Question

Suppose, $z_1,z_2,z_3$ are the vertices of an equilateral triangle inscribed in the circle $âˆ\pounds zâˆ\pounds =2$. If $z_1=1+i\sqrt{3}$ then $z_2$ and $z_3$ are equal to

• A. $−2,1−i\sqrt{3}$
• B. $2,1−i\sqrt{3}$
• C. $−2,1+i\sqrt{3}$
• D. None of these

Answer 1

Answer: (A)

$A\left(z_1\right),B\left(z_2\right),C\left(z_3\right)$ lie on $âˆ\pounds zâˆ\pounds =2$ whose centre is at $O(0,0)$ and radius $2.$
$z_1=1+\sqrt{3}i$ hence $âˆ\pounds zâˆ\pounds =2$ and $\text{Arg}\left(z_1\right)=\frac{Ï€}{3}$

In turn $âˆ\pounds z_2âˆ\pounds =âˆ\pounds z_3âˆ\pounds =2$ and $\text{Arg}\left(z_2\right)=\text{Arg}\left(z_1\right)+120^{∘}=180^{∘}$
$∴z_2=−2$
Further, $\text{Arg}\left(z_3\right)=\text{Arg}\left(z_2\right)+120^{∘}=300^{∘}$
Hence, $z_3=2\left[\cos \left(2π−\frac{π}{3}\right)+i\sin \left(2π−\frac{π}{3}\right)\right]$
$=2\left[\cos \frac{π}{3}−i\sin \frac{π}{3}\right]=2\left(\frac{1}{2}−\frac{i\sqrt{3}}{2}\right)$
$=1−\sqrt{3}i$
Thus, $z_2=−2$
and $z_3=1−i\sqrt{3}$