QJEEM24Mae193: JEE Main 2024 Mathematics - Class 11 JEEM24Mae193 Solution: Let and . Then the area of the region is :

$S_1=\{z \in C:|z| \leq 5\}$

$S_2=\operatorname{Im}\left(\frac{z+1-\sqrt{3} i}{1-\sqrt{3} i}\right) \geq 0$

Take $z=x+i y$

$=\frac{x+i y+1-\sqrt{3} i}{1-\sqrt{3} i} \times \frac{1+\sqrt{3} i}{1+\sqrt{3} i}$

$\begin{aligned}= & \frac{x+i y+1-\sqrt{3} i+\sqrt{3} i x-\sqrt{3} y+\sqrt{3} i+3}{1+3} \\& =y+\sqrt{3} x \geq 0 \\S_3= & x \geq 0\end{aligned}$

$\begin{aligned}& y+\sqrt{3} x=0 \\& y=-\sqrt{3} x \\& \text { Slope }=-\sqrt{3} \\& 90+\theta=120^{\circ} \\& \theta=30^{\circ} \\& 90-\theta=60^{\circ}\end{aligned}$

In first quadrant we have angle $\frac{\pi}{2}$

so, total angle $90^{\circ}+60^{\circ}=150^{\circ}$

So, area $\frac{\pi r^2}{2 \pi} \times \frac{5 \pi}{6}=\frac{125 \pi}{12}$