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
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B
1
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C
2
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D
|a1|+|a2|+...+|an|
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Solution
The correct options are B1 C2 D|a1|+|a2|+...+|an| |λ1a1ω+λ2a2ω2+...+λnanωn| ≤|λ1a1ω|+|λ2a2ω2|+...+|λnanωn| =|λ1||a1||ω|+|λ2||a2||ω2|+...+|λn||an||ωn| As |ω|=1 and λi≥0, we get =λ1|a1|+λ2|a2|+...+λn|an|...(1) <λ1+λ2+...+λn =1
Also, as λ1+λ2+...+λn=1 and λi≥0, we can infer that none of the λi can exceed 1, thus 0≤λi≤1. ∴ from (1) |λ1a1ω+λ2a2ω2+...+λnanωn|<|a1|+|a2|+...|an|