Series of Complex Stuff

Algebra Level 4

$\displaystyle \sum_{p=1}^{32} (3p+2) \left[ \displaystyle \sum_{q=1}^{10} \left( \sin\left(\frac{2q\pi}{11} \right)-i\cos\left( \frac{2q\pi}{11} \right) \right)\right]^{4}=\, ?$

Note : $i=\sqrt{-1}$

Practice the set Target JEE_Advanced - 2015 and boost up your preparation.

The answer is 1648.

1 solution

Tanishq Varshney
Apr 1, 2015

the expression is $\displaystyle \sum_{p=1}^{32} (3p+2) (i)^{4 \pi}$

For the above step see this

$i^4=1$

the expression becomes

$\displaystyle \sum_{p=1}^{32} (3p+2)$

$\frac{3p(p+1)}{2}+2p~ where ~p=32$

$48\times 33+64$

$\boxed{1648}$

but i^(4*pi) is a complex number which is equal to 0.62968 + 0.77685i

Surya Prakash Bugatha - 5 years, 11 months ago

$(i^{4})^{\pi}$ = $1^{\pi}$ =1

Tanishq Varshney - 5 years, 11 months ago

Actually , $i^{4n}=1$ for all integers 'n'. But $\pi$ is not integer. To clarify this, $i^{2}=i^{4*(0.5)}=(i^{4})^{0.5}=1^{0.5}=1$

Surya Prakash Bugatha - 5 years, 11 months ago

did same..

Dev Sharma - 5 years, 5 months ago

@Chew-Seong Cheong sir please resolve this question as well .

Ujjwal Mani Tripathi - 4 years, 5 months ago

Series of Complex Stuff

Practice the set Target JEE_Advanced - 2015 and boost up your preparation.

The answer is 1648.

1 solution

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