Tag: demoivre's theorem

Questions Related to demoivre's theorem

The value of $\displaystyle \left ( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right )^{8}$

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. -1

  2. 1

  3. 0

  4. None of these

Show answer
Correct Option: A
Explanation:

$z={ \left( \sin { \frac { \pi  }{ 8 } +i } \cos { \frac { \pi  }{ 8 }  }  \right)  }^{ 8 }={ \left[ i\left( \cos { \frac { \pi  }{ 8 } -i\sin { \frac { \pi  }{ 8 }  }  }  \right)  \right]  }]^8$
     ...{$\because \quad { i }^{ 8 }=1$}
$\Rightarrow z=\cos { \pi -i\sin { \pi  }  } =-1$        ...{De Moivre's Theorem}
Hence, option 'A' is correct.

If $z=\cos 2\theta +i\sin 2\theta $ then which is correct 

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $\displaystyle \sum _{r=0}^{n}C _{r}\cos2r\theta =2^{n} \cos ^{n}\theta \cos n\theta $

  2. $\displaystyle \sum _{r=1}^{n}C _{r}\cos2r\theta =2^{n} \sin ^{n}\theta \cos n\theta $

  3. $\sum _{ r=0 }^{ n } C _{ r }\sin  2r\theta =2^{ n }\cos ^{ n } \theta \sin  n\theta $

  4. $\displaystyle \sum _{r=0}^{n}C _{r}\sin2r\theta =2^{n} \sin ^{n}\theta \sin n\theta $

Show answer
Correct Option: A,C
Explanation:

By Binomial Theorem
${ \left( 1+z \right)  }^{ n }={ C } _{ 0 }+{ C } _{ 1 }z+{ C } _{ 2 }{ z }^{ 2 }+{ C } _{ 3 }{ z }^{ \ 3 }+....+{ C } _{ n }{ z }^{ n }=\sum _{ r=0 }^{ n }{ { C } _{ r }{ z }^{ r } } $      ...(1)

Substituting $z=\cos { 2\theta  } +i\sin { 2\theta  }  $ in eq. (1), we get

${ \left( 1+\cos { 2\theta  } +i\sin { 2\theta  }  \right)  }^{ n }=\sum _{ r=0 }^{ n }{ { C } _{ r }\left( \cos { 2\theta  } +i\sin { 2\theta  }  \right) ^{ r } } $

$\Rightarrow \sum _{ r=0 }^{ n }{ { C } _{ r }\left( \cos { 2r\theta  } +i\sin { 2r\theta  }  \right)  }$      ...{De Moivre's Theorem}

$\Rightarrow { \left[ 2\cos { \theta  } \left( \cos { \theta  } +i\sin { \theta  }  \right)  \right]  }^{ n }=\sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } +i\sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  } $

$\Rightarrow \sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } +i\sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  } ={ 2 }^{ n }\cos ^{ n }{ \theta  } { \left( \cos { n\theta  } +i\sin { n\theta  }  \right)  }$         ...{De Moivre's Theorem}

$\Rightarrow \sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } +i\left( \sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  }  \right) ={ 2 }^{ n }\cos ^{ n }{ \theta  } \cos { n\theta  } +i\left( { 2 }^{ n }\cos ^{ n }{ \theta  } \sin { n\theta  }  \right) $

On comparing real and Imaginary parts, we get
$\sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } ={ 2 }^{ n }\cos ^{ n }{ \theta  } \cos { n\theta  } \quad & \quad \sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  } ={ 2 }^{ n }\cos ^{ n }{ \theta  } \sin { n\theta  } $
Hence, option 'A' and 'C' are correct.

Put in the form  A +iB

$\displaystyle \frac{\left ( \cos 2\theta -i\sin 2\theta  \right )^{7}\left ( \cos 3\theta +i\sin 3\theta  \right )^{-5}}{\left ( \cos 4\theta +i\sin 4\theta  \right )^{12}\left ( \cos 5\theta +i\sin 5\theta  \right )^{-6}}$

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $\displaystyle\cos 47\theta +i\sin47\theta.$

  2. $\displaystyle\cos 47\theta -i\sin47\theta.$

  3. $\displaystyle\cos 41\theta +i\sin41\theta.$

  4. $\displaystyle\cos 41\theta -i\sin41\theta.$

Show answer
Correct Option: B
Explanation:

Using De-Moivre's Theorem, the given expression
$\displaystyle = \frac{\left ( \cos \theta -i\sin \theta  \right )^{14}\left ( \cos \theta +i\sin \theta  \right )^{-15}}{\left ( \cos \theta +i\sin \theta  \right )^{48}\left ( \cos \theta +i\sin \theta  \right )^{-30}}$
$\displaystyle=\frac{\left ( e^{i\theta } \right )^{-29}}{\left ( e^{i\theta } \right )^{18}}=\left ( e^{i\theta } \right )^{-47}$
$\displaystyle=\left ( \cos \theta +i\sin \theta  \right )-^{47}=\cos 47\theta -\sin47\theta.$ 

Ans: B

If $z = \left(\displaystyle\frac{\sqrt3}{2}+\displaystyle\frac{i}{2}\right)^5 + \left(\displaystyle\frac{\sqrt3}{2}-\displaystyle\frac{i}{2}\right)^5,$ then

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $Re(z) = 0$

  2. $Im(z) = 0$

  3. $Re(z) > 0, \space Im(z) > 0$

  4. $Re(z) > 0, \space Im(z) < 0$

Show answer
Correct Option: B
Explanation:

As we know that,
$\dfrac { \sqrt { 3 }  }{ 2 } +\dfrac { i }{ 2 } =\cos { \dfrac { \pi  }{ 6 }  } +i\sin { \dfrac { \pi  }{ 6 }  } $
and $\dfrac { \sqrt { 3 }  }{ 2 } -\dfrac { i }{ 2 } =\cos { \dfrac { \pi  }{ 6 }  } -i\sin { \dfrac { \pi  }{ 6 }  } $

$z=\left( \dfrac { \sqrt { 3 }  }{ 2 } +\dfrac { i }{ 2 }  \right) ^{ 5 }+\left( \dfrac { \sqrt { 3 }  }{ 2 } -\dfrac { i }{ 2 }  \right) ^{ 5 }$

$\Rightarrow z=\left( \cos { \dfrac { \pi  }{ 6 }  } +i\sin { \dfrac { \pi  }{ 6 }  }  \right) ^{ 5 }+\left( \cos { \dfrac { \pi  }{ 6 }  } -i\sin { \dfrac { \pi  }{ 6 }  }  \right) ^{ 5 }$         ......{ De Moivre's Theorem}

$\Rightarrow z=\cos { \dfrac { 5\pi  }{ 6 }  } +i\sin { \dfrac { 5\pi  }{ 6 }  } +\cos { \dfrac { 5\pi  }{ 6 }  } -i\sin { \dfrac { 5\pi  }{ 6 }  } $

$\Rightarrow z=-\dfrac { \sqrt { 3 }  }{ 2 } $
Therefore, $Im(z)=0$

Ans: B

If $z + z^{-1} = 1$, then $z^{100} + z^{-100}$ is equal to

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $i$

  2. $-i$

  3. $1$

  4. $-1$

Show answer
Correct Option: D
Explanation:
If $z+z^{-1}=1,$ then $z^{100}+z^{-100}$
$\Rightarrow z+z^{-1}=1,$ when we multiply by $z$
$\Rightarrow z^2+1=z$
$\Rightarrow z^2-z+1=0$
By solving, $z=\dfrac { 1\pm \sqrt { 1-4 }  }{ 2 } =\dfrac { 1\pm i\sqrt { 3 }  }{ 5 } $
In polar form : $z=re^{iQ},$
$\Rightarrow r^2={ \left( \dfrac { 1 }{ 2 }  \right)  }^{ 2 }{ \left( \dfrac { \sqrt { 3 }  }{ 2 }  \right)  }^{ 2 }=\dfrac{1}{4}+\dfrac{3}{4}=1$
$\therefore r=1$
$\Rightarrow \tan \theta \dfrac { \pm \dfrac { \sqrt { 3 }  }{ 2 }  }{ \dfrac { 1 }{ 2 }  } =\pm \sqrt { 3 } $ i.e, $\theta =\pm \dfrac { \pi  }{ 3 } $ or $\pm \dfrac { 2\pi  }{ 3 } $
$\therefore z={ e }^{ \pm i{ \pi  }/{ 3 } }$ and $z^{-1}={ e }^{ \pm i{ \pi  }/{ 3 } }$
then $z^{100}={ e }^{ \pm i{ 100  }/{ 3 }\pi }=z^{\pm i\left(16\pi+\pi+1/3\pi\right)}$
$={ e }^{ \pm i{ \pi  }/{ 3 } }=-z$
$\therefore z^{-100}=-z^{-1}$
$\Rightarrow z^{100}+z^{-100}=-z-z^{-1}=-\left(z+z^{-1}\right)=-1$
Hence, the answer is $-1.$

The modulus and amplitude of the complex number $[e^{3-i \tfrac{\pi}{4}}]^3$ are respectively.

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $e^9, \dfrac{\pi}{2}$

  2. $e^9, \dfrac{-\pi}{2}$

  3. $e^6, \dfrac{-3\pi}{4}$

  4. $e^9, \dfrac{-3\pi}{4}$

Show answer
Correct Option: D
Explanation:

$z=(e^{3-\tfrac{i\pi}{4}})^{3}$
$=(e^{3}.e^{-\tfrac{i\pi}{4}})^{3}$
$=e^{9}.e^{-\tfrac{3i\pi}{4}}$
$=|z|e^{i\arg(z)}$ 
By comparing RHS and LHS we get
$|z|=e^{9}$ and $\arg(z)=\dfrac{-3\pi}{4}$.

If $\displaystyle\alpha =\cos { \left( \frac { 8\pi  }{ 11 }  \right)  } +i\sin { \left( \frac { 8\pi  }{ 11 }  \right)  } ,$ then $Re\left( \alpha +{ \alpha  }^{ 2 }+{ \alpha  }^{ 3 }+{ \alpha  }^{ 4 }+{ \alpha  }^{ 5 } \right) $ is equal to

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $\displaystyle\frac { 1 }{ 2 } $

  2. $\displaystyle-\frac { 1 }{ 2 } $

  3. $0$

  4. None of these

Show answer
Correct Option: B
Explanation:

$\displaystyle\alpha =\cos { \left( \frac { 8\pi  }{ 11 }  \right)  } +i\sin { \left( \frac { 8\pi  }{ 11 }  \right)  } $


We know, $z=\cos\theta+i\sin\theta=e^{i\theta}$
$\therefore$  $\alpha=e^{i\frac{8\pi}{11}}$
$\Rightarrow$  $\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5=\dfrac{\alpha(\alpha^5-1)}{\alpha-1}$             ........................... as forming G.P.

                                                  $=\dfrac{\alpha^6-\alpha}{\alpha-1}$

                                                  $=\dfrac{\left(e^{i\frac{8\pi}{11}}\right)^6-e^{i\frac{8\pi}{11}}}{e^{i\frac{8\pi}{11}}-1}$           ---- ( 1 )

$e^{-\frac{48\pi}{11}}=\cos\dfrac{48\pi}{11}+i\sin\dfrac{48\pi}{11}$

          $=\cos\left(4\pi+\dfrac{4\pi}{11}\right)+i\sin\left(4\pi+\dfrac{4\pi}{11}\right)$

          $=\cos\dfrac{4\pi}{11}+i\sin\dfrac{4\pi}{11}$

          $=e^{i\frac{4\pi}{11}}$

Substituting above value in ( 1 ) we get,
$\Rightarrow$  $\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5=\dfrac{e^{i\frac{4\pi}{11}}-e^{i\frac{8\pi}{11}}}{e^{i\frac{8\pi}{11}}-1}$

                                                  $=\dfrac{t-t^2}{t^2-1}$                       [ Let $e^{i\frac{4\pi}{11}}=t]$

                                                  $=\dfrac{-t(1-t)}{(t-1)(t+1)}$

                                                  $=\dfrac{-t}{t+1}$

                                                  $=\dfrac{-(\cos\frac{4\pi}{11}+i\sin\dfrac{4\pi}{11})}{\cos\dfrac{4\pi}{11}+i\sin\dfrac{4\pi}{11}+1}$

Let $a=\cos\dfrac{4\pi}{11}=a$ and $b=\sin\dfrac{4\pi}{11}$

                                                  $=\left(\dfrac{a+ib}{(a+1)+ib}\right)\times\dfrac{(a+1)-ib}{(a+1)-ib}$

                                                   $=-\dfrac{(1+ib)(a+1)-ib}{[(a+1)+ib][(a+1)-ib]}$

                                                   $=-\dfrac{(a(a+1)+b^2)+i(b(a+1)-ab)}{(a+1)^2+b^2}$

                                                   $=-\dfrac{a(a+1)+b^2}{(a+1)^2+b^2}$           [ Taking real part only ]

                                                   $=-\dfrac{a^2+b^2+a}{a^2+b^2+1+2a}$

                                                   $=-\dfrac{1+a}{2+2a}$

                                                   $=-\dfrac{(1+a)}{2(1+a)}$

                                                   $=\dfrac{-1}{2}$

If $x = \cos  \theta + i  \sin  \theta$ the value of $x^n + \dfrac{1}{x^n}$ is

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $2 \cos n \theta$

  2. $2 i \sin n \theta$

  3. $2 \sin n \theta$

  4. $2 i \cos n \theta$

Show answer
Correct Option: A
Explanation:
$x=\cos \theta+i\sin \theta$
Applying Euler's form
$x=\cos \theta+i\sin \theta=e^{i\theta}$.
Hence 
$x^{n}=e^{in\theta}$. 
Similarly 
$\dfrac{1}{x}=\bar{x}=\cos \theta-i\sin \theta=e^{-i\theta}$
Hence 
$\dfrac{1}{x^{n}}=e^{-in\theta}$.
Hence 
$x^{n}+\dfrac{1}{x^{n}}=e^{in\theta}+e^{-in\theta}$
$=\cos n\theta+i\sin n\theta+\cos n\theta-i\sin n\theta$
$=2\cos n\theta$.

If $\alpha, \beta$ are the roots of the equation $u^2-2u+2=0$ and if $\cot\theta=x+1$, then $[(x+\alpha)^n-(x+\beta)^m]/[\alpha-\beta]$ is equal to

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $\displaystyle \frac {\sin n\theta}{\sin^n\theta}$

  2. $\displaystyle \frac {\cos n\theta}{\cos^n\theta}$

  3. $\displaystyle \frac {\sin n\theta}{\cos^n\theta}$

  4. $\displaystyle \frac {\cos n\theta}{\sin^n\theta}$

Show answer
Correct Option: A
Explanation:

${ u }^{ 2 }-2u+2=0$
$\Longrightarrow \quad u=1\pm i$
So,$\alpha =1+i\quad and\quad \beta =1-i$
Now given that,$x=\cot { \theta  } -1$
so,$\displaystyle \frac { { (x+\alpha ) }^{ n }-{ (x+\beta ) }^{ n } }{ \alpha -\beta  } =\frac { { (\cot { \theta  } -1+1+i) }^{ n }-{ (\cot { \theta  } -1 }+1-i)^{ n } }{ 2i } \ \ $
$=\displaystyle \frac { { (\cot { \theta  } +i) }^{ n }-(\cot { \theta  } -i)^{ n } }{ 2i } =\frac { { (\cos { \theta  } +i\sin { \theta  } ) }^{ n }-{ (\cos { \theta  } -\sin { \theta  } ) }^{ n } }{ ({ \sin { \theta  }  })^{ n }(2i) } \ \ $
$=\displaystyle \frac { { e }^{ (in\theta ) }-{ e }^{ -(in\theta ) } }{ ({ \sin { \theta ) }  }^{ n }2i } \ \ $
=$\displaystyle \frac { (\cos { (n\theta ) } +i\sin { (n\theta )) } -(\cos { (n\theta ) } -i\sin { (n\theta )) }  }{ ({ \sin { \theta ) }  }^{ n }2i } =\frac { \sin { (n\theta ) }  }{ { (\sin { \theta ) }  }^{ n } } \ \ $

If $z _{1}$ and $\bar {z} _{1}$ represent adjacent of a regular polygon of $n$ sides with centre at the origin & if $\dfrac{Im\ z _{1}}{Re\ z _{1}}=\sqrt{2}-1$ then the value of $n$ is equal to:

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $8$

  2. $12$

  3. $16$

  4. $24$

Show answer
Correct Option: A