Tag: complex numbers

Questions Related to complex numbers

If  $z _ { 1 }$  and  $z  _ { 2 }$  be the  $n ^ { th }$  roots of unity which subtend right angle at the origin. Then  $n$  must be of the form

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

  1. $4 k + 1$

  2. $4 k + 2$

  3. $4 k + 3$

  4. $4 k$

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Correct Option: A

The value of the expression $\left( \omega -1 \right) \left( \omega -{ \omega  }^{ 2 } \right) \left( \omega -{ \omega  }^{ 3 } \right) ...\left( \omega -{ \omega  }^{ n-1 } \right) ,$ where $\omega$ is the nth root of unity, is 

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

  1. $n{ \omega  }^{ n-1 }$

  2. $n{ \omega  }^{ n }$

  3. $\left( n-1 \right) { \omega  }^{ n }$

  4. $\left( n-1 \right) { \omega  }^{ n-1 }$

Show answer
Correct Option: A
Explanation:

We have, $\displaystyle { x }^{ n }-1=\left( x-1 \right) \left( x-\omega  \right) \left( x-{ \omega  }^{ 2 } \right) ...\left( x-{ \omega  }^{ n-1 } \right) $


$\displaystyle \Rightarrow \frac { { x }^{ n }-1 }{ x-\omega  } =\left( x-1 \right)  \left( x-{ \omega  }^{ 2 } \right) ...\left( x-{ \omega  }^{ n-1 } \right) $


Putting $x=\omega $ on both sides, we have

$\displaystyle \left( \omega -1 \right) \left( \omega -{ \omega  }^{ 2 } \right) ...\left( \omega -{ \omega  }^{ n-1 } \right) =\lim _{ x\rightarrow \omega  }{ \frac { { x }^{ n }-1 }{ x-\omega  }  } \left( \frac { 0 }{ 0 } form \right) $

$\displaystyle =\lim _{ x\rightarrow \omega  }{ \frac { n{ x }^{ n-1 } }{ 1 }  } =n{ \omega  }^{ n-1 }$

If $1,\ \alpha _{1},\ \alpha _{2},\ \alpha _{3},\ \alpha _{4},\ \alpha _{5},\ \alpha _{6}$ are sevan $7^{th}$ root of unity then $|(3-\alpha _{1})(3-\alpha _{3})(3-\alpha _{5})|$ is 

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

  1. $\sqrt {2186}$

  2. $\sqrt {1093}$

  3. $\sqrt {1023}$

  4. $\sqrt {511}$

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Correct Option: B

The maximum number of real root of the equation $\displaystyle x^{2n} - 1 = 0$ is

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

  1. $\displaystyle 2$

  2. $\displaystyle 3$

  3. $\displaystyle n$

  4. $\displaystyle 2n$

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Correct Option: A
Explanation:

$x^{2n}=1$
Now if  $n$ is odd we have
$x^{n}=\pm1 $
$x^{n}=1$ and $x^{n}=-1$
$x^{n}=-1$
$x=-1$
Now if $n$ is odd
$x^{n}-1=0$
$(x-1)(1+x+x^{2}+..x^{n-1})=0$
Hence $x=1$ and  the equation $1+x+x^{2}+..x^{n-1}=0$ gives $nth$  roots of unity.
Hence at most $2$ real roots.
Similarly if $n$  is even.
Then
$x^{n}=\pm1 $
$x^{n}=1$ and $x^{n}=-1$
Now 
$x^{n}=-1$ will given imaginary roots.
$x^{n}=1$ can be further simplified in to
$x^{\frac{n}{2}}=\pm1 $ and so on.
Hence we will get remaining pairs of imaginary roots ans two real roots $1$ and  $-1$  at the end.
Hence at-most $2$ real roots.

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

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

  1. -$\alpha ^2$

  2. 0

  3. $\alpha ^2$

  4. $\alpha $

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Correct Option: A
Explanation:

$\alpha$ is non real root of $x^6 = 1$

one possible complex value of 
$\alpha = \cos (2 \pi / 6) + i \sin (2\pi / 6)$
$\alpha = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i$
$\alpha^2 = \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i \right) \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i \right) = \dfrac{1}{4} + \dfrac{\sqrt{3}}{4} i + \dfrac{\sqrt{3}}{4} i - \dfrac{3}{4} = \dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i$
$\alpha^3 = \left(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i \right) \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i \right) = \dfrac{-1}{4} = \dfrac{\sqrt{3}}{4} i + \dfrac{\sqrt{3}}{4} - \dfrac{3}{4}= -1$
$\dfrac{\alpha^5 + \alpha^3 + \alpha + 1}{\alpha^2 + 1} = \alpha^3 + \dfrac{\alpha + 1}{\alpha^2 + 1}$
$= -1 + \dfrac{\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i + 1 \right)}{\left(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i + 1 \right)}$
$= \dfrac{\dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i - 1 + \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i + 1}{(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i + 1)}$
$= \dfrac{1}{\alpha^2 + 1} = \dfrac{\alpha^2}{\alpha^4 + \alpha^2}$
$= \alpha^4 = \alpha^2 . \alpha^2 = \left(\dfrac{\sqrt{3}}{2} i - \dfrac{1}{2} \right) \left(\dfrac{\sqrt{3}}{2} i - \dfrac{1}{2} \right) = -\dfrac{\sqrt{3}}{2} i - \dfrac{1}{2}$
$= \dfrac{\alpha^2}{\alpha^4 + \alpha^2} = \dfrac{\alpha^2}{\left(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i - \dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i\right)} = - \alpha^2$
Considering option C as $- \alpha^2$ , it is correct

The roots of the equation  $z^{5}+z^{4}+z^{3}+z^{2}+z+1=0$   are given by

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

  1. $-1$

  2. $\displaystyle -\frac{1}{2}+\frac{i\sqrt{3}}{2}$

  3. $\displaystyle \frac{1}{2}+\frac{i\sqrt{3}}{2}$

  4. $\displaystyle \frac{-1-i\sqrt{3}}{2}$

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

$\displaystyle { z }^{ 5 }+{ z }^{ 4 }+{ z }^{ 3 }+{ z }^{ 2 }+z+1=0\ \Rightarrow \frac { { z }^{ 6 }-1 }{ z-1 } =0\ \Rightarrow z\neq 1\quad & \quad { z }^{ 6 }=1=\cos { 0 } +i\sin { 0 } \ \Rightarrow z={ \left( \cos { 0 } +i\sin { 0 }  \right)  }^{ \frac { 1 }{ 6 }  }=\cos { \frac { 2k\pi  }{ 6 }  } +i\sin { \frac { 2k\pi  }{ 6 }  } \ \Rightarrow z=\cos { \frac { k\pi  }{ 3 }  } +i\sin { \frac { k\pi  }{ 3 }  } $
where $k=0,1,2,3,4,5$.

For $k=0$,
$z=1$ but $z\neq 1$

For $k=1$,
$\displaystyle z=\cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  } =\frac { 1+i\sqrt { 3 }  }{ 2 } $

For  $k=2$,
$\displaystyle z=\cos { \frac { 2\pi  }{ 3 }  } +i\sin { \frac { 2\pi  }{ 3 }  } =\frac { -1+i\sqrt { 3 }  }{ 2 } $

For  $k=3$,
$\displaystyle z=\cos { \frac { 3\pi  }{ 3 }  } +i\sin { \frac { 3\pi  }{ 3 }  } =-1$

For  $k=4$,
$\displaystyle z=\cos { \frac { 4\pi  }{ 3 }  } +i\sin { \frac { 4\pi  }{ 3 }  } =\frac { -1-i\sqrt { 3 }  }{ 2 } $

For $k=4$,
$\displaystyle z=\cos { \frac { 5\pi  }{ 3 }  } +i\sin { \frac { 5\pi  }{ 3 }  } =\frac { 1-i\sqrt { 3 }  }{ 2 } $

Hence, all the options A,B,C and D are correct.

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

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

  1. 4

  2. 2

  3. 1

  4. none of these

Show answer
Correct Option: A
Explanation:
$\alpha =\sqrt [ 5 ]{ 1 } ,$   ${ 2 }^{ \left| 1+\alpha +{ \alpha  }^{ 2 }+{ \alpha  }^{  3}-{ \alpha  }^{ -1 } -\alpha^{-2}\right|  }$

$\Rightarrow \alpha =1,$

$\Rightarrow { 2 }^{ \left| 1+1+1+1-1-1 \right|  }=2^2=4.$

${ 2 }^{ \left| 1+\alpha +{ \alpha  }^{ 2 }+{ \alpha  }^{  3}-{ \alpha  }^{ -1 } -\alpha^{-2}\right|  }=4$

Hence, the answer is $4.$

If $1,\omega ,\omega ^{2},....\omega ^{n-1}$ are $n,n^{th}$ roots ofunity then the value of $\left ( 13-\omega  \right )\left ( 13-\omega ^{n-1} \right )$ equals

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

  1. $ \displaystyle \frac{13^{n}+1}{3}$

  2. $ \displaystyle\frac{13^{n}-1}{3}$

  3. $ \displaystyle 13^{n}-1$

  4. None of these

Show answer
Correct Option: D
Explanation:

$(13-w)(13-w^{n-1})$
$=(13-w)(13-\dfrac{w^{n}}{w})$
$=(13-w)(13-\overline{w})$
$=169-13(w+\overline{w})+w\overline{w}$
$=169-13(Re(w))+1$
$=170-13(Re(w))$
Hence answer is none of these.

If $\displaystyle w\neq 1 $ is $n^{th}$ root of unity, then value of $\displaystyle \sum _{k=0}^{n-1}\left | z _{1}+w^{k}z _{2} \right |^{2} $ is

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

  1. $\displaystyle n\left ( \left | z _{1} \right |^{2}+\left | z _{2} \right |^{2} \right )$

  2. $\displaystyle \left | z _{1} \right |^{2}+\left | z _{2} \right |^{2}$

  3. $\displaystyle \left ( \left | z _{1} \right |+\left | z _{2} \right | \right )^{2}$

  4. $\displaystyle n\left ( \left | z _{1} \right |+\left | z _{2} \right | \right )^{2}$

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Correct Option: A

If  $\omega$  be a complex  $n ^ { t h }$  root of unity, then  $\sum _ { r = 1 } ^ { n } ( a r + b ) \omega ^ { r - 1 }$  is

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

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

  2. $\dfrac { n b } { 1 - n }$

  3. $\dfrac { n a } { \omega - 1 }$

  4. none of these

Show answer
Correct Option: A