COMPASSLearning

Talk to us Register Login

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Complex Numbers

If

α

and

β

be the roots of the equation

x^{2} - 2 x + 2 = 0

, then the least value of

n

for which

(\frac{α}{β})^{n} = 1

is:

(A)4✓ Correct
(B)2
(C)5
(D)3

Answer: (A)

Solution

Complete the square in

x^{2} - 2 x + 2 = 0

to get

(x - 1)^{2} = - 1

, so the roots are

x = 1 \pm i

. Then

\frac{α}{β} = \frac{1 + i}{1 - i} = \frac{( 1 + i ) ^{2}}{( 1 - i ) ( 1 + i )} = \frac{2 i}{2} = i

. Requiring

(\frac{α}{β})^{n} = i^{n} = 1

, and recalling

i^{4} = 1

, the least positive

n

is

4

.

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Complex Numbers →

More JEE Main questions on Complex Numbers

If $x^{2} + x + 1 = 0$ , then the value of $(x + \frac{1}{x})^{4} + (x^{2} + \frac{1}{x ^{2}})^{4} + (x^{3} + \frac{1}{x ^{3}})^{4} + \dots + (x^{25} + \frac{1}{x ^{25}})^{4}$ is :
JEE Main 2026View solution →
Let z be the complex number satisfying $∣ z - 5∣ \leq 3$ and having maximum positive principal argument.
Then $34 \frac{5 z - 12}{5 i z + 16}^{2}$ is equal to:
JEE Main 2026View solution →
Let $α = \frac{- 1 + i 3}{2}$ and $β = \frac{- 1 - i 3}{2}, i = - 1$ . If $(7 - 7 α + 9 β)^{20} + (9 + 7 α - 7 β)^{20} + (- 7 + 9 α + 7 β)^{20} + (14 + 7 α + 7 β)^{20} = m^{10}$ , then m is
JEE Main 2026View solution →
Let $S = {z \in C : 4 z^{2} + \overline{z} = 0}$ . Then $\sum_{z \in S} ∣ z ∣^{2}$ is equal to :
JEE Main 2026View solution →
Let $S = {z : 3 \leq ∣2 z - 3 (1 + i) ∣ \leq 7}$ be a set of complex numbers. Then $Min_{z \in S} (z + \frac{1}{2} (5 + 3 i))$ is equal to:
JEE Main 2026View solution →
If $z = \frac{3}{2} + \frac{i}{2}, i = - 1$ , then $(z^{201} - i)^{8}$ is equal to
JEE Main 2026View solution →