Tag: demoivre's theorem

Quizzes (2) Questions (102)

Questions Related to demoivre's theorem

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

If $1,{ \alpha  } _{ 1 },{ \alpha  } _{ 2 },{ \alpha  } _{ 3 }$ and $\alpha _4$ be the roots of $x^5-1=0$, then $\displaystyle \frac { \omega -{ \alpha  } _{ 1 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 1 } } .\frac { \omega -{ \alpha  } _{ 2 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 2 } } .\frac { \omega -{ \alpha  } _{ 3 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 3 } } .\frac { \omega -{ \alpha  } _{ 4 } }{ { \omega  }^{ 2 }-{ \alpha  } _{ 4 } } =$ 

  1. $1$
  2. $\omega$
  3. ${ \omega }^{ 2 }$

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

Since $1,{ \alpha  } _{ 1 },{ \alpha  } _{ 2 },{ \alpha  } _{ 3 },{ \alpha  } _{ 4 }$ are roots of the equation ${ x }^{ 5 }-1=0$

Thus ${ x }^{ 5 }-1=\left( x-1 \right) \left( x-{ \alpha  } _{ 1 } \right) \left( x-{ \alpha  } _{ 2 } \right) \left( x-{ \alpha  } _{ 3 } \right) \left( x-{ \alpha  } _{ 4 } \right)$
$ \Rightarrow \displaystyle\frac { { x }^{ 5 }-1 }{ \left( x-1 \right)  } =\left( x-{ \alpha  } _{ 1 } \right) \left( x-{ \alpha  } _{ 2 } \right) \left( x-{ \alpha  } _{ 3 } \right) \left( x-{ \alpha  } _{ 4 } \right) $........1
Putting $x=w$ in 1 we get,
$\Rightarrow \displaystyle\frac { { w }^{ 5 }-1 }{ \left( w-1 \right)  } =\left( w-{ \alpha  } _{ 1 } \right) \left( w-{ \alpha  } _{ 2 } \right) \left( w-{ \alpha  } _{ 3 } \right) \left( w-{ \alpha  } _{ 4 } \right) $
$\Rightarrow \displaystyle\frac { { w }^{ 2 }-1 }{ \left( w-1 \right)  } =\left( w-{ \alpha  } _{ 1 } \right) \left( w-{ \alpha  } _{ 2 } \right) \left( w-{ \alpha  } _{ 3 } \right) \left( w-{ \alpha  } _{ 4 } \right) $............2
Putting $x={ w }^{ 2 }$ in 1 we get
$\Rightarrow \displaystyle\frac { { w }^{ 10 }-1 }{ \left( { w }^{ 2 }-1 \right)  } =\left( { w }^{ 2 }-{ \alpha  } _{ 1 } \right) \left( { w }^{ 2 }-{ \alpha  } _{ 2 } \right) \left( { w }^{ 2 }-{ \alpha  } _{ 3 } \right) \left( { w }^{ 2 }-{ \alpha  } _{ 4 } \right) $
$\Rightarrow \displaystyle\frac { { w }-1 }{ \left( { w }^{ 2 }-1 \right)  } =\left( { w }^{ 2 }-{ \alpha  } _{ 1 } \right) \left( { w }^{ 2 }-{ \alpha  } _{ 2 } \right) \left( { w }^{ 2 }-{ \alpha  } _{ 3 } \right) \left( { w }^{ 2 }-{ \alpha  } _{ 4 } \right) $..........3
Dividing 2 by 3 we get, 
$\Rightarrow \displaystyle\frac { \left( { w }-{ \alpha  } _{ 1 } \right) \left( { w }-{ \alpha  } _{ 2 } \right) \left( { w }-{ \alpha  } _{ 3 } \right) \left( { w }-{ \alpha  } _{ 4 } \right)  }{ \left( { w }^{ 2 }-{ \alpha  } _{ 1 } \right) \left( { w }^{ 2 }-{ \alpha  } _{ 2 } \right) \left( { w }^{ 2 }-{ \alpha  } _{ 3 } \right) \left( { w }^{ 2 }-{ \alpha  } _{ 4 } \right)  } =\frac { { \left( { w }^{ 2 }-1 \right)  }^{ 2 } }{ { \left( { w }-1 \right)  }^{ 2 } } $
$=\displaystyle\frac { { w }^{ 4 }+1-2{ w }^{ 2 } }{ { w }^{ 2 }+1-2w } =\frac { { w }+1-2{ w }^{ 2 } }{ { w }^{ 2 }+1-2w } =\frac { -{ w }^{ 2 }-2{ w }^{ 2 } }{ -w-2w } =w$

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

Let, $z _1$ and $z _2$ be $n$th roots of unity which subtend a right angle at the origin. Then n must be of the from 

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

$z _{1}$ and $z _{2}$ subtend right angle at the origin. 

Then 

$arg \dfrac{z _{1}}{z _{2}} = \dfrac{\pi}{2}$

$\therefore \dfrac{z _{1}}{z _{2}} = \cos \dfrac{\pi}{2} +i\sin\dfrac{\pi}{2} 0+i =i$

$\left (\dfrac{z _{1}}{z _{2}}\right )^n = i^n = 1$    ( nth root of unity )

$\therefore n = 4k$    (as $i^4 =1$)

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 cube root of unity, then the equation $\left|z-\omega\right|^{2}+\right|z-\omega^{2}\right|^{2}=\lambda$ will represent a circle if

  1. $\lambda \epsilon\left(0,\dfrac{3}{2}\right)$
  2. $\lambda \epsilon\left[\dfrac{3}{2},\infty\right)$
  3. $\lambda \epsilon\left(0,3\right)$
  4. $\lambda \epsilon\left[1,\infty\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation |z - w|^2 + |z - w^2|^2 = lambda represents a circle if the constant term after simplification is positive. For w being the cube root of unity, this simplifies to a circle for certain values of lambda.

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

Let $\displaystyle z _{1}$ and $\displaystyle z _{2}$ be the $n^{th}$ roots of unity, which are ends of a line segment that subtends a right angle at the origin. Then, $n$ must be of the form

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

The $n^{th}$ roots of unity lie on a circle, where the angle between any 2 consecutive roots is $\dfrac { 2\pi  }{ n } $.
Hence, $ \dfrac{2r\pi}{n} = \dfrac{\pi}{2}$
$\Rightarrow 4r=n$
Hence, option D is correct.

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

If $z _{1},z _{2}$be two $nth$ roots of unity such that they represent two point $A,B$ in the Argand plane where $\angle AOB=60^{\circ}$ and $O$ is the orgin then the positive integer $n$ is of the form 

  1. $4k,k\:\epsilon\:N$
  2. $4k+3,k\:\epsilon\:N$
  3. $6k,k\:\epsilon\:N$
  4. $6k+5,k\:\epsilon\:N$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From above concept


Let consider $k=0$ for $z _{1}$

So second root will be in $60^{\circ}$ 

Thus $z _{2}=cos\dfrac{\pi}{3}+isin\dfrac{\pi}{3}$

Hence $\dfrac{\pi}{3}=\dfrac{2k\pi}{n}$

$\Rightarrow n=6k$

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

If ${ z } _{ 1 },{ z } _{ 2 }$ are two complex numbers and ${ \omega  }^{ k },k=0,1,...,n-1$ are the nth roots of unity, then $\displaystyle \sum _{ k=0 }^{ n-1 }{ { \left| { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right|  }^{ 2 } } $

  1. $<n\left( { \left| { z } _{ 1 } \right| }^{ 2 }+{ \left| { z } _{ 2 } \right| }^{ 2 } \right) $
  2. $=n\left( { \left| { z } _{ 1 } \right| }^{ 2 }+{ \left| { z } _{ 2 } \right| }^{ 2 } \right) $
  3. $>n\left( { \left| { z } _{ 1 } \right| }^{ 2 }+{ \left| { z } _{ 2 } \right| }^{ 2 } \right) $

  4. can't say
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have, $\displaystyle { \left| { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right|  }^{ 2 }=\left( { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right) \left( \overline { { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } }  \right) $


$\displaystyle =\left( { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right) \left( \overline { { z } _{ 1 } } +\overline { { z } _{ 2 } } { \omega  }^{ -k } \right) \quad \quad \quad \left[ { \omega  }^{ k }={ e }^{ i(2\pi k/n) }\Rightarrow { \omega  }^{ \overline { k }  }={ e }^{ -i(2\pi k/n) }={ \omega  }^{ -k } \right] $


$={ \left| { z } _{ 1 } \right|  }^{ 2 }+{ \left| { z } _{ 2 } \right|  }^{ 2 }+\overline { { z } _{ 1 } } { z } _{ 2 }{ \omega  }^{ k }+{ z } _{ 1 }\overline { { z } _{ 2 } } { \omega  }^{ -k }$

Therefore, we have

$\displaystyle \sum _{ k=0 }^{ n-1 }{ { \left| { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right|  }^{ 2 } } =n\left( { \left| { z } _{ 1 } \right|  }^{ 2 }+{ \left| { z } _{ 2 } \right|  }^{ 2 } \right) +\overline { { z } _{ 1 } } { z } _{ 2 }\sum _{ k=0 }^{ n-1 }{ { \omega  }^{ k } } +{ z } _{ 1 }\overline { { z } _{ 2 } } \sum _{ k=0 }^{ n-1 }{ { \omega  }^{ -k } } $

$\displaystyle =n\left( { \left| { z } _{ 1 } \right|  }^{ 2 }+{ \left| { z } _{ 2 } \right|  }^{ 2 } \right) \left[ \sum _{ k=0 }^{ n-1 }{ { \omega  }^{ k } } =\sum _{ k=0 }^{ n-1 }{ { \omega  }^{ -k } }  \right] $

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

If $\displaystyle \alpha$ is a non-real root of $\displaystyle x^{5}+1=0$ then $\displaystyle \alpha ^{10n+2}+\alpha ^{5n+2}+\alpha ^{5n}$, where n is an odd positive integer,has the value

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

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

$x^{5}=-1$
Hence
$\alpha^{5}=-1$
Therefore
$\alpha^{10n+2}+\alpha^{5n+2}+\alpha^{5n}$
$=(\alpha^{5n})^{2}\alpha^{2}+(\alpha^{5n}).\alpha^{2}+\alpha^{5n}$
$=(-1)^{2}\alpha^{2}+(-1)\alpha^{2}+\alpha^{5n}$ .... Since n is odd
$=-\alpha^{2}+\alpha^{2}-1$
$=-1$