Question

If $\omega$ is a complex cube root of unity, then value of expression $\cos \left[\left\{\right(1-\omega \left)\right(1-{\omega }^2)+...+(10-\omega )(10-{\omega }^2\left)\right\}\frac{\pi }{900}\right]$:

• A. $1$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (B)

$0$

We know that,
$(a-b\omega )(a-b{\omega }^2)=a^2+ab+b^2$

for $a=1;b=1$

$(1-\omega )(1-{\omega }^2)=1+1+1$

for $a=2;b=1$

$=2^2+2+1$

Continuing the trend, summation becomes,

$=(1^2+2^2---{10}^2)+(1+2---10)+1\times 10$

$=\frac{10\left(11\right)\left(21\right)}{6}+\frac{10\left(11\right)}{2}+10$

$=385+55+10$

$=450$

So, $\cos [\left\{\right(1-\omega \left)\right(1-{\omega }^2)--(10-\omega \left)\right(10-{\omega }^2\left)\right\}\times \frac{\pi }{900}]$

$=\cos [450\times \frac{\pi }{900}]$

$=0$