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

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

  1. $0$
  2. $1$
  3. $-1$
  4. $1+\omega$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
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
$\Pi _{r=1}^{n-1}(\omega-z _r)=\dfrac {\omega^n-1}{\omega-1}=\left{\begin{matrix}0 & if \ n=3k, k\epsilon Z \ 1, & if\  n=3k+1, k\epsilon Z \ 1+\omega, & if\   n=3k+2, k\epsilon Z\end{matrix}\right.$

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

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

  1. $\displaystyle \frac{n(n+1)a}{2}$
  2. $\displaystyle \frac{nb}{1-n}$
  3. $\displaystyle \frac{na}{\omega - 1}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $-16$
  2. $16$
  3. $-1$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The 5th roots of unity are the roots of the equation x^5 - 1 = 0, which can be factored as (x - 1)(x^4 + x^3 + x^2 + x + 1) = 0. For the non-unity roots a_1, a_2, a_3, a_4, the polynomial can also be expressed as (x - 1)(x - a_1)(x - a_2)(x - a_3)(x - a_4) = x^5 - 1. Substituting x = -1 into both sides gives (-1 - a_1)(-1 - a_2)(-1 - a_3)(-1 - a_4) = (-1)^5 - 1 = -2, which factors out to give the product (1 + a_1)...(1 + a_4) = 1.

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

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

  1. $4$
  2. $3$
  3. $5$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A complex number is both an nth root and an mth root of unity if and only if it is an LCM(n, m)th root of unity, or more precisely, the common roots are the kth roots of unity where k is the greatest common divisor of n and m. Here, the common roots of the 15th and 25th roots of unity are the gcd(15, 25) = 5th roots of unity. Thus, there are 5 common roots.

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

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

  1. min$(p, q)$
  2. max$(p, q)$
  3. $1$

  4. zero
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. $\pm 2,\pm 2i$
  2. $\pm 4,\pm 4i$
  3. $\pm 1,\pm i$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 ${ x }^{ 2 }=-4\Longrightarrow x=\pm \sqrt { -4 } =\pm 2i$

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

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

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

  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 )$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $k=1,2,3..6$.

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

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

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

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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  $\displaystyle a= \frac{1}{1-a _{1}}, \frac{1}{1-a _{2}}.....\frac{1}{1-a _{n-1}}$

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

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

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

  1. $0$
  2. $1$
  3. $-1$
  4. $(-1)^n$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $0$
$\Rightarrow 1({ \alpha  } _{ 1 }+{ \alpha  } _{ 2 }+....,{ +\alpha  } _{ n })+\sum _{ 1\le i<j\le n-1 }^{  }{ { \alpha  } _{ i }{ \alpha  } _{ j } } =0$
$\Rightarrow \sum _{ 1\le i<j\le n-1 }^{  }{ { \alpha  } _{ i }{ \alpha  } _{ j } } =-({ \alpha  } _{ 1 }+{ \alpha  } _{ 2 }+....,{ +\alpha  } _{ n })=-(-1)=1$

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

$\alpha _{1},\alpha _{2},\alpha _{3},\alpha _{4},.........\alpha _{100},$ 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



  1. 20

  2. 0
  3. $(20)^{1/20}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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