JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Complex Numbers

$S_1$ is the interior of a circle of radius 4 centered at origin: $|z|<4$.

For $S_2$, let $z=x+iy$:

$\mathrm{Im}\left[\frac{(x-1)+i(y+\sqrt{3})}{1-i\sqrt{3}}\right]>0$

Multiplying numerator and denominator by the conjugate:

$\frac{[(x-1)+i(y+\sqrt{3})][1+i\sqrt{3}]}{4}$

The imaginary part is:

$\frac{\sqrt{3}(x-1)+(y+\sqrt{3})}{4}>0$

$\sqrt{3}x+y>0$

This is the half-plane above the line $y=-\sqrt{3}x$ (which passes through origin with slope $-\sqrt{3}$, making angle $150°$ with positive $x$-axis).

$S_3$ is the right half-plane: $x>0$.

$S$ is the intersection: a sector of the circle in the first and second quadrants, bounded by the positive $x$-axis and the line $y=-\sqrt{3}x$.

The angle of this sector is from $0°$ to $150°$, which is $5\pi /6$ radians.

Area = $\frac{1}{2}{r}^2\theta =\frac{1}{2}(4{)}^2\cdot \frac{5\pi }{6}=\frac{20\pi }{3}$