Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : MAY 2015

Circle,$ |z-2-i|=2$ has center $(2,1)$ and radius=2

Let $I=\int\dfrac {z+2}{z^3-2z^2}dz$

For singularity

$$z^3-2z^2=0$$

$\therefore z^2(z-2)=0\\ \therefore z=0 \space \space or \space \space z=2$

$Z_0=0$ lies outside while $Z_0=2$ lies inside the given circle

$\therefore $ The given integral is not analytic at $z=2$

Now $I=\int\dfrac {\frac {(z+2)}{z^2}}{z-2}dz$

let $f(z)=\dfrac {z+2}{z^2}$

$\therefore I=2\pi i\space f(z_0)$ [cauchy's integral formula

$$=2\pi i\times \dfrac {z_0+2}{(z_0)^2}$$

$=2\pi i\times \dfrac {2+2}{2^2}\\ =2\pi i$