Tag: de moivre’s theorem and its applications

Quizzes (0) Questions (38)

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

The number of solutions of equation $z^{10}-z^{5}+1=0$ are 

  1. only two solution

  2. No solution

  3. only five solution

  4. exactly 10
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${ z }^{ 10 }-{ z }^{ 5 }+1=0$

Let ${ z }^{ 5 }=w$
$\Rightarrow { w }^{ 2 }-w+1=0$

$\Rightarrow w=\frac { 1\pm \sqrt { 3 }  }{ 2 } =\cos { \frac { \Pi  }{ 3 }  } \pm \sin { \frac { \Pi  }{ 3 }  } =cis\left( \pm \frac { \Pi  }{ 3 }  \right) $


$\Rightarrow { z }^{ 5 }=cis\left( \pm \frac { \Pi  }{ 3 }  \right) $

Case 1:
${ z }^{ 5 }=cis\left( \frac { \Pi  }{ 3 }  \right) $

$\Rightarrow z={ \left( cis\left( \frac { \Pi  }{ 3 }  \right)  \right)  }^{ \frac { 1 }{ 5 }  }=cis\left( \frac { 2k\Pi +\Pi  }{ 15 }  \right) \ $        ...{De Moivre's Theorem}

Where k=0,1,2,3,4.
Therefore number of solutions are 5.

Case 2:
${ z }^{ 5 }=cis\left( -\frac { \Pi  }{ 3 }  \right) $

$\Rightarrow z={ \left( cis\left( -\frac { \Pi  }{ 3 }  \right)  \right)  }^{ \frac { 1 }{ 5 }  }=cis\left( \frac { 2k\Pi -\Pi  }{ 15 }  \right) $       ...{De Moivre's Theorem}

Where k=0,1,2,3,4.
Therefore number of solutions are 5.

From case 1 & case 2 total number of solutions of equation ${ z }^{ 10 }-{ z }^{ 5 }+1=0$ are 10.

Ans: D

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

If $\displaystyle z=1+\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}$, then

  1. $\displaystyle Re(z^{5})=\frac{\sqrt{3}}{2}$
  2. $\displaystyle Re(z^{5})=\frac{1}{2}$
  3. $\displaystyle Im(z^{5})=\frac{1}{2}$
  4. $\displaystyle Im(z^{5})=\frac{\sqrt{3}}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$z=1+\cos { \frac { 2\pi  }{ 3 }  } +i\sin { \frac { 2\pi  }{ 3 }  } =2\cos ^{ 2 }{ \frac { \pi  }{ 3 }  } +2i\sin { \frac { \pi  }{ 3 } \cos { \frac { \pi  }{ 3 }  }  } $

$\displaystyle \Rightarrow z=2\cos { \frac { \pi  }{ 3 }  } \left( \cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  }  \right) =\cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  } $

$\displaystyle \Rightarrow { z }^{ 5 }={ \left( \cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  }  \right)  }^{ 5 }=\cos { \frac { 5\pi  }{ 3 }  } +i\sin { \frac { 5\pi  }{ 3 }  } $        ...{De Moivre's Theorem}
 
$\displaystyle \Rightarrow { z }^{ 5 }=\frac { 1-i\sqrt { 3 }  }{ 2 } $

$\displaystyle \therefore \quad Re\left( { z }^{ 5 } \right) =\frac { 1 }{ 2 } \quad & \quad Im\left( { z }^{ 5 } \right) =\frac { -\sqrt { 3 }  }{ 2 } $
Hence, option B is correct.

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

Construct an equation whose roots are $n^{th}$ powers of the roots of the equation $\displaystyle x^{2}-2x\cos \theta +1= 0.$

  1. $\displaystyle x^{2}-2n\cos n\theta x+1= 0$
  2. $\displaystyle x^{2}-2n\cos \theta x+1= 0$
  3. $\displaystyle x^{2}-2\cos n\theta x+1= 0$
  4. $\displaystyle x^{2}-2\cos ^{n}\theta x+1= 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know, $\displaystyle \alpha = \cos \theta +i\sin \theta , \beta = \cos \theta -i\sin \theta $
$\displaystyle \alpha ^{n}= \cos n\theta +i\sin n\theta ,$
$\displaystyle \beta ^{n}= \cos n\theta -i\sin n\theta $
$\displaystyle S= 2\cos n\theta , P= 1 \therefore x^{2}-Sx+P= 0$
or $\displaystyle x^{2}-2\cos n\theta x+1= 0$ is the required equation.

Ans: C

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

If $z = \left(\displaystyle\frac{\sqrt3}{2} + \displaystyle\frac{i}{2}\right)^{2009}+\left(\displaystyle\frac{\sqrt3}{2} - \displaystyle\frac{i}{2}\right)^{2009}$, then 

  1. $Im(z) = 0$
  2. $Re(z) > 0$
  3. $Im(z) > 0$
  4. $Re(z) < 0, Im(z) > 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 }  } $

$z=\left( \dfrac { \sqrt { 3 }  }{ 2 } +\dfrac { i }{ 2 }  \right) ^{ 2009 }+\left( \dfrac { \sqrt { 3 }  }{ 2 } -\dfrac { i }{ 2 }  \right) ^{ 2009 }$

$\Rightarrow z=\left( \cos { \dfrac { \pi  }{ 6 }  } +i\sin { \dfrac { \pi  }{ 6 }  }  \right) ^{ 2009 }+\left( \cos { \dfrac { \pi  }{ 6 }  } -i\sin { \dfrac { \pi  }{ 6 }  }  \right) ^{ 2009 }$         ......{ De Moivre's Theorem}

$\Rightarrow z=\cos { \dfrac { 2009\pi  }{ 6 }  } +i\sin { \dfrac { 2009\pi  }{ 6 }  } +\cos { \dfrac { 2009\pi  }{ 6 }  } -i\sin { \dfrac { 2009\pi  }{ 6 }  } $

$\Rightarrow z=2\cos { \dfrac { 2009\pi  }{ 6 }  } $


Therefore, $Im(z)=0$

Ans: B

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

The roots of $\displaystyle \left ( -64a^{4} \right )^{\tfrac14}$ are

  1. $\displaystyle \pm 2a\left ( 1\pm i \right ).$
  2. $\displaystyle \pm a\left ( 1\pm i \right ).$
  3. $\displaystyle \pm 2a\left ( 1\pm 2i \right ).$
  4. $\displaystyle \pm a\left ( 1\pm 2i \right ).$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\displaystyle \left ( -64a^{4} \right )^{\tfrac14}= \left ( 2\sqrt{2} \right )a\left ( -1 \right )^{\tfrac14}$
We know that $\displaystyle -1= \cos \pi +i\sin \pi $
Now put $\displaystyle -1= r\cos \theta , 0= r\sin \theta $
$\displaystyle \therefore \left ( -64a^{4} \right )^{\tfrac14}= 2\sqrt{2}a.\left [ \cos \pi +i\sin \pi  \right ]^{\tfrac14}$
$\displaystyle = 2\sqrt{2a}\left [ \cos \left ( 2n\pi +\pi  \right )+i\sin \left ( 2n\pi +\pi  \right ) \right ]^{\tfrac14}$
$\displaystyle = 2\sqrt{2a}\left [ \cos \cfrac{2n\pi +\pi }{4}+i\sin \cfrac{2n\pi +\pi }{4} \right ],$
where n=0, 1, 2 and 3.Hence the required roots are
$\displaystyle 2\sqrt{2}a\left [ \cos \left ( \cfrac{\pi}{4} \right )+i\sin \left ( \cfrac{\pi}{4} \right ) \right ],$
$\displaystyle 2\sqrt{2}a\left [ \cos \left ( 3\cfrac{\pi}{4} \right )+i\sin \left ( 3\cfrac{\pi}{4} \right ) \right ],$
$\displaystyle 2\sqrt{2}a\left [ \cos \left ( 5\cfrac{\pi}{4} \right )+i\sin \left ( 5\cfrac{\pi}{4} \right ) \right ],$
$\displaystyle 2\sqrt{2}a\left [ \cos \left ( 7\cfrac{\pi}{4} \right )+i\sin \left ( 7\cfrac{\pi}{4} \right ) \right ],$
Thus the roots on putting the values are
$\displaystyle 2\sqrt{2}a\left ( \dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}} \right ), 2\sqrt{2}a\left (\dfrac{-1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}} \right ),$
$\displaystyle 2\sqrt{2}a\left ( \dfrac{-1}{\sqrt{2}}-\dfrac{i}{\sqrt{2}} \right ), 2\sqrt{2}a\left (\dfrac{1}{\sqrt{2}}-\dfrac{i}{\sqrt{2}} \right ).$
Hence the roots are $\displaystyle \pm 2a\left ( 1\pm i \right ).$

Ans: $A$
Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

The value of $(iz+z^5+z^8)$ when $z=\dfrac{\sqrt{3}+i}{2}$ is?

  1. $0$
  2. $-1$
  3. $\dfrac{-\sqrt{3}+i}{2}$
  4. $z$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$z=\dfrac{\sqrt{3}+i}{2}=\cos \dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}=cis\dfrac{\pi}{6}$


$iz=icis\dfrac{\pi}{6}=cis\dfrac{\pi}{3}$

$z^5=cis\dfrac{5\pi}{6}$

$z^8=cis\dfrac{8\pi}{6}=-cis\dfrac{\pi}{3}$

$\implies \left(cis\dfrac{5\pi}{6}\right)=\dfrac{-\sqrt{3}+i}{2}$ 

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

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

  1. -1

  2. 1

  3. 0

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

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

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

  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 $
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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 &amp; \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.

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

Put in the form  A +iB

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

  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. $\displaystyle\cos 41\theta -i\sin41\theta.$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

  1. $Re(z) = 0$
  2. $Im(z) = 0$
  3. $Re(z) > 0, \space Im(z) > 0$
  4. $Re(z) > 0, \space Im(z) < 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
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