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nth root of unity - (Concept)

We solve the nth root of unity the same as the cube root, only the value becomes n instead of 3 .
Let $z$ be the nth root of unity

$
\begin{aligned}
& \text { So, } z^n=1 \\
& z=(1)^{1 / n}=1+i(0) \\
& =(\cos 0+i \sin 0)^{1 / n} \\
& =(\cos (2 k \pi+0)+i \sin (2 k \pi+0))^{1 / n}, \text { where } k=\text { integer } \\
& F(\cos (2 k \pi)+i \sin (2 k \pi))^{1 / n}
\end{aligned}
$

Using the De-moivre theorem, it can be written as 

$\mathrm{z}=\left(\cos \frac{2 \mathrm{k} \pi}{\mathrm{n}}+\mathrm{i} \sin \frac{2 \mathrm{k} \pi}{\mathrm{n}}\right)$

Now for $k=0,1,2, \ldots,(n-1)$, we get $n$ different solutions, nth roots of unity are represented by $\mathrm{a}^{\mathrm{k}}$ where $\mathrm{k}=0,1,2, \ldots,(n-1)$.

Let, $\alpha=\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}$

Then, nth root of unity are $a^k(k=0,1,2,3$, $\qquad$ $(n-1))$
l.e. $1, a, a^2, a^3, a^4$ $\qquad$ $a^{n-1}$

So these roots of unity are in geometric progression with a common ratio $\alpha=e^{2 i \pi / n}$
(We will study geometric progression and common ratio in the chapter Sequences and Series later)
The sum of nth roots of unity
As all these numbers are roots of the polynomial equation

$
z^n-1=0
$
So, using the sum of roots relation, we can see that the sum of roots $=0$ (As there is no term with $z^{\mathrm{n}-1}$ in the equation)
Product of nth roots of unity
As all these numbers are roots of the polynomial equation

$
z^n-1=0
$
So, using the product of roots relation, we can see that the product of roots will be 1 if $n$ is odd and it will be ( -1 ) if $n$ is even.

Exam Chapter
JEE MAIN Complex numbers and quadratic equations
Algebra (Arihant)
Page No. : 27
Line : 1

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