Let S_1=z C:|z| 5,S_2=z C:Im(z+1-√3i1-√3i) 0 and S_3=z C:Re(z) 0.

Question

Let ${S}_{1}={z\in C:|z|\le 5},{S}_{2}={z\in C:Im(\frac{z+1-\sqrt{3}i}{1-\sqrt{3}i})\ge 0}$ and ${S}_{3}={z\in C:Re(z)\ge 0}$. Then the area of the region ${S}_{1}\cap {S}_{2}\cap {S}_{3}$ is :

Accepted Answer

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

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

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

Solutions · 1