Tag: the nth roots of unity

Questions Related to the nth roots of unity

If $\alpha $ is a non-real root of $x^6=1$, then $\displaystyle \frac{\alpha ^5+\alpha ^3+\alpha +1}{\alpha ^2+1}=$

the nth roots of unity complex numbers maths

  1. $\alpha ^2$

  2. $0$

  3. $-\alpha ^2$

  4. $\alpha $

Show answer
Correct Option: C
Explanation:


$1+\alpha+...+\alpha^{5}=0$ [sum of n roots of unity]
$\Rightarrow 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 }=-\left( { \alpha  }^{ 2 }+{ \alpha  }^{ 4 } \right) $
$\displaystyle \Rightarrow 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 }=-{ \alpha  }^{ 2 }\left( { 1+\alpha  }^{ 2 } \right) $
$\displaystyle \Rightarrow \frac { 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 } }{ { 1+\alpha  }^{ 2 } } =-{ \alpha  }^{ 2 }$

lf $\alpha$ be the $n^{th}$ root of unity then the sum of the series $1+2\alpha+3\alpha^{2}+\ldots.+n\alpha^{n-1}$ equals?

the nth roots of unity complex numbers maths

  1. $\displaystyle \frac{-n}{1-\alpha}$

  2. $\displaystyle \frac{-n}{(1-\alpha)^{2}}$

  3. $\displaystyle \frac{n}{(1-\alpha)}$

  4. $\displaystyle \frac{n}{(1-\alpha)^{2}}$

Show answer
Correct Option: A
Explanation:

$s=1+2\alpha+3\alpha^{2}- -n\alpha^{n-1}$
$s\alpha=\alpha+2\alpha^{2}+3\alpha^{3}- - n\alpha
\So, s-s\alpha=1+\alpha+\alpha^{2}- -\alpha^{n-1}-n\alpha^{n}$
(Sum of roots of unity)
$\Rightarrow s (1-\alpha)=-n\alpha^{n}$
$\Rightarrow s=\dfrac{-n\alpha^{n}}{1-\alpha}=\dfrac{-n}{1-\alpha}$

as $\alpha^{n}=1$
$\alpha$ being of unity root

If $(2 + i \sqrt 3)$ is a root of the equation $x^2 + px + q = 0$, where p and q are real, then (p, q) equals to

the nth roots of unity complex numbers maths

  1. $(4, 7)$

  2. $(-4, -7)$

  3. $(-4, 7)$

  4. $(4, -7)$

Show answer
Correct Option: C
Explanation:

Since $2 + i \sqrt 3$ is one root, then other root will be $2 - i \sqrt 3$.
$\therefore x^2 + px + q = 0$ is given equatiion
$\therefore$ Sum of roots $= 2 + i \sqrt 3 + 2 - i \sqrt 3 = p$
$\therefore p = - 4$
Product of roots $q = 4 + 3 = 7$

In the multiplicative group of $n^{th}$ roots of unity the inverse of ${ \omega  }^{ k },\left( k<n \right) $ is

the nth roots of unity complex numbers maths

  1. ${ \omega }^{ { 1 }/{ k } }$

  2. ${ \omega }^{ -1 }$

  3. ${ \omega }^{ n-k }$

  4. ${ \omega }^{ { n }/{ k } }$

Show answer
Correct Option: C
Explanation:

Since $\omega $ is the $n^{th}$ root of unity, hence $\omega^{n}=1$.
Now inverse of $\omega ^{k}$ where $k<n$ is
$=\dfrac{1}{\omega ^{k}}$
$=\dfrac{\omega^{n}}{\omega^{k}}$
$=\omega^{n-k}$.
Hence inverse of $\omega^{k}$
$={\omega^{-k}}$
$=\omega^{n-k}$. 

The 4th roots of unity in the argand plane form a

the nth roots of unity complex numbers maths

  1. Square

  2. Rectangle

  3. Parallelogram

  4. Rhombus

Show answer
Correct Option: A
Explanation:

The fourth roots of unity are $1,i,-1, -i $. Hence, they form a square in the Argand plane. 
We can see that by plotting the points on the graph with coordinate system according to that of complex numbers.

If $\omega, \omega^2, \omega^3, ........ \omega^{n - 1}$ are nth roots of unity then $(1- \omega) (1- \omega^2) ....... (1 - \omega^{n  -1})$ equals:

the nth roots of unity complex numbers maths

  1. $0$

  2. $1$

  3. $n$

  4. $n^2$

Show answer
Correct Option: C
Explanation:

Since $1, \omega, \omega^2, \omega^3, ......... \omega^{n - 1}$ are nth roots of unity, therefore, we have the identity
$(x - 1)(x - \omega)(x - \omega^2) ...... (x - \omega^{n - 1})$
$= x^n - 1$
or $(x - \omega)(x - \omega^2) ...... (x - \omega^{n - 1}) = \displaystyle \frac{x^n - 1}{x - 1}$
$= x^{n - 1} + x^{n - 2} + ...... + x + 1$
Putting x = 1 on both sides, we get $(1 - \omega) (1 - \omega^2) (1 - \omega^{n - 1}) = n$

Which of the following is incorrect regarding $n^{th}$ roots of unity?

the nth roots of unity complex numbers maths

  1. The number of distinct roots is $n$

  2. The roots are in G.P. with common ratio $c = \dfrac{2\pi}{n}$

  3. The arguments are in A.P. with common difference $\dfrac{2\pi}{n}$

  4. Product of the roots is $0$ and the sum of the roots is $\pm 1$

Show answer
Correct Option: D
Explanation:
Consider ${x}^{n}=1$ , where $x$ is $n^{th}$ root of unity
Here in ${x}^{n}-1=0$ , the product of roots is $\pm1$ and the sum of roots is $0$
Therefore option $D$ is incorrect 

If $2 + i$ and $\sqrt {5} - 2i$ are the roots of the equation $(x^{2} + ax + b)(x^{2} + cx + d) = 0$, where $a, b, c, d$ are real constants, then product of all roots of the equation is

the nth roots of unity complex numbers maths

  1. $40$

  2. $9\sqrt {5}$

  3. $45$

  4. $35$

Show answer
Correct Option: C
Explanation:

$2 - i$ and $\sqrt {5} + 2i$ are other roots.
So, Product is $(2 + i)(2 - i)(\sqrt {5} + 2i)(\sqrt {5} - 2i)$
$= 5\times 9 = 45$

$1 , z _1, z _2, z _3, ..., z _{n-1}$ are the $n$th roots of unity, then the value of $\displaystyle\frac{1}{(3-z _1)} +\displaystyle\frac{1}{(3-z _2)} + ... +\displaystyle\frac{1}{(3-z _{n-1})}$ is equal to  

the nth roots of unity complex numbers maths

  1. $\displaystyle \frac { n{ 3 }^{ n-1 } }{ { 3 }^{ n }-1 } -\displaystyle \frac { 1 }{ 2 } $

  2. $\displaystyle \frac { n{ 3 }^{ n-1 } }{ { 3 }^{ n }-1 } +1$

  3. $\displaystyle \frac { n{ 3 }^{ n-1 } }{ { 3 }^{ n }-1 } -1$

  4. none of these

Show answer
Correct Option: A
Explanation:

If $\alpha _1, \alpha _2, ... , \alpha _m$ are the roots of polynomial equation
$\quad f(x) = a _0x^m + a _1x^{m-1} + ... + a _{m-1}x + a _m = 0$
Then,
$\quad f(x) = a _0(x-\alpha _1) ... (x-\alpha _m).$


and $\quad \displaystyle\frac{f'(x)}{f(x)} = \displaystyle\frac{1}{x-\alpha _1}+...+\displaystyle\frac{1}{x-\alpha _m}$

The equation in question is ${ x }^{ n }-1=0$

 $f(x)={ x }^{ n }-1=(x-1)(x-{ z } _{ 1 })...(x-{ z } _{ n-1 })$


Thus, $\dfrac { f'(x) }{ f(x) } =\dfrac { n{ x }^{ n-1 } }{ { x }^{ n }-1 } =\dfrac { 1 }{ x-1 } +\dfrac { 1 }{ x-{ z } _{ 1 }  } +...+\dfrac { 1 }{ x-{ z } _{ n-1 } } $

Substituting $x=3$:

$\dfrac { 1 }{ 3-1 } +\dfrac { 1 }{ 3-{ z } _{ 1 }  } +...+\dfrac { 1 }{ 3-{ z } _{ n-1 } } =\dfrac { n{ 3 }^{ n-1 } }{ { 3 }^{ n }-1 } $

Hence, $\dfrac { 1 }{ 3-{ z } _{ 1 }  } +...+\dfrac { 1 }{ 3-{ z } _{ n-1 } } =\dfrac { n{ 3 }^{ n-1 } }{ { 3 }^{ n }-1 } - \dfrac { 1 }{ 2 }$

Hence,  (A) is correct.

If $1,\omega,\omega^{2},...,\omega^{n-1}$ are $n^{th}$ roots of unity, then the value of $(5-\omega)(5-\omega^{2})...(5-\omega^{n-1})=$

the nth roots of unity complex numbers maths

  1. $\displaystyle \frac{5^{n}-2}{4}$

  2. $\displaystyle \frac{5^{n}+2}{4}$

  3. $\displaystyle \frac{5^{n}+1}{4}$

  4. $\displaystyle \frac{5^{n}-1}{4}$

Show answer
Correct Option: D
Explanation:

${ x }^{ n }-1=0$ has n roots (of unity).
Thus, ${ x }^{ n }-1=(x-1)(x-\omega )(x-{ \omega  }^{ 2 })...(x-{ \omega  }^{ n-1 })$.
Substitute $x=5$: 
${ 5 }^{ n }-1=(5-1)(5-\omega )(5-{ \omega  }^{ 2 })...(5-{ \omega  }^{ n-1 })$
=> $(5-\omega )(5-{ \omega  }^{ 2 })...(5-{ \omega  }^{ n-1 })=\dfrac { { 5 }^{ n }-1 }{ 4 } $
Hence, option D is correct.