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Let $\mathrm{z}_{1}, \mathrm{z}_{2}$ and $\mathrm{z}_{3}$ be non-zero complex numbers satisfying $\mathrm{z}^{2}=\mathrm{i} \overline{\mathrm{z}}$, where $\mathrm{i}=\sqrt{-1}$.
Consider the following statements:
1- $\quad \mathrm{z}_{1} \mathrm{z}_{2} \mathrm{z}_{3}$ is purely imaginary.
2- $\quad z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}$ is purelyreal. Which of the above statements is/are correct?
Options:
Consider the following statements:
1- $\quad \mathrm{z}_{1} \mathrm{z}_{2} \mathrm{z}_{3}$ is purely imaginary.
2- $\quad z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}$ is purelyreal. Which of the above statements is/are correct?
Solution:
2224 Upvotes
Verified Answer
The correct answer is:
Both 1 and 2
$z_{1} z_{2} z_{3}=(-i)\left(\frac{\sqrt{3}+i}{2}\right)\left(\frac{-\sqrt{3}+i}{2}\right)$
$=\frac{-i}{4}\left(i^{2}-(\sqrt{3})^{2}\right)$
$=\frac{-i}{4}(-3-1)=i$
Hence $z_{1} z_{2} z_{3}$ is purely imaginary. $z_{1} z_{2}=-i\left(\frac{\sqrt{3}+i}{2}\right)=\frac{-\sqrt{3} i+1}{2}$
$\begin{aligned} z_{2} z_{3} &=\frac{(\sqrt{3}+i)(-\sqrt{3}+i)}{4} \\ &=\left(\frac{-3-\sqrt{3} i+\sqrt{3} i+i^{2}}{4}\right)=-1 \end{aligned}$
$z_{3} z_{1}=\frac{(-\sqrt{3}+i)(-i)}{2}=\frac{+\sqrt{3} i+1}{2}$
$z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}=\left(\frac{-\sqrt{3} i+1}{2}\right)+(-1)+\left(\frac{\sqrt{3} i+1}{2}\right)$
$=\left(\frac{-\sqrt{3} i+1+\sqrt{3} i+1}{2}\right)-1$
$=0 \in R$
Hence $z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}=0$ is purelyreal. Hence both statements are correct.
$=\frac{-i}{4}\left(i^{2}-(\sqrt{3})^{2}\right)$
$=\frac{-i}{4}(-3-1)=i$
Hence $z_{1} z_{2} z_{3}$ is purely imaginary. $z_{1} z_{2}=-i\left(\frac{\sqrt{3}+i}{2}\right)=\frac{-\sqrt{3} i+1}{2}$
$\begin{aligned} z_{2} z_{3} &=\frac{(\sqrt{3}+i)(-\sqrt{3}+i)}{4} \\ &=\left(\frac{-3-\sqrt{3} i+\sqrt{3} i+i^{2}}{4}\right)=-1 \end{aligned}$
$z_{3} z_{1}=\frac{(-\sqrt{3}+i)(-i)}{2}=\frac{+\sqrt{3} i+1}{2}$
$z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}=\left(\frac{-\sqrt{3} i+1}{2}\right)+(-1)+\left(\frac{\sqrt{3} i+1}{2}\right)$
$=\left(\frac{-\sqrt{3} i+1+\sqrt{3} i+1}{2}\right)-1$
$=0 \in R$
Hence $z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}=0$ is purelyreal. Hence both statements are correct.
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