Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : DEC 2014

Let $I=\int\dfrac {z+3}{z^2+2z+5}dz$

For singularity, $z^2+2z+5=0\\ \therefore z=-1\pm 2i$

let $\alpha =-1+2i$ & $\beta=-1-2i$

Case 1: Circle $|Z|=1$ has center $(0, 0)$ and radius=1. Here, $Z_0=\alpha (-1+2)$ lies outside the circle $|Z|=1$

By Cauchy’s integral theorem,

$$\int f(z).dz=0$$

$$\therefore \int\dfrac {z+3}{z^2+2z+5}.dz=0$$

Case 2: Circle $|Z+1-i|=2$ has center $(1,-1)$ and radius =2. Here, $Z_0=β(-1,-2)$ lies outside and $Z_0=α(-1,2)$ lies inside the circle $|Z+1-i|=2$

$Z_0=α$ is a simple pole.

$R_1$=Residue of $f(x)$ at $"z=\alpha "$

$$=\lim\limits_{z\to z_0} (z-z_0)\times f(z)$$

$=\lim\limits_{z\to\infty}(z-\alpha)\times \dfrac {z+3}{(z-\alpha)(z-\beta)}\\ =\dfrac {\alpha+3}{\alpha-\beta}\\ =\dfrac {-1+2i+3}{(-1+2i)-(-1-2i)}\\ =\dfrac {2+2i}{4i}$

By Cauchy’s Residue Theorem,

$$\int f(z).dz=2\pi i [R_1+R_2]$$

$\therefore \int \dfrac {z+3}{z^2+2z+5}.dz\\ =2\pi i [\dfrac {2+2i}{4i}]\\ =\pi[1+i]$