Question

Let $z_1,z_2,z_3$ be three vertices of an equilateral triangle circumscribing the circle $|z|=\frac{1}{2}$. If $z_1=\frac{1}{2}+\frac{i\sqrt{3}}{2}$ and $z_1,z_2,z_3$ are in anticlockwise sense, then $z_2$ is

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

Answer 1

The correct option is D $-1$

Given $z_1,z_2,z_3$ are the vertices of an equilateral triangle.

$|z|=1$

$\Rightarrow$ Centre $(0,0)$ and radius $r=\frac{1}{2}$

Let $z_1=\frac{1}{2}+\frac{i\sqrt{3}}{2}$

$\Rightarrow \tan \alpha =\frac{\sqrt{3}/2}{1/2}=\sqrt{3}$

$\Rightarrow \alpha ={60}^o$

$\Rightarrow z_3=\frac{1}{2}-\frac{i\sqrt{3}}{2}$ from the figure

Since $z_3$ lies in iv quadrant

$z_2$ lies on the circle Ref. image

$|z_1|=\sqrt{{\left[\frac{1}{2}\right]}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$=\sqrt{\frac{1}{4}+\frac{3}{4}}=1$

$\therefore |z_1|=|z_2|=|z_3|=1$

$\Rightarrow z_2$ lies in negative direction of x-axis

$\therefore z_2=-1$.