The value of $\mathop \sum \limits_{r = 1}^8 \left( {\sin \frac{{2\;r\;\pi }}{9} + i\cos \frac{{2\;r\;\pi }}{9}} \right),$ is

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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The value of

\sum \limits_{r = 1}^8 \left( {\sin \frac{{2\;r\;\pi }}{9} + i\cos \frac{{2\;r\;\pi }}{9}} \right),

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Correct option: D

- i

Option D (

- i

) is correct because it is the simplified value of the given summation after applying the formula for the sum of a geometric series and using properties of complex numbers.

More Questions from COMPLEX NUMBERS

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The number of non-zero integer solutions of the equation

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

is If

|z|=1

and

z \neq \pm 1

, then for any complex number

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

Let

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

, then

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

is less than The roots of the cubic equation

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

, where

a \neq 0

represents the vertices of a triangle. If

\alpha, \beta

and

\gamma

are the roots, then find the value of

|\alpha-\beta|

. Number of solutions of the equation

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

where

z \in C

is If

2x = - 1 + \sqrt 3 i

, then the value of

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

is equal to