Tag: de moivre’s theorem and its applications

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Questions Related to de moivre’s theorem and its applications

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

If $\left ( 2+z \right )^{6}+\left ( 2-z \right )^{6}=0$ and $\omega =\dfrac{2+z}{2-z}$

  1. $\displaystyle \omega =e^{i}\tfrac{\left (2p+1 \right )\pi }{6},p=0,1,2,3,4,5$
  2. $\displaystyle z=\frac{2\left ( \omega -1 \right )}{\omega +1}$
  3. $\displaystyle \omega = ( -1 )^(\frac{1}{6})$

  4. All of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle { \left( 2+z \right)  }^{ 6 }{ +\left( 2-z \right)  }^{ 6 }=0\quad \quad & \quad w=\frac { 2+z }{ 2-z } $

$\displaystyle \Rightarrow { \left( \frac { 2+z }{ 2-z }  \right)  }^{ 6 }=-1\ \Rightarrow { w }^{ 6 }=-1$

$\displaystyle { \therefore \quad w=\left( -1 \right)  }^{ \frac { 1 }{ 6 }  }$

$\displaystyle \because \quad \frac { 2+z }{ 2-z } =w\ \Rightarrow 2\left( w-1 \right) =z\left( w+1 \right) $

$\displaystyle \therefore \quad z=\frac { 2\left( w-1 \right)  }{ w+1 } $

$\displaystyle { \because \quad w=\left( -1 \right)  }^{ \frac { 1 }{ 6 }  }$

$\displaystyle w={ \left( \cos { \pi  } +i\sin { \pi  }  \right)  }^{ \frac { 1 }{ 6 }  }=\cos { \left( \frac { 2p\pi +\pi  }{ 6 }  \right) +i } \sin { \left( \frac { 2p\pi +\pi  }{ 6 }  \right)  } $       ..{De Moivre's Theorem}

Where$ p=0,1,2,3,4,5.$

$\displaystyle \Rightarrow w={ e }^{ i\frac { \left( 2p+1 \right) \pi  }{ 6 }  }$
Hence, option 'D' is correct.

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

Given $z$ is a complex number with modulus $1$. Then the equation $\dfrac{(1+ia)}{(1-ia)}$ = $z$ has

  1. all roots real and distinct

  2. two real and one imaginary

  3. three roots real and one imaginary

  4. one root real and three imaginary
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle { \left( \frac { 1+ia }{ 1-ia }  \right)  }^{ 4 }=z\quad $         ...(1)
$\displaystyle & \quad \left| z \right| =1$
$\displaystyle z=cisA=\cos { A } +i\sin { A } $
Substitute $z$  in equation (1)
$\displaystyle { \left( \frac { 1+ia }{ 1-ia }  \right)  }={ cisA }^{ \frac { 1 }{ 4 }  }=cis\frac { 2k\pi +A }{ 4 } $       ...{De Moivre's Theorem}
where $ k=0,1,2,3$

Let $\displaystyle B=\frac { 2k\pi +A }{ 4 } $

$\displaystyle \Longrightarrow ia=\frac { -1+cisB }{ 1+cisB } =\frac { \sin { \frac { B }{ 2 } \left( i\cos { \frac { B }{ 2 }  } -\sin { \frac { B }{ 2 }  }  \right)  }  }{ \cos { \frac { B }{ 2 } \left( \cos { \frac { B }{ 2 }  } +i\sin { \frac { B }{ 2 }  }  \right)  }  } $

$\displaystyle \Longrightarrow ia=\frac { i\sin { \frac { B }{ 2 } \left( \cos { \frac { B }{ 2 }  } +i\sin { \frac { B }{ 2 }  }  \right)  }  }{ \cos { \frac { B }{ 2 } \left( \cos { \frac { B }{ 2 }  } +i\sin { \frac { B }{ 2 }  }  \right)  }  } $

$\displaystyle \Longrightarrow a=\tan { \frac { B }{ 2 }  } $

Therefore roots are real and distinct.

Ans: A

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

If n is a natural number$ \ge$ 2, such that $z^n = (z+ 1)^n$, then 

  1. roots of equation lie on a straight line parallel to the y-axis

  2. roots of equation lie on a straight line parallel to the x-axis

  3. sum of the real parts of the roots is -[(n-1)/2]

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

$\displaystyle { z }^{ n }={ \left( z+1 \right)  }^{ n }$   where, $n\ge 2$     ...(1)

$\displaystyle \Rightarrow { \left( \frac { z+1 }{ z }  \right)  }^{ n }=1=\cos { 0 } +i\sin { 0 } $

$\displaystyle \Rightarrow \frac { z+1 }{ z } ={ \left( \cos { 0 } +i\sin { 0 }  \right)  }^{ \frac { 1 }{ n }  }$

$\displaystyle \Rightarrow \frac { z+1 }{ z } =\cos { \frac { 2k\Pi  }{ n }  } +i\sin { \frac { 2k\Pi  }{ n }  } $      ...{De Moivre's Theorem}

$\displaystyle \Rightarrow z=\frac { -1 }{ 1-\cos { \frac { 2k\Pi  }{ n }  } -i\sin { \frac { 2k\Pi  }{ n }  }  } \quad =\frac { -1 }{ 2\sin { \frac { k\Pi  }{ n } \left( \sin { \frac { k\Pi  }{ n } -i } \cos { \frac { k\Pi  }{ n }  }  \right)  }  } =\frac { -1\left( \sin { \frac { k\Pi  }{ n } +i } \cos { \frac { k\Pi  }{ n }  }  \right)  }{ 2\sin { \frac { k\Pi  }{ n }  }  } $

$\displaystyle \Rightarrow z=\frac { -\left( 1+i\cot { \frac { k\Pi  }{ n }  }  \right)  }{ 2 } $

Where$ k=1,2,3,....,n-1 $     ..{Since at k=0, z is not defined}

$\because \quad Re\left( z \right) $ is constant.
Therfore roots of ${ z }^{ n }={ \left( z+1 \right)  }^{ n }$ lie on straight line parellel to y-axis.

$\displaystyle \because \quad z=\frac { -\left( 1+i\cot { \frac { k\Pi  }{ n }  }  \right)  }{ 2 } $ and $k=1,2,3,....,(n-1)$

Sum of $\displaystyle Re\left( z \right) $= $-\frac { \left( n-1 \right)  }{ 2 } $.

Ans: A,C

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

For positive integers $\displaystyle n _{1}$ and $\displaystyle n _{2}$ the value of the expression $\displaystyle (1+i)^{n _{1}}+(1+i^{3})^{n _{1}}+(1+i^{5})^{n _{2}}+(1+i^{2})^{n _{2}}$ where
$\displaystyle i= \sqrt{-1}$ is a real number iff

  1. $\displaystyle n _{1}= n _{2}$
  2. $\displaystyle n _{2}= n _{2}-1$
  3. $\displaystyle n _{1}= n _{2}+1$
  4. $\displaystyle \forall n _{1}$ and $\displaystyle n _{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(1+i)^{n _{1}}  = $ $  ^{n _{1}}C _{0} + ^{n _{1}}C _{1} i +^{n _{1}}C _{2} i^2 + ......+^{n _{1}}C _{n _{1}} i^{n _{1}}$ --------(1)


$(1+i^3)^{n _{1}}  =(1-i)^{n _{1}}=  $ $ ^{n _{1}}C _{0} - ^{n _{1}}C _{1} i +^{n _{1}}C _{2} i^2 - ......+^{n _{1}}C _{n _{1}} i^{n _{1}}$--------(2)

$(1+i^5)^{n _{2}}  =(1+i)^{n _{2}}=  $ $ ^{n _{2}}C _{0} + ^{n _{2}}C _{1} i +^{n _{2}}C _{2} i^2 + ......+^{n _{2}}C _{n _{2}} i^{n _{2}}$--------(3)

$(1+i^7)^{n _{2}}  =(1-i)^{n _{2}}=  $ $ ^{n _{2}}C _{0} - ^{n _{2}}C _{1} i +^{n _{2}}C _{2} i^2 - ......+^{n _{2}}C _{n _{2}} i^{n _{2}}$--------(4)

Adding (1),(2),(3) and (4),

$(1+i)^{n _{1}} +(1+i^3)^{n _{1}} +(1+i^5)^{n _{2}} +(1+i^7)^{n _{2}}$ 
$= 2(^{n _{1}}C _{0} +^{n _{1}}C _{2}i^2 +^{n _{1}} C _{4}i^4 +...........)+2(^{n _{2}}C _{0} +^{n _{2}}C _{2}i^2 +^{n _{2}} C _4 i^4+...........)$

$ = 2(^{n _{1}}C _{0} -^{n _{1}}C _{2} +^{n _{1}} C _{4} +...........)+2(^{n _{2}}C _{0} -^{n _{2}}C _{2} +^{n _{2}} C _4 +...........)$
$\Rightarrow$  As there are only even powers of $i$, the expression is real for all  positive integers $n _{1}$ and $n _{2}$

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

If ${ x }^{ 6 }={ \left( 4-3i \right)  }^{ 5 }$, then the product of all of its roots is (where $\displaystyle \theta =-\tan ^{ -1 }{ \frac { 3 }{ 4 }  } $)

  1. ${ 5 }^{ 5 }\left( \cos { 5\theta } +i\sin { 5\theta } \right) $
  2. $-{ 5 }^{ 5 }\left( \cos { 5\theta } +i\sin { 5\theta } \right) $
  3. ${ 5 }^{ 5 }\left( \cos { 5\theta } -i\sin { 5\theta } \right) $
  4. $-{ 5 }^{ 5 }\left( \cos { 5\theta } -i\sin { 5\theta } \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle { x }^{ 6 }={ \left( 4-3i \right)  }^{ 5 }\Rightarrow { x }^{ 6 }={ 5 }^{ 6 }\left( \frac { 4 }{ 5 } -\frac { 3i }{ 5 }  \right) ={ 5 }^{ 5 }{ \left( \cos { \theta  } +i\sin { \theta  }  \right)  }^{ 5 }$

where $\displaystyle \theta =-\tan ^{ -1 }{ \frac { 3 }{ 4 }  } ={ 5 }^{ 5 }\left( \cos { 5\theta  } +i\sin { 5\theta  }  \right) $
$\displaystyle x={ 5 }^{ 5/6 }{ \left( \cos { 5\theta  } +i\sin { 5\theta  }  \right)  }^{ 1/6 }={ 5 }^{ 5/6 }\left[ \cos { \left( \frac { 2k\pi +5\theta  }{ 6 }  \right)  } +i\sin { \left( \frac { 2k\pi +5\theta  }{ 6 }  \right)  }  \right] $
${ x } _{ 1 }{ x } _{ 2 }{ x } _{ 3 }...{ x } _{ 6 }={ 5 }^{ 5 }\left( \cos { \left( 5\pi +5\theta  \right)  } +i\sin { \left( 5\pi +5\theta  \right)  }  \right) \ ={ 5 }^{ 5 }\left( -\cos { 5\theta  } -i\sin { 5\theta  }  \right) =-{ 5 }^{ 5 }\left( \cos { 5\theta  } +i\sin { 5\theta  }  \right) $

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

If $C _{o},C _{1},C _{2}...C _{n}$ are the Binomial coefficient in the expansion of $\left ( 1+x \right )^{n}$ then which is not correct

  1. $C _{0}-C _{2}+C _{4}-C _{6}+...=2\tfrac{n}{2}\cos \frac{n\pi }{4}$
  2. $C _{1}-C _{3}+C _{5}+...=2\tfrac{n}{2}\sin \frac{n\pi }{4}$
  3. $C _{1}+C _{5}+C _{9}+C _{13}+...=\tfrac{1}{2}\left ( 2^{n-1}+2\tfrac{n}{2}\sin \frac{n\pi }{4} \right )$

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

By Binomial theorem
${ \left( 1+x \right)  }^{ n }={ C } _{ 0 }+{ C } _{ 1 }x+{ C } _{ 2 }{ x }^{ 2 }+{ C } _{ 3 }{ x }^{ \ 3 }+{ C } _{ 4 }{ x }^{ 4 }+...$      ...(1)

Substitute $x=i$ in equation (1), we get 
$\Rightarrow { \left( 1+i \right)  }^{ n }={ C } _{ 0 }+{ C } _{ 1 }i-{ C } _{ 2 }-{ C } _{ 3 }i+{ C } _{ 4 }+.....$       ...(2)

Substitute $x=-i$ in equation (1), we get
${ \left( 1-i \right)  }^{ n }={ C } _{ 0 }-{ C } _{ 1 }i-{ C } _{ 2 }+{ C } _{ 3 }i+{ C } _{ 4 }+.....$      ...(3)

Adding (2) & (3), we have
$\Rightarrow 2\left( { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+.... \right) ={ \left( 1+i \right)  }^{ n }+{ \left( 1-i \right)  }^{ n }$

$\Rightarrow { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+....=\dfrac { { \left( 1+i \right)  }^{ n }+{ \left( 1-i \right)  }^{ n } }{ 2 } ={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { { \left( \cos { \dfrac { \pi  }{ 4 }  } +i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n }+{ \left( \cos { \dfrac { \pi  }{ 4 }  } -i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n } }{ 2 }  \right) $

$\Rightarrow { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+....={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { \cos { \dfrac { n\pi  }{ 4 }  } +i\sin { \dfrac { n\pi  }{ 4 } +\cos { \dfrac { n\pi  }{ 4 }  } -i\sin { \dfrac { n\pi  }{ 4 }  }  }  }{ 2 }  \right) $

$\Rightarrow { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+....={ 2 }^{ \dfrac { n }{ 2 }  }\cos { \dfrac { n\pi  }{ 4 }  } $

Subtracting (2) & (3), we have
$\Rightarrow 2i\left( { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+.... \right) ={ \left( 1+i \right)  }^{ n }-{ \left( 1-i \right)  }^{ n }$

$\Rightarrow { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+....=\dfrac { { \left( 1+i \right)  }^{ n }-{ \left( 1-i \right)  }^{ n } }{ 2i } ={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { { \left( \cos { \dfrac { \pi  }{ 4 }  } +i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n }-{ \left( \cos { \dfrac { \pi  }{ 4 }  } -i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n } }{ 2i }  \right) $

$\Rightarrow { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+....={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { \cos { \dfrac { n\pi  }{ 4 }  } +i\sin { \dfrac { n\pi  }{ 4 } -\cos { \dfrac { n\pi  }{ 4 }  } +i\sin { \dfrac { n\pi  }{ 4 }  }  }  }{ 2i }  \right) $

$\Rightarrow { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+.....={ 2 }^{ \dfrac { n }{ 2 }  }\sin { \dfrac { n\pi  }{ 4 }  } $        ...(4)

Substitute $x=1$ in equation (1), we get
$\Rightarrow { C } _{ 0 }+{ C } _{ 1 }{ +C } _{ 2 }+{ C } _{ 3 }+{ C } _{ 4 }+.....\quad ={ 2 }^{ n }$          ...(5)

Substitute $x=-1$ in equation (1), we have
$\ \Rightarrow { C } _{ 0 }{ -C } _{ 1 }{ +C } _{ 2 }-{ C } _{ 3 }+{ C } _{ 4 }+.....\quad =0$       ...(6)

Subtracting (5) & (6), we get
$\Rightarrow 2\left( { C } _{ 1 }+{ C } _{ 3 }+{ C } _{ 5 }+..... \right) ={ 2 }^{ n }$
$\Rightarrow { C } _{ 1 }+{ C } _{ 3 }+{ C } _{ 5 }+.....\quad ={ 2 }^{ n-1 }$       ...(7)

Adding (4) & (7), we get
$\Rightarrow 2\left( { C } _{ 1 }+{ C } _{ 5 }+{ C } _{ 9 }+..... \right) ={ 2 }^{ n-1 }+{ 2 }^{ \dfrac { n }{ 2 }  }\sin { \dfrac { n\pi  }{ 4 }  } $

$\Rightarrow { C } _{ 1 }+{ C } _{ 5 }+{ C } _{ 9 }+.....=\dfrac { 1 }{ 2 } \left( { 2 }^{ n-1 }+{ 2 }^{ \dfrac { n }{ 2 }  }\sin { \dfrac { n\pi  }{ 4 }  }  \right) $
Hence, option 'D' is correct.

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

If $ x+\dfrac{1}{x}=2\cos \theta \   and \ y+\dfrac{1}{y}=2\cos \phi$  then which of the following is not correct?

  1. $\displaystyle \frac{x}{y} +\frac{y}{x}=2\cos \left ( \theta -\phi \right )$
  2. $x^{m}y^{n}=\cos \left ( m\theta +n\phi \right )+i\sin \left ( m\theta +n\phi \right )$
  3. $x^{m}y^{n}+x^{-m}y^{-n}=2\cos \left ( m\theta +n\phi \right )$

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

$x+\dfrac { 1 }{ x } =2\cos { \theta  } \quad & \quad y+\dfrac { 1 }{ y } =2\cos { \phi  } \ $

$\Rightarrow x=\cos { \theta  } +i\sin { \theta  } =cis\theta \ \quad & \quad y=\cos { \phi  } +i\sin { \phi  } =cis\phi \ $

$\dfrac { x }{ y } +\dfrac { y }{ x } =\dfrac { cis\theta  }{ cis\phi  } +\dfrac { cis\phi  }{ cis\theta  } =cis\left( \theta -\phi  \right) +cis\left( -\theta +\phi  \right) $

$\therefore \quad \dfrac { x }{ y } +\dfrac { y }{ x } =2\cos { \left( \theta -\phi  \right)  } $

${ x }^{ m }{ y }^{ n }={ \left( cis\theta  \right)  }^{ m }{ \left( cis\phi  \right)  }^{ n }=\left( cism\theta  \right) \left( cisn\phi  \right) $          ...{ De Moivre's Theorem}

$\therefore \quad { x }^{ m }{ y }^{ n }=cis\left( m\theta +n\phi  \right) =\cos { \left( m\theta +n\phi  \right)  } +i\sin { \left( m\theta +n\phi  \right)  } $

${ x }^{ -m }{ y }^{ -n }={ \left( cis\theta  \right)  }^{ -m }{ \left( cis\phi  \right)  }^{ -n }=\left( cis\left( -m\theta  \right)  \right) \left( cis\left( -n\phi  \right)  \right) \ \therefore \quad { x }^{ -m }{ y }^{ -n }=cis\left( -m\theta -n\phi  \right) =\cos { \left( m\theta +n\phi  \right)  } -i\sin { \left( m\theta +n\phi  \right)  } \ $

$\therefore \quad { x }^{ m }{ y }^{ n }+{ x }^{ -m }{ y }^{ -n }=2\cos { \left( m\theta +n\phi  \right)  } $

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


Let $\mathrm{z}=\cos\theta+\mathrm{i}\sin\theta$. Then the value of $\displaystyle \sum _{\mathrm{m}=1}^{15}{\rm Im}(\mathrm{z}^{2\mathrm{m}-1})$ at $\theta =2^{\mathrm{o}}$ is

  1. $\displaystyle \frac{1}{\sin 2^{\mathrm{o}}}$
  2. $\displaystyle \frac{1}{3\sin 2^{\mathrm{o}}}$
  3. $\displaystyle \frac{1}{2\sin 2^{\mathrm{o}}}$
  4. $\displaystyle \frac{1}{4\sin 2^{\mathrm{o}}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\mathrm{z}=\cos\theta+\mathrm{i}\sin\theta$
$\Rightarrow z^{2m-1} = \cos(2m-1)\theta+i\sin (2m-1)\theta$


Let $X = \displaystyle \sum _{\mathrm{m}=1}^{15}{\rm Im}(\mathrm{z}^{2\mathrm{m}-1})$

$\therefore \mathrm{X}=\sin\theta+\sin 3\theta+\ldots+\sin 29\theta$


$ 2(sin\theta)X=2\sin\theta\sin\theta + 2\sin\theta\sin3\theta...........+2\sin\theta\sin29\theta$
$\Rightarrow  2(\sin\theta)\mathrm{X}=1-\cos 2\theta+\cos 2\theta-\cos 4\theta+\ldots+\cos 28\theta-\cos 30\theta$

$\displaystyle \therefore  \mathrm{X}=\frac{1-\cos 30\theta}{2\sin\theta}=\frac{1}{4\sin 2^{\mathrm{o}}}$