Question

If$z_1,z_2,z_3$ are the vertices of an equilaterial triangle circumscribing the circle$|z|=1$. If$z_1=1+\sqrt{3}i$ and $z_1,z_2,z_3$ are in the anticlockwise sense then $z_2$

• A. $1-\sqrt{3}i$
• B. $2$
• C. $\frac{1}{2}\left(\sqrt{3}i\right)$
• D. none of these

Answer 1

The correct option is D none of these

$|z|=1$

Represents a circle centered at the origin.
In an equilateral triangle, we know that the circumcentre and centroid coincide.

Hence the centroid is the origin.

Therefore

$z_1+z_2+z_3=0$

$z_1=-2w$

Therefore
$z_2+z_3=2w$ ...(i)

Now they must satisfy the condition,

$z_1^2+z_2^2+z_3^2=z_1{z}_2+z_2{z}_3+z_3{z}_1$ since they are the vertices of an equilateral triangle.

Hence

$4w^2+(z_2+z_3{)}^2-2z_2{z}_3=2w(z_2+z_3)+z_2{z}_3$

$4w^2+(z_2+z_3{)}^2-2z_2{z}_3=4w^2+z_2{z}_3$

$(z_2+z_3{)}^2=3z_2{z}_3$

$4w^2=3z_2{z}_3$

$z_3=\frac{4w^2}{3z_2}$

Substituting in 1, we get

$3z_2^2+4w^2=6w{z}_2$

$3t^2-6wt+4w^2=0$

$t=\frac{6w\pm \sqrt{36w^2-48w^2}}{6}$

$=w\pm \frac{4iw}{6}$

$z_2=w(1\pm \frac{2i}{3})$

Hence answer is none of these.