Question

Let $S=S_1\cap S_2\cap S_3,$ where
$S_1=\{z\in C:|z|<4\},$ $S_2=\{z\in C:\text{Im}\left(\frac{z-1+\sqrt{3}i}{1-\sqrt{3}i}\right)>0\}$ and $S_3=\{z\in C:\text{Re}z>0\}$

Area of $S=$

• A. $\frac{10\pi }{3}$
• B. $\frac{20\pi }{3}$
• C. $\frac{16\pi }{3}$
• D. $\frac{32\pi }{3}$

Answer 1

The correct option is B $\frac{20\pi }{3}$

Let $z=x+iy$
$S_1:x^2+y^2<16$

$S_2:\text{Im}\left(\frac{(x-1)+i(y+\sqrt{3})}{1-i\sqrt{3}}\right)>0$
$\Rightarrow \text{Im}(\frac{(x-1)+i(y+\sqrt{3})}{1-i\sqrt{3}}\times \frac{1+i\sqrt{3}}{1+i\sqrt{3}})>0$
$\Rightarrow \sqrt{3}(x-1)+y+\sqrt{3}>0\Rightarrow \sqrt{3}x+y>0$

$S_3:x>0$
$S=S_1\cap S_2\cap S_3$
Shaded area represents $S.$



Area of $S$
$=\frac{1}{2}{r}^2\theta =\frac{1}{2}\times 4^2\times \frac{5\pi }{6}=\frac{20\pi }{3}$