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 find $z_2$ and $z_3$

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

Answer 1

The correct option is A -2 & 1 - i $\sqrt{3}$

$z_2=\overline{z_1}$
$=1-\sqrt{3}i$
Now since they are vertices of an equilateral triangle.
Hence
$z_1^2+z_2^2+z_3^2=z_1.z_2+z_2.z_3+z_3.z_1$
$-2-2\sqrt{3}i-2+2\sqrt{3}i+z_2^2=4+2z_2$
$-4+z_2^2=4+2z_2$
$z_2^2-2z_2-8=0$
$z^2-4z+2z-8=0$
$(z-4)(z+2)=0$
Hence
$z=4$ and $z=-2$
Hence
The vertices are $-2,(1-\sqrt{3}i)$.