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

Let λ1+λ2+...+λn=1 where λi>0 for i=1,2,...,n. If |ai|<1 and ω is a complex cube root of unity, then |λ1a1ω+λ2a2ω2+...+λnanωn| cannot exceed

A
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
|a1|+|a2|+...+|an|
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
B 1
C 2
D |a1|+|a2|+...+|an|
|λ1a1ω+λ2a2ω2+...+λnanωn|
|λ1a1ω|+|λ2a2ω2|+...+|λnanωn|
=|λ1||a1||ω|+|λ2||a2||ω2|+...+|λn||an||ωn|
As |ω|=1 and λi0, we get
=λ1|a1|+λ2|a2|+...+λn|an| ...(1)
<λ1+λ2+...+λn
=1

Also, as λ1+λ2+...+λn=1 and λi0, we can infer that none of the λi can exceed 1, thus 0λi1.
from (1)
|λ1a1ω+λ2a2ω2+...+λnanωn|<|a1|+|a2|+...|an|

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integrated Rate Equations
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon