Q75 / 170 · Manipal (MET) 2019 · Mathematics
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$1$
- Correct answerD$-1$
Solution
Given that, $|z|=\frac{1}{2}$ and $z_1=\frac{1}{2}+\frac{\sqrt{3} i}{2}$
Here, one of the number must be a conjugate of $z_1=\frac{1}{2}+\frac{\sqrt{3} i}{2}$ i.e. $z_2=\frac{1}{2}-\frac{\sqrt{3} i}{2}$ and $z_2=z_1 e^{-i 2 \pi / 3}$ $\begin{aligned} \Rightarrow \quad z_2 & =\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)\left[\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right] \\ & =-2 \times \frac{1}{2}=-1 \\ & {\left[\because(1+i \sqrt{3})\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)=-2\right] } \end{aligned}$
Here, one of the number must be a conjugate of $z_1=\frac{1}{2}+\frac{\sqrt{3} i}{2}$ i.e. $z_2=\frac{1}{2}-\frac{\sqrt{3} i}{2}$ and $z_2=z_1 e^{-i 2 \pi / 3}$ $\begin{aligned} \Rightarrow \quad z_2 & =\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)\left[\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right] \\ & =-2 \times \frac{1}{2}=-1 \\ & {\left[\because(1+i \sqrt{3})\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)=-2\right] } \end{aligned}$
Hence, the correct answer is option: 'D'.