Question:

Let $z_1$ and $z_2$ be nth roots of unity which subtend a right angled at the origin, then n must be of the form (where, k is an integer)

Updated On: Jun 14, 2022
  • 4k+1
  • 4k+2
  • 4k+3
  • 4k
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The Correct Option is D

Solution and Explanation

Since, arg $\frac{z_1}{z_2}=\frac{\pi}{2}$
$\Rightarrow \, \, \frac{z_1}{z_2}=cos\frac{\pi}{2}+ i sin \frac{\pi}{2}=i$
$\therefore \, \, \, \, \, \, \frac{z_1^n}{z_2^n}=(i)^n \, \Rightarrow i^n=1 \, \, \, \, \, \, \, \, \, \, \, \, \, [\because \, |z_2|=|z_1|=1] $
$\Rightarrow \, \, \, \, \, \, \, \, n=4k $
Alternate Solution
$Since ,\, \, \, arg \frac{z_2}{z_1}=\frac{\pi}{2}$
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{z_2}{z_1}=\bigg|\frac{z_2}{z_1}\bigg|e^{i\frac{\pi}{2}}$
$\Rightarrow \, \, \, \, \, \frac{z_2}{z_1}=i \, \, \, \, \, \, \, \, \, \, \, \, \, [\because \, |z_1|=|z_2|=1] $
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \bigg(\frac{z_2}{z_1}\bigg)^n=i^n$
$\therefore \, \, \, z_1 \, and \, \, z_2$ are nth roots of unity
$z_1^n=z_2^n=1 \, \Rightarrow \, \, \bigg(\frac{z_2}{z_1}\bigg)^n =1 \, \, \, \Rightarrow \, \, i^n=1$
$\Rightarrow \, \, \, \, \, $ n=4k, where k is an integer.
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Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.