Question

The value of $\sum\limits_{n = 1}^{10} {\left( {\sin \dfrac{{2n\pi }}{{11}} - i\cos \dfrac{{2n\pi }}{{11}}} \right)} $ is
A. $ - 1$
B. $0$
C. $ - i$
D. $i$

Answer 1

Hint: First multiply and divide the expression by $ - i$ to remove the imaginary part from the cosine part. After that use the formula $\cos \dfrac{{2n\pi }}{k} + i\sin \dfrac{{2n\pi }}{k} = {e^{i\dfrac{{2n\pi }}{k}}}$. Now expand the summation, it will form a G.P. After that use the sum of the G.P. formula $S = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$ to find the sum of the terms. Then do the simplification to get the desired result.

Complete step-by-step solution:
The given expression is,
$ \Rightarrow \sum\limits_{n = 1}^{10} {\left( {\sin \dfrac{{2n\pi }}{{11}} - i\cos \dfrac{{2n\pi }}{{11}}} \right)} $
Multiply and divide the expression by $ - i$ to remove the imaginary part from the cosine,
$ \Rightarrow \sum\limits_{n = 1}^{10} {\left( {\sin \dfrac{{2n\pi }}{{11}} - i\cos \dfrac{{2n\pi }}{{11}}} \right)} \times \dfrac{i}{i}$
Multiply the numerator of the expression,
$ \Rightarrow \dfrac{1}{i}\sum\limits_{n = 1}^{10} {\left( {i\sin \dfrac{{2n\pi }}{{11}} - {i^2}\cos \dfrac{{2n\pi }}{{11}}} \right)} $
As we know that ${i^2} = - 1$. Then the expression will be,
$ \Rightarrow \dfrac{1}{i}\sum\limits_{n = 1}^{10} {\left( {i\sin \dfrac{{2n\pi }}{{11}} + \cos \dfrac{{2n\pi }}{{11}}} \right)} $
We know that,
$\cos \dfrac{{2n\pi }}{k} + i\sin \dfrac{{2n\pi }}{k} = {e^{i\dfrac{{2n\pi }}{k}}}$
Also, we know that,
${i^4} = 1$
Then the expression can be written as,
$ \Rightarrow \dfrac{{{i^4}}}{i}\sum\limits_{n = 1}^{10} {\left( {{e^{i\dfrac{{2n\pi }}{{11}}}}} \right)} $
Cancel out the common factors,
$ \Rightarrow {i^3}\sum\limits_{n = 1}^{10} {\left( {{e^{i\dfrac{{2n\pi }}{{11}}}}} \right)} $
Factor ${i^3}$ to ${i^2} \times i$,
$ \Rightarrow {i^2} \times i\sum\limits_{n = 1}^{10} {\left( {{e^{i\dfrac{{2n\pi }}{{11}}}}} \right)} $
As we know that ${i^2} = - 1$. Then the expression will be,
$ \Rightarrow - i\sum\limits_{n = 1}^{10} {\left( {{e^{i\dfrac{{2n\pi }}{{11}}}}} \right)} $
Expand the expression,
\[ \Rightarrow - i\left( {{e^{i\dfrac{{2\pi }}{{11}}}} + {e^{i\dfrac{{4\pi }}{{11}}}} + {e^{i\dfrac{{6\pi }}{{11}}}} + {e^{i\dfrac{{8\pi }}{{11}}}} + {e^{i\dfrac{{10\pi }}{{11}}}} + {e^{i\dfrac{{12\pi }}{{11}}}} + {e^{i\dfrac{{14\pi }}{{11}}}} + {e^{i\dfrac{{16\pi }}{{11}}}} + {e^{i\dfrac{{18\pi }}{{11}}}} + {e^{i\dfrac{{20\pi }}{{11}}}} + {e^{i\dfrac{{22\pi }}{{11}}}}} \right)\]
The above expression is in form of GP where $a = {e^{i\dfrac{{2\pi }}{{11}}}}$ and $r = {e^{i\dfrac{{2\pi }}{{11}}}}$.
The formula of the sum of GP is given by,
$\dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$
Substitute the values in the expression,
$ \Rightarrow - i\left\{ {{e^{i\dfrac{{2\pi }}{{11}}}}\dfrac{{\left( {1 - {e^{i\dfrac{{20\pi }}{{11}}}}} \right)}}{{\left( {1 - {e^{i\dfrac{{2\pi }}{{11}}}}} \right)}}} \right\}$
Multiply the terms inside the bracket,
$ \Rightarrow - i\left\{ {\dfrac{{\left( {{e^{i\dfrac{{2\pi }}{{11}}}} - {e^{i\dfrac{{22\pi }}{{11}}}}} \right)}}{{\left( {1 - {e^{i\dfrac{{2\pi }}{{11}}}}} \right)}}} \right\}$
As we know that ${e^{i2\pi }} = 1$. Then,
$ \Rightarrow - i\left\{ {\dfrac{{\left( {{e^{i\dfrac{{2\pi }}{{11}}}} - 1} \right)}}{{\left( {1 - {e^{i\dfrac{{2\pi }}{{11}}}}} \right)}}} \right\}$
Take -1 common from the numerator,
$ \Rightarrow - i\left\{ {\dfrac{{ - 1\left( {1 - {e^{i\dfrac{{2\pi }}{{11}}}}} \right)}}{{\left( {1 - {e^{i\dfrac{{2\pi }}{{11}}}}} \right)}}} \right\}$
Cancel out the common factor,
$ \Rightarrow i$
Thus, the value of $\sum\limits_{n = 1}^{10} {\left( {\sin \dfrac{{2n\pi }}{{11}} - i\cos \dfrac{{2n\pi }}{{11}}} \right)} $ is $i$.

Hence, the option (D) is the correct answer.

Note: The complex numbers are the field C of numbers of the form $x + iy$, where x and y are real numbers and i is the imaginary unit equal to the square root of -1. When a single letter z is used to denote a complex number. It is denoted as $z = x + iy$.
A G.P is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called the common ratio.