wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If ω is a cube root of unity, and the value of
(1ω)(1ω2)(1ω4)(1ω8)=α, and
a+bω+cω2b+cω+aω2+a+bω+cω2c+aω+bω2=β then

A
α+β=8
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
αβ=9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
αβ=9
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
α+β=9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct options are
A α+β=8
C αβ=9
(1ω)(1ω2)(1ω4)(1ω8)
=(1ω)(1ω2)(1ω)(1ω2) [ω3=1]
=(1ω2)2(1ω)2
=(1+ω42w2)(1+ω22ω) [ω3=1]
=(1+ω2ω2)(ω2ω) [1+ω+ω2=1]
=(3ω2)(3ω)
=9=α(1)

Now,
a+bω+cω2b+cω+aω2+a+bω+cω2c+aω+bω2
=aω3+bω+cω2b+cω+aω2+a+bω+cω2cω3+aω+bω2 [ω3=1]=ω+ω2 [1+ω+ω2=1]=1=β(2)

From (1) and (2)
α+β=8
and, α×β=9

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon