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Tardigrade
Question
Mathematics
If ω (≠ 1) be a cube root of unity and (1 + ω2)n = (1 + ω4)n, then the least positive value of n is
Q. If
ω
(
=
1
)
be a cube root of unity and
(
1
+
ω
2
)
n
=
(
1
+
ω
4
)
n
,
then the least positive value of
n
is
2773
244
IIT JEE
IIT JEE 2004
Complex Numbers and Quadratic Equations
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A
2
25%
B
3
53%
C
5
15%
D
6
7%
Solution:
Given,
(
1
+
ω
2
)
n
=
(
1
+
ω
4
)
n
⇒
(
−
ω
)
n
=
(
−
ω
2
)
n
[
∵
w
3
=
1
and
1
+
ω
+
ω
2
=
0
]
⇒
ω
n
=
1
⇒
n
=
3
is the least positive value of
n
.