Question

Value of \[\sum\limits_{K = 1}^{10} {\left( {\sin \left( {\dfrac{{2K\pi }}{{11}}} \right) + i\cos \left( {\dfrac{{2K\pi }}{{11}}} \right)} \right)} \]is

Answer 1

Hint: The given question refers to the concepts of summation and the complex number. To find the value of the given function, first take out ‘i’ common from the equation. Then by assuming $\cos \theta + i\sin \theta = {e^{i\theta }}$, we get an equation in the form of G.P. On solving it, we will get the required answer.
Formula Used ${S_n} = a\dfrac{{\left( {1 - {r^n}} \right)}}{{1 - r}}$

Complete step-by-step answer:
$\sum\limits_{K = 1}^{10} {\left( {\sin \dfrac{{2K\pi }}{{11}} + i\cos \dfrac{{2K\pi }}{{11}}} \right)} $
Take out ‘I’ common from the equation
$\sum\limits_{K = 1}^{10} {i\left( {\dfrac{1}{i}\sin \dfrac{{2K\pi }}{{11}} + \cos \dfrac{{2K\pi }}{{11}}} \right)} $
When we multiply and divide $\dfrac{1}{i}$ with i we get $\dfrac{i}{{{i^2}}}$=$ - i$
Now we write as,
$\sum\limits_{K = 1}^{10} {} $$i\left( {\cos \dfrac{{2K\pi }}{{11}} - i\sin \dfrac{{2K\pi }}{{11}}} \right)$ _(I)
$\because $z = $\cos \theta + i\sin \theta $
$\therefore $$z = {e^{i\theta }}$
Now, put the value of z in equation (I)
=$\sum\limits_{K = 1}^{10} i .{e^{ - i\dfrac{{2K\pi }}{{11}}}}$
Let us assume ${e^{ - i\dfrac{{2\pi }}{r}}} = r$, take i as a common
Then, $i\sum\limits_{K = 1}^{10} {{r^k}} $
Now, the above equation is in the form of G.P.
The formula of G.P. is ${S_n} = a\dfrac{{\left( {1 - {r^n}} \right)}}{{1 - r}}$
=$i\dfrac{{r\left( {1 - {r^{10}}} \right)}}{{1 - r}}$
$\therefore i\dfrac{{\left( {r - {r^{11}}} \right)}}{{1 - r}}$ _(II)
We have assume ${r^{11}}$=${\left( {{e^{ - i\dfrac{{2\pi }}{{11}}}}} \right)^{11}}$
$ = {e^{ - 2\pi }}$
$ = \cos 2\pi - i\sin 2\pi $
$\because \cos 2\pi = 1$and $\sin 2\pi $=0
$\therefore 1 - 0$$ = 1$
Now, put the value of ${r^{11}}$in equation (II)
$ = i\dfrac{{\left( {r - 1} \right)}}{{1 - r}}$
$ = - i\dfrac{{\left( {1 - r} \right)}}{{1 - r}}$
$ = - i$

Note: The value $\cos \theta + i\sin \theta $ is also written as ${e^{i\theta }}$ which is called the Euler’s form of complex number. We can write it as ${e^{i\theta }}$ because when we derive the function $\cos \theta + i\sin \theta $ then its answer will be found to be ${e^{i\theta }}$.