Tag: n th root of unity

Questions Related to n th root of unity

If $1, z _1, z _2, z _3, ...., z _{n-1}$ be the nth roots of unity and $\omega$ be a non-real complex cube root of unity, then the product
$\Pi _{r=1}^{n-1}(\omega-z _r)$ can be equal to

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $0$

  2. $1$

  3. $-1$

  4. $1+\omega$

Show answer
Correct Option: A,B,D
Explanation:

$x^n-1=(x-1)(x-z _1)(x-z _2)....(x-z _{n-1})$
$\Rightarrow \dfrac {x^n-1}{x-1}=(x-z _1)(x-z _2)....(x-z _{n-1})$
Putting $x=\omega$, we have
$\Pi _{r=1}^{n-1}(\omega-z _r)=\dfrac {\omega^n-1}{\omega-1}=\left{\begin{matrix}0 & if \ n=3k, k\epsilon Z \ 1, & if\  n=3k+1, k\epsilon Z \ 1+\omega, & if\   n=3k+2, k\epsilon Z\end{matrix}\right.$

If $\omega$ is a complex $n$th root of unity, then $\displaystyle \sum _{r=1}^{n} (ar + b)\omega^{r-1}$ is equal to

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

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

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

  3. $\displaystyle \frac{na}{\omega - 1}$

  4. $none\ of\ these$

Show answer
Correct Option: C
Explanation:

Upon expanding, we get
$(a+b)+(2a+b)w+(3a+b)w^{2}+...(na+b)w^{n-1}$
$=a(1+2w+3w^{2}+...nw^{n-1})+b(1+w+w^{2}+...w^{n-1})$
$=a(1+2w+3w^{2}+...nw^{n-1})+b(\cfrac{1-w^{n}}{1-w})$
$=a(1+2w+3w^{2}+...nw^{n-1})+0$

Let
$S=a(1+2w+3w^{2}+...nw^{n-1})$
$Sw=a(w+2w^{2}+3w^{3}+...(n-1)w^{n-1}-nw^{n})$
$S(1-w)=a(1+w+w^{2}....w^{n-1})-anw^{n}$
$S(1-w)=a(\cfrac{1-w^{n}}{1-w})-anw^{n}$
$S(1-w)=a(0)-an$
$S=-\cfrac{an}{1-w}=\dfrac{an}{w-1}$
Hence, option 'C' is correct.

$\begin{array} { l } { 1 , a _ { 1 } , \ldots , a _ { 4 } \text { are the } 5 ^ { \text { th } } \text { roots of unity. The value } } \ { \text { of } \left( 1 + a _ { 1 } \right) \dots \left( 1 + a _ { 4 } \right) \text { is } } \end{array}$ ?

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $-16$

  2. $16$

  3. $-1$

  4. $1$

Show answer
Correct Option: A

The no. of common roots of $15th$ roots of unity which are also $25th$ the roots of unity is

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $4$

  2. $3$

  3. $5$

  4. $2$

Show answer
Correct Option: A

If $p$ and $q$ are distinct prime numbers, then the number of distinct imaginary numbers which are $p$th as well as $q$th roots of unity are

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. min$(p, q)$

  2. max$(p, q)$

  3. $1$

  4. zero

Show answer
Correct Option: D
Explanation:

It is given that, $p$ and $q$ are prime numbers.
Hence the only common $pth$ and $qth$ root of unity will be the number 1.
Thus there will be no common imaginary $pth$ and $qth$ root of unity.
Hence answer is zero.

The value of ${ \left( 16 \right)  }^{ 1/4 }$ are

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $\pm 2,\pm 2i$

  2. $\pm 4,\pm 4i$

  3. $\pm 1,\pm i$

  4. None of these

Show answer
Correct Option: A
Explanation:

Let $x={ \left( 16 \right)  }^{ { 1 }/{ 4 } }$

${ x }^{ 4 }=16$

${ x }^{ 4 }-16=0$

${ x }^{ 4 }-{ 2 }^{ 4 }=0$

$\left( { x }^{ 2 }-{ 2 }^{ 2 } \right) \left( { x }^{ 2 }+{ 2 }^{ 2 } \right) =0$

$\left( x-2 \right) \left( x+2 \right) \left( { x }^{ 2 }+4 \right) =0$

$x-2=0,x+2=0,{ x }^{ 2 }+4=0$

$x=2,-2$ or ${ x }^{ 2 }=-4\Longrightarrow x=\pm \sqrt { -4 } =\pm 2i$

$\therefore x=\pm 2,\pm 2i$

Find all those roots of the equation $z^{12} - 56z^6 - 512 = 0$ whose imaginary part is positive.

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $2, 2 \left ( cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2,$

    $ 2^{2/3} \left ( cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{2/3} \left ( cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{2/3} \left ( cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$

  2. $2, 2 \left ( cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2, $

    $2^{1/3} \left ( cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{1/3} \left ( cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{1/3} \left ( cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$

  3. $2, 2 \left ( cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2,$

    $ 2^{1/3} \left ( -cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{1/3} \left ( -cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{1/3} \left ( -cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$

  4. $2, 2 \left ( -cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( -cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2,$

    $ 2^{2/3} \left ( cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{2/3} \left ( cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{2/3} \left ( cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$

Show answer
Correct Option: B
Explanation:

$(z^{6}-28)^{2}-784-512=0$
$(z^{6}-28)^{2}=1296$
$z^{6}-28=\pm36$
$z^{6}=64$ and $z^{6}=-8$
$z^{3}=\pm8$
$z=2$ and $z=-2$ ...(i)
$z^{6}=2^{3}.e^{i(2k-1)\pi}$
$z=2^{\frac{1}{2}}(e^{i\frac{(2k-1)\pi}{6}})$ where $k=1,2,3..6$.

If $\displaystyle 1,a _{1},a _{2}...,a _{n-1} $ are $\displaystyle n^{th}$ roots of unity, then $\displaystyle \frac{1}{1-a _{1}}+\frac{1}{1-a _{2}}+...+\frac{1}{1-a _{n-1}}$ equals

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

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

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

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

  4. None of these

Show answer
Correct Option: B
Explanation:

Given $\displaystyle z^{n}= 1,z= 1,a _{1}, a _{2},..., a _{n-1}$
Let $\displaystyle a= \frac{1}{1-z}\Rightarrow z= 1-\frac{1}{a}$
 $\displaystyle \therefore \left ( 1-\frac{1}{a} \right )^{n}= 1$
$\displaystyle \Rightarrow \left ( a-1 \right )^{n}-a^{n}= 0$
$\displaystyle \Rightarrow -C _{1}a^{n-1}+C _{2}a^{n-2}+...+\left ( -1 \right )^{n}= 0$ where  $\displaystyle a= \frac{1}{1-a _{1}}, \frac{1}{1-a _{2}}.....\frac{1}{1-a _{n-1}}$

$\displaystyle \Rightarrow \frac{1}{1-a _{1}}+\frac{1}{1-a _{2}}+.....+\frac{1}{1-a _{n-1}}= \frac{^{n}c _{2}}{n}= \frac{n-1}{2}$

If $n\ge 3$ and $1,\alpha _1, \alpha _2, ... , \alpha _{n-1}$ are $nth$ roots of unity, then the value of $\displaystyle\sum _{1 \le i < j \le n-1}{\alpha _i\alpha _j}$ is

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $0$

  2. $1$

  3. $-1$

  4. $(-1)^n$

Show answer
Correct Option: B
Explanation:

Given that $ 1,{ \alpha  } _{ 1 },{ \alpha  } _{ 2 },....,{ \alpha  } _{ n }$ are $n$th roots of unity

$ \Rightarrow x^{n}=1$
So the sum of roots is $0$
$\Rightarrow 1+{ \alpha  } _{ 1 }+{ \alpha  } _{ 2 }+....,{ +\alpha  } _{ n }=0$
Sum of product of roots taken two at a time is $0$
$\Rightarrow 1({ \alpha  } _{ 1 }+{ \alpha  } _{ 2 }+....,{ +\alpha  } _{ n })+\sum _{ 1\le i<j\le n-1 }^{  }{ { \alpha  } _{ i }{ \alpha  } _{ j } } =0$
$\Rightarrow \sum _{ 1\le i<j\le n-1 }^{  }{ { \alpha  } _{ i }{ \alpha  } _{ j } } =-({ \alpha  } _{ 1 }+{ \alpha  } _{ 2 }+....,{ +\alpha  } _{ n })=-(-1)=1$

$\alpha _{1},\alpha _{2},\alpha _{3},\alpha _{4},.........\alpha _{100},$ are all the $100^{th}$ roots of unity. Then the numerical value of $\sum _{1 \leq i}^{ }  \sum _{j \leq 100}^{ } (\alpha _{i}\alpha _{j})^{5}$ is



finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. 20

  2. 0

  3. $(20)^{1/20}$

  4. None of these

Show answer
Correct Option: B
Explanation:

$\sum _{1 \leq i}^{ }  \sum _{j \leq 100}^{ } \alpha _{i}^{5} \alpha _{j}^{5}=(\alpha _{1}^{5}+\alpha _{2}^{5}.......+\alpha _{100}^{5})^{2}-(\alpha _{1}^{10}+\alpha _{2}^{10}.......+\alpha _{100}^{10})$
=0-0=0