formula at $z = 0. f(z)$ will be $\dfrac {1-z}{(z-1)(z-2)}$ putting $z = 0.$

$$\phi_c\dfrac {f(z)}zdz=2\pi i \space f(z_0)$$

$=2\pi i(\dfrac 12)\\ =\pi i$

Where $f(z) =\dfrac {1-2z}{(z-1)(z-2)}$ & $z_0 = 0$

Now applying the same for for $z = 1$

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

By Cauchy integral formula

$$\phi_c\dfrac {1-2z}{z(z-1)(z-2)}=\phi \dfrac {\frac {1-2z}{z(z-2)}dz}{(z-1)}$$

$=\phi\dfrac {f(z)}{(z-1)}dz\\ 2\pi i\space f(z_0)\\ where \space \space f(z)=\dfrac {1-2}{z(z-2)} \space and \space z_0=1\\ =2\pi i\times \dfrac {(1-2)}{1(1-2)}\\ =2\pi i$

Now adding both value we get

$$\phi_c\dfrac {1-2z}{z(z-1)(z-2)}dz=\pi i+2\pi i=3\pi i\space \space\space for |z|=1.5$$