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

Value of ∑K=110(sin⁡(2Kπ11)+icos⁡(2Kπ11))\sum\limits_{K = 1}^{10} {\left( {\sin \left( {\dfrac{{2K\pi }}{{11}}} \right) + i\cos \left( {\dfrac{{2K\pi }}{{11}}} \right)} \right)} is

Explanation

Solution

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⁡θ+isin⁡θ=eiθ\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 Sn=a(1−rn)1−r{S_n} = a\dfrac{{\left( {1 - {r^n}} \right)}}{{1 - r}}

Complete step-by-step answer:
∑K=110(sin⁡2Kπ11+icos⁡2Kπ11)\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
∑K=110i(1isin⁡2Kπ11+cos⁡2Kπ11)\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 1i\dfrac{1}{i} with i we get ii2\dfrac{i}{{{i^2}}}=−i - i
Now we write as,
∑K=110\sum\limits_{K = 1}^{10} {} i(cos⁡2Kπ11−isin⁡2Kπ11)i\left( {\cos \dfrac{{2K\pi }}{{11}} - i\sin \dfrac{{2K\pi }}{{11}}} \right) _(I)
∵\because z = cos⁡θ+isin⁡θ\cos \theta + i\sin \theta
∴\therefore z=eiθz = {e^{i\theta }}
Now, put the value of z in equation (I)
=∑K=110i.e−i2Kπ11\sum\limits_{K = 1}^{10} i .{e^{ - i\dfrac{{2K\pi }}{{11}}}}
Let us assume e−i2πr=r{e^{ - i\dfrac{{2\pi }}{r}}} = r, take i as a common
Then, i∑K=110rki\sum\limits_{K = 1}^{10} {{r^k}}
Now, the above equation is in the form of G.P.
The formula of G.P. is Sn=a(1−rn)1−r{S_n} = a\dfrac{{\left( {1 - {r^n}} \right)}}{{1 - r}}
=ir(1−r10)1−ri\dfrac{{r\left( {1 - {r^{10}}} \right)}}{{1 - r}}
∴i(r−r11)1−r\therefore i\dfrac{{\left( {r - {r^{11}}} \right)}}{{1 - r}} _(II)
We have assume r11{r^{11}}=(e−i2π11)11{\left( {{e^{ - i\dfrac{{2\pi }}{{11}}}}} \right)^{11}}
=e−2π= {e^{ - 2\pi }}
=cos⁡2π−isin⁡2π= \cos 2\pi - i\sin 2\pi
∵cos⁡2π=1\because \cos 2\pi = 1and sin⁡2π\sin 2\pi =0
∴1−0\therefore 1 - 0 =1 = 1
Now, put the value of r11{r^{11}}in equation (II)
=i(r−1)1−r= i\dfrac{{\left( {r - 1} \right)}}{{1 - r}}
=−i(1−r)1−r= - i\dfrac{{\left( {1 - r} \right)}}{{1 - r}}
=−i= - i

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