Question

Let ${\lambda }_1+{\lambda }_2+...+{\lambda }_n=1$ where ${\lambda }_i>0$ for $i=1,2,...,n$. If $|a_i|<1$ and $\omega$ is a complex cube root of unity, then $|{\lambda }_1{a}_1\omega +{\lambda }_2{a}_2{\omega }^2+...+{\lambda }_n{a}_n{\omega }^n|$ cannot exceed

• A. $0$
• B. $1$
• C. $2$
• D. $|a_1|+|a_2|+...+|a_n|$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $1$
C $2$
D $|a_1|+|a_2|+...+|a_n|$

$|{\lambda }_1{a}_1\omega +{\lambda }_2{a}_2{\omega }^2+...+{\lambda }_n{a}_n{\omega }^n|$
$\le |{\lambda }_1{a}_1\omega |+|{\lambda }_2{a}_2{\omega }^2|+...+|{\lambda }_n{a}_n{\omega }^n|$
$=|{\lambda }_1||a_1||\omega |+|{\lambda }_2||a_2||{\omega }^2|+...+|{\lambda }^n||a_n||{\omega }^n|$
As $|\omega |=1$ and ${\lambda }_i\ge 0$, we get
$={\lambda }_1|a_1|+{\lambda }_2|a_2|+...+{\lambda }_n|a_n|...\left(1\right)$
$<{\lambda }_1+{\lambda }_2+...+{\lambda }_n$
$=1$

Also, as ${\lambda }_1+{\lambda }_2+...+{\lambda }_n=1$ and ${\lambda }_i\ge 0$, we can infer that none of the ${\lambda }_i$ can exceed $1$, thus $0\le {\lambda }_i\le 1$.
$\therefore$ from $\left(1\right)$
$|{\lambda }_1{a}_1\omega +{\lambda }_2{a}_2{\omega }^2+...+{\lambda }_n{a}_n{\omega }^n|<|a_1|+|a_2|+...|a_n|$