Question

The value of complex intergral,
$I=\underset{c}{\oint }\frac{z+1dz}{z^3+z^2+(z+1)}dzisk\pi iwherecisacircle|z|=2$

Then the value of k is __________

• A. 0

Answer 1

The correct option is A 0


$I=\int \frac{(z+1)dz}{z^3+z^2+z+1}$
$=\int \frac{(z+1)dz}{(z+i)(z-i)(z+1)}$
$=\int \frac{dz}{(z+i)(z-i)}=\int f\left(z\right)dz$

So, f(z) has singular points $z=\pm i$ which lies inside c,
Now,

$\underset{{}_{z\to i}}{R}esf\left(z\right)=\underset{z\to i}{lim}\frac{z+i}{(z+i)(z-i)}=\frac{-1}{2i}=\frac{i}{2}...\left(i\right)$

$and\underset{{}_{z\to i}}{R}esf\left(z\right)=\underset{z\to i}{lim}\frac{z-i}{(z+i)(z-i)}=\frac{1}{2i}=\frac{-i}{2}...\left(ii\right)$

$So,I=2\pi i[\frac{i}{2}-\frac{i}{2}]=0$

$\therefore k=0$