Question

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

• A. $z_2=1+i\sqrt{3},z_3=-2$
• B. $z_2=2-i\sqrt{3},z_3=-1$
• C. $z_2=1-i\sqrt{3},z_3=-2$
• D. $z_2=2+i\sqrt{3},z_3=-1$

Answer 1

The correct option is C $z_2=1-i\sqrt{3},z_3=-2$

Since the vertices will appear in conjugate form hence two vertices are
$(1+\sqrt{3}i),(1-\sqrt{3}i)$... (i)
Now
$z_1^2+z_2^2+z_3^2=z_1{z}_2+z_2{z}_3+z_3{z}_1$
For an equilateral triangle.
Hence
$1-3+2\sqrt{3}i+1-3-2\sqrt{3}i+z_3^2=4+z_3\left(2\right)$
$-4+z_3^2=4+z_3\left(2\right)$
$z^2-2z-8=0$
or
$(z+2)(z-4)=0$
$z=-2$ or $z=4$
Now Z lies on $|z|=2$
Hence
$z_3=-2$.