Tag: complex numbers

Quizzes (9) Questions (142)

Questions Related to complex numbers

Multiple choice the nth roots of unity complex numbers maths

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}) $

  1. $3$
  2. $0$
  3. $n$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=3$
Multiple choice the nth roots of unity complex numbers maths

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

  1. $2$
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the system of equations z^3 = -(conjugate(w)^7) and z^5 * w^11 = 1 involves substituting magnitudes and arguments. The system yields two distinct solutions for the complex number w.

Multiple choice the nth roots of unity complex numbers maths

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 :

  1. ${ n } _{ 1 }={ n } _{ 2 }+1$
  2. ${ n } _{ 1 }={ n } _{ 2 }-1$
  3. ${ n } _{ 1 }={ n } _{ 2 }$
  4. ${ n } _{ 1 }>0,{ n } _{ 2 }>0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression (1+i)^n1 + (1+i)^n1 + (1+i)^n2 + (1+i)^n2 simplifies to 2 * ((1+i)^n1 + (1+i)^n2). For this to be real, the imaginary parts must cancel or be zero, which occurs when n1 = n2 + 1.

Multiple choice the nth roots of unity complex numbers maths

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

  1. $i$
  2. $-i$
  3. $0$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum is a geometric series of complex numbers. Using the identity for the sum of roots of unity, the sum evaluates to i.

Multiple choice the nth roots of unity complex numbers maths

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-

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The general term is (k-1)(k-w)(k-w^2) = (k-1)(k^2 + k + 1) = k^3 - 1. Summing this from k=2 to n gives the result (n(n+1)/2)^2.

Multiple choice the nth roots of unity complex numbers maths

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

  1. n

  2. 0

  3. 1

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots of unity are solutions to z^n - 1 = 0. The expression (1-a1)(1-a2)...(1-an-1) is the evaluation of (z^n - 1)/(z - 1) at z=1, which equals n.

Multiple choice the nth roots of unity complex numbers maths

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: $z _1+z _2+z _3+z _4=0$.

  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.
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice the nth roots of unity complex numbers maths

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} $

  1. $\omega^2$
  2. $1$
  3. $\omega$
  4. $(\omega-\alpha _1)(\omega-\alpha _2)(\omega-\alpha _3)(\omega-\alpha _4)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

Multiple choice the nth roots of unity complex numbers maths

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

  1. $\displaystyle \frac {-n}{(1-\alpha)^2}$
  2. $\displaystyle \frac {-n}{1-\alpha}$
  3. $\displaystyle \frac {-2n}{1-\alpha}$
  4. $\displaystyle \frac {-2n}{(1-\alpha)^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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,
$S(1-\alpha)=-n$
$S=\dfrac{-n}{1-\alpha}$

Multiple choice the nth roots of unity complex numbers maths

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

  1. $\displaystyle\ 4$
  2. $\displaystyle\ 2$
  3. $\displaystyle\ 1$

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