JEE Advanced 2013 · Paper 2, Mathematics question 49

S₂. With $w=z-(1-\sqrt3i)$ and $1-\sqrt3i=2e^{-i\pi/3}$: $\operatorname{Im}\left(we^{i\pi/3}\right)>0$, i.e. $\arg w\in\left(-\frac\pi3,\frac{2\pi}3\right)$. Since $1-\sqrt3i$ lies on the line $y=-\sqrt3x$ through the origin, S₂ is the half-plane $y>-\sqrt3x$.

S. The disc $|z|<4$ cut by $x>0$ and $y>-\sqrt3x$: a sector from $-\frac\pi3$ to $\frac\pi2$, angle $\frac{5\pi}6$.

Area $\frac12\cdot16\cdot\frac{5\pi}6=\dfrac{20\pi}3$.

(B)