The value of $\mathop \sum \limits_{k = 1}^{10} \left( {\sin \frac{{2k\pi }}{{11}} + i\cos \frac{{2k\pi }}{{11}}} \right)$ is

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

The value of

\sum \limits_{k = 1}^{10} \left( {\sin \frac{{2k\pi }}{{11}} + i\cos \frac{{2k\pi }}{{11}}} \right)

View answer and explanation

Correct option: C

- i

Option C (

- i

) is correct because the summation simplifies to

-i

using geometric series and Euler's formula.

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{\left| {1 - i} \right|^x} = {2^x}

|z|=1

z \neq \pm 1

z, \frac{z}{1-z^2}

\left|z_r-r\right| \leq r, \forall r=1,2,3, \ldots, n

\left|\sum\limits_{r=1}^n z_r\right|

(z+a b)^3=a^3

a \neq 0

\alpha, \beta

\gamma

|\alpha-\beta|

{z^2} + {\left| z \right|^2} = 0,

z \in C

2x = - 1 + \sqrt 3 i

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}