Question

Suppose $z_1,z_2,z_3$ are the vertices of an equilateral triangle inscribed in the circle $|z|=2$. If $z_1=1+i\sqrt{3}$, then the other two vertices are

• A. $1+i\sqrt{3}$
• B. $1-i\sqrt{3}$
• C. $2$
• D. $-2$

Answer 1

Answer: (B), (D)

$-2$

Given $z_1=1+i\sqrt{3}=-2{\omega }^2$

Rotating $OA$ about $O$ by an angle $2\pi /3$, we have
$\frac{z-0}{-2{\omega }^2-0}=\frac{|z-0|}{|-2{\omega }^2-0|}{e}^{\pm i2\pi /3}$
$\Rightarrow z=-2{\omega }^2\left(\omega \right),-2{\omega }^2\left({\omega }^2\right)$
$\Rightarrow z=-2{\omega }^3,-2\omega$
$\Rightarrow z=-2,1-i\sqrt{3}$