NIMCET 2023 Question Paper with solutions

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Question : 28 of 120

Marks: +1, -0

The value of

∑↙{k=1}↖{n} (x_k)

, where

x_k=\cos (\;{2 π k}/{n}) +i sin \; (\;{2 π k}/{n})

is equal to:

$-1$
$2$
$0$
$1$

Solution:

We should know that

\cos θ+i sin \;θ=e^{i θ}

\;x_k=\;{\cos 2 π k}/{n}+\;{i sin \;2 π k}/{n}=e^{i \;{2 k π}/{n}}

\;∑↙{k=1}↖{n} x_k=∑↙{k=1}↖{n} e^{i \;{2 k π}/{n}}=e^{i \;{2 π}/{n}}+e^{i \;{4 π}/{n}}+e↖{i \;{6 π}/{n}}+. . .+e^{{i \;{2 n π}/{n}}

Let

e^{i \;{2 π}/{n}}=α

Hence, this series

=α+α^2+α^3+. . .+α^n

\;=\;{α (1-α^n) }/{1-α}

\;=\;{e^{i {2 π n}/{n}} (1-e^{ ({i 2 π}/{n}) n}) }/{ 1-e^{{i2π}/n}}

[e^{i 2 π}=\cos 2 π+i sin \;2 π=1]

\;=\;{e^{{i 2 π}/{n}} (1-e^{i 2 π}) }/{1-e^{\;{i 2 π}/{n}}}=0 .

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