Tag: the nth roots of unity

Questions Related to the nth roots of unity

If $1,\alpha, \alpha^2,.....,\alpha^{n - 1}$ be the $n^{th}$ roots of unity, then $(1-\alpha)(1-\alpha^2).....(1-\alpha^{n-1}) $

the nth roots of unity complex numbers maths

  1. $3$

  2. $0$

  3. $n$

  4. $1$

Show answer
Correct Option: A
Explanation:
Basically $1,a^1,a^2......a^{n-1}$ all these are the roots of this equation $x^3 – 1 =0$
So we can write
$x^3 -1= (x-1)(x-a _1)(x-a _2).....(x-a _{n-1})$

$\dfrac{x^3 -1}{(x-1)}= (x-a^1)(x-a^2).....(x-a^{n-1})$

$x^2 + x +1= (x-a^1)(x-a^2).....(x-a^{n-1})$

Put $x=1$ on both the sides now

Ans $=3$

Find the number of values of complex numbers $\omega$ satisfying the system of equations ${ z }^{ 3 }=-{ \left( \overline { \omega  }  \right)  }^{ 7 }$ and ${ z }^{ 5 }.{ \omega  }^{ 11 }=1$

the nth roots of unity complex numbers maths

  1. $2$

  2. $4$

  3. $6$

  4. $8$

Show answer
Correct Option: A

For positive integers ${ n } _{ 1 },{ n } _{ 2 }$ the value of the expression; ${ (1+i) }^{ { n } _{ 1 } }+{ (1+i) }^{ { n } _{ 1 } }+{ (1+i) }^{ { n } _{ 2 } }+{ (1+i) }^{ { n } _{ 2 } }$, where $i=\sqrt { -1 } $, is a real number if :

the nth roots of unity complex numbers maths

  1. ${ n } _{ 1 }={ n } _{ 2 }+1$

  2. ${ n } _{ 1 }={ n } _{ 2 }-1$

  3. ${ n } _{ 1 }={ n } _{ 2 }$

  4. ${ n } _{ 1 }>0,{ n } _{ 2 }>0$

Show answer
Correct Option: A

The value of $\sum _{ n=1 }^{ 10 }{ \left( sin\frac { 2n\pi  }{ 11 } -icos\frac { 2n\pi  }{ 11 }  \right)  } $

the nth roots of unity complex numbers maths

  1. $i$

  2. $-i$

  3. $0$

  4. none of these

Show answer
Correct Option: A

The value of the expression $1+(2-\omega )+(2-{ \omega  }^{ 2 })+2+(3-\omega )+(3-{ \omega  }^{ 2 })+..........+(n-1)(n-\omega )(n-{ \omega  }^{ 2 })$ where $\omega $ is an imaginary cube root of unity is-

the nth roots of unity complex numbers maths

  1. ${ (\frac { n(n+1) }{ 2 } ) }^{ 2 }$

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

  3. ${ (\frac { n(n+1) }{ 2 } ) }^{ 2 }+n$

  4. None of above

Show answer
Correct Option: A

If 1,${ a } _{ 1 }{ a } _{ 2,........, }{ a } _{ n-1 }$ are the ${ n }^{ th }$ roots of unity, then $\left( 1-{ a } _{ 1 } \right) \left( 1-{ a } _{ 2 } \right) ....\left( 1-{ a } _{ n-1 } \right) $ is equal to

the nth roots of unity complex numbers maths

  1. n

  2. 0

  3. 1

  4. none of these

Show answer
Correct Option: A

Let the four roots of unity be $z _1, z _2, z _3$, and $z _4$, respectively.
Statement 1: $z _1^2+z _2^2+z _3^2+z _4^2=0$
Statement 2: $z _1+z _2+z _3+z _4=0$.

the nth roots of unity complex numbers maths

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.

Show answer
Correct Option: B
Explanation:

$x^{4}=1$
$x^{2}=\pm1$
$x^{2}=1$ and $x^{2}=-1$
$x=\pm1$ and $x=\pm i$
Hence the four roots are
$1,-1,i,-i$.
Now
$z _{1}=1=-z _{2}$
$z _{3}=i=-z _{4}$
Hence
$z _{1}^{2}+z _{2}^{2}+z _{3}^{2}+z _{4}^{2}$
$=1+1+(i)^{2}+(-i)^{2}$
$=2-2$
$=0$ ...(i)
And also
$z _{1}+z _{2}+z _{3}+z _{4}$
$=1-1+i-i$
$=0$ ...(ii)
Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

If $\alpha _1, \alpha _2, \alpha _3, \alpha _4$ be the roots of $x^5 - 1 = 0$ then find $\displaystyle \frac{\omega - \alpha _1}{\omega^2 - \alpha _1} \cdot \frac{\omega - \alpha _2}{\omega^2 - \alpha _2} \cdot \frac{\omega - \alpha _3}{\omega^2 - \alpha _3} \cdot \frac{\omega - \alpha _4}{\omega^2 - \alpha _4} $

the nth roots of unity complex numbers maths

  1. $\omega^2$

  2. $1$

  3. $\omega$

  4. $(\omega-\alpha _1)(\omega-\alpha _2)(\omega-\alpha _3)(\omega-\alpha _4)$

Show answer
Correct Option: C
Explanation:

$x^5 - 1 = 0$ has roots $1, \alpha _1, \alpha _2, \alpha _3, \alpha _4$
$\therefore (x^5 - 1) = (x- 1) (x - \alpha _1) (x - \alpha _2) (x - \alpha _3) (x- \alpha _4)$
$\Rightarrow \displaystyle \frac{x^5 -1}{x - 1} = (x - \alpha _1) (x- \alpha _2) (x - \alpha _3) (x - \alpha _4)$           ........   (1)
Putting $x = \omega$ (1) we have
$\displaystyle \frac{\omega^5 - 1}{\omega - 1} = (\omega - \alpha _1) (\omega - \alpha _2) (\omega - \alpha _3) (\omega - \alpha _4)$
$\displaystyle \frac{\omega^2 - 1}{\omega - 1} = (\omega - \alpha _1) (\omega - \alpha _2) (\omega - \alpha _3) (\omega - \alpha _4)$         ....... (2)
and putting $x = \omega^2$ in (1) we have
$\displaystyle \frac{\omega^{10} - 1}{\omega^2 - 1} = (\omega^2- \alpha _1) (\omega^2 - \alpha _2) (\omega^2 - \alpha _3) (\omega^2 - \alpha _4)$
$\Rightarrow \displaystyle \frac{\omega - 1}{\omega^2 - 1} = (\omega^2- \alpha _1) (\omega^2 - \alpha _2) (\omega^2 - \alpha _3) (\omega^2 - \alpha _4)$           ....... (3)
Dividing (2) by (3)
then $\displaystyle \frac{\omega - \alpha _1}{\omega^2 - \alpha _1} \cdot \frac{\omega - \alpha _2}{\omega^2 - \alpha _2} \cdot \frac{\omega - \alpha _3}{\omega^2 - \alpha _3} \cdot \frac{\omega - \alpha _4}{\omega^2 - \alpha _4} \cdot = \frac{(\omega^2 - 1)^2}{(\omega - 1)^2}$
                                                                                  $= \displaystyle \frac{\omega^4 + 1 - 2 \omega^2}{\omega^2 + 1 - 2 \omega}$
                                                                                   $= \displaystyle \frac{\omega + 1 - 2 \omega^2}{\omega^2 + 1 - 2 \omega}$
                                                                                   $= \displaystyle \frac{- \omega^2 - 2 \omega^2}{- \omega - 2 \omega}$
                                                                                    $= \displaystyle \frac{- 3 \omega^2}{- 3 \omega}$
                                                                                    $= \omega$

Ans: C

If $\alpha$ is the n$^{th}$ root of unity, then $1+2\alpha+3\alpha^2+.... $ to $n$ terms equal to

the nth roots of unity complex numbers maths

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

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

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

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

Show answer
Correct Option: B
Explanation:

$S=1+2\alpha+3\alpha^{2}+...n\alpha^{n-1}$
$\alpha S=\:\:\alpha+2\alpha^{2}+3\alpha^{3}+...(n-1)\alpha^{n-1}+n\alpha^{n}$
$S(1-\alpha)=1+\alpha+\alpha^{2}+\alpha^{3}+...\alpha^{n-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{1-\alpha^{n}}{1-\alpha}-n\alpha^{n}$
Now $\alpha^{n}=1$ since it is the $n^{th}$ root of unity.
Therefore,
$S(1-\alpha)=-n$
$S=\dfrac{-n}{1-\alpha}$

If $\displaystyle\ \alpha$ is nonreal and $\displaystyle\ \alpha=\sqrt[5]{1}$ then the value of $\displaystyle\ 2^{|1+\alpha+\alpha^{2}+\alpha^{3} +\alpha^{-1}|}$ is equal to

the nth roots of unity complex numbers maths

  1. $\displaystyle\ 4$

  2. $\displaystyle\ 2$

  3. $\displaystyle\ 1$

  4. None of these

Show answer
Correct Option: C
Explanation:

We have $1 + \alpha +\alpha^2 +\alpha^3 +\alpha^4 = 0$


$ 1 + \alpha +\alpha^2 + \alpha^3 +\alpha^{-1} $
$= -\alpha^4 + \dfrac{1}{\alpha} $
$ = \dfrac{1 -\alpha^5}{\alpha} $
$ =0 $
Hence, 
$2^{| 1+ \alpha + \alpha^2 +\alpha^3 + \alpha^{-1} | } =1 $