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 $z_{1}, z_{2} z_{3}$ were in anticlockwise sense, then $z_{2}$ is

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

Answer 1

Answer: (D)

Given that, $|z|=\frac{1}{2}$ and $z_{1}=\frac{1}{2}+\frac{i \sqrt{3}}{2}$
Here, one of the number must be a conjugate of
$z_{1}=\frac{1}{2}+\frac{i \sqrt{3}}{2}$
i.e. $z_{2}=\frac{1}{2}-\frac{i \sqrt{3}}{2}$
and $z_{2}=z_{1} e^{-i 2 \pi / 3}$
$\Rightarrow z_{2}=\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)\left[\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right]$
$=-2 \times \frac{1}{2}=-1$
$\left.\left[\because(1+i \sqrt{3}) \cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)=-2\right]$