First slide

Algebra of complex numbers

Question

Let $z_{1}$ and $z_{2}$ be two non-zero complex numbers such that $\frac{z_{1}}{z_{2}} + \frac{z_{2}}{z_{1}} = 1,$ then the origin and points represented by $z_{1}$ and $z_{2}$

Moderate

A

lie on a straight line
B

form a right triangle
C

form an equilateral triangle
D

none of these

Solution

Let $z = \frac{z_{1}}{z_{2}},$ and $z + \frac{1}{z} = 1 \Rightarrow z^{2} - z + 1 = 0$
$\Rightarrow z = \frac{1 \pm \sqrt{3} i}{2} \Rightarrow \frac{z_{1}}{z_{2}} = \frac{1 \pm \sqrt{3} i}{2}$
If $z_{1}$ and $z_{2}$ are represented by $A$ and $B$ respectively and $O$ be the origin, then
$\begin{array}{l} \frac{O A}{O B} = \frac{|z_{1}|}{|z_{2}|} = |\frac{1 \pm \sqrt{3 i}}{2}| = \sqrt{\frac{1}{4} + \frac{3}{4} = 1} \\ \Rightarrow O A = O B \end{array}$
Also, $\frac{A B}{O B} = \frac{|z_{2} - z_{1}|}{|z_{2}|} = |1 - \frac{z_{1}}{z_{2}}|$

$= |1 - (\frac{1}{2} \pm \frac{\sqrt{3}}{2} i)| = |\frac{1}{2} \mp \frac{\sqrt{3}}{2} i| = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$
$\Rightarrow A B = O B$
Thus, $O A = O B = A B .$
$∴ ∆ O A B$ is an equilateral triangle.

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