JEE Advanced 2013 Paper 2Question 1
Area of $S=$
Complex Numbers
Solution
$\operatorname{Im}\frac{z-(1-\sqrt3 i)}{1-\sqrt3 i}>0$: the line through $1-\sqrt3 i$ with direction $1-\sqrt3 i$ passes through $0$, namely $y=-\sqrt3x$, and $S_2$ is the side containing $(1,0)$. With $x>0$ and $|z|<4$, $S$ is the sector of radius $4$ from angle $-\frac\pi3$ to $\frac\pi2$ (angle $\frac{5\pi}6$): area $=\frac12\cdot16\cdot\frac{5\pi}6=\frac{20\pi}3$.