Tag: complex numbers

Quizzes (9) Questions (142)

Questions Related to complex numbers

Multiple choice the nth roots of unity complex numbers maths

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

  1. $\alpha ^2$
  2. $0$
  3. $-\alpha ^2$
  4. $\alpha $
Reveal answer Fill a bubble to check yourself
C Correct answer
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) $
$\displaystyle \Rightarrow \frac { 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 } }{ { 1+\alpha  }^{ 2 } } =-{ \alpha  }^{ 2 }$

Multiple choice the nth roots of unity complex numbers maths

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?

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

Multiple choice the nth roots of unity complex numbers maths

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

  1. $(4, 7)$
  2. $(-4, -7)$
  3. $(-4, 7)$
  4. $(4, -7)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $q = 4 + 3 = 7$

Multiple choice the nth roots of unity complex numbers maths

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

  1. ${ \omega }^{ { 1 }/{ k } }$
  2. ${ \omega }^{ -1 }$
  3. ${ \omega }^{ n-k }$
  4. ${ \omega }^{ { n }/{ k } }$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\omega^{k}$
$={\omega^{-k}}$
$=\omega^{n-k}$. 

Multiple choice the nth roots of unity complex numbers maths

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

  1. Square

  2. Rectangle

  3. Parallelogram

  4. Rhombus
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice the nth roots of unity complex numbers maths

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:

  1. $0$
  2. $1$
  3. $n$
  4. $n^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $(1 - \omega) (1 - \omega^2) (1 - \omega^{n - 1}) = n$

Multiple choice the nth roots of unity complex numbers maths

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

  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 $\pm 1$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 
Multiple choice the nth roots of unity complex numbers maths

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

  1. $40$
  2. $9\sqrt {5}$
  3. $45$
  4. $35$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Multiple choice the nth roots of unity complex numbers maths

$1 , z _1, z _2, z _3, ..., z _{n-1}$ 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  

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

Multiple choice the nth roots of unity complex numbers maths

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

  1. $\displaystyle \frac{5^{n}-2}{4}$
  2. $\displaystyle \frac{5^{n}+2}{4}$
  3. $\displaystyle \frac{5^{n}+1}{4}$
  4. $\displaystyle \frac{5^{n}-1}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.