Tag: de moivre’s theorem and its applications

Questions Related to de moivre’s theorem and its applications

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

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

  1. only two solution

  2. No solution

  3. only five solution

  4. exactly 10

Show answer
Correct Option: D
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

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

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

  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. $\displaystyle Im(z^{5})=\frac{\sqrt{3}}{2}$

Show answer
Correct Option: B
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.

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

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

  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. $\displaystyle x^{2}-2\cos ^{n}\theta x+1= 0$

Show answer
Correct Option: C
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

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 

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

  1. $Im(z) = 0$

  2. $Re(z) > 0$

  3. $Im(z) > 0$

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

Show answer
Correct Option: A
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

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

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

  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. $\displaystyle \pm a\left ( 1\pm 2i \right ).$

Show answer
Correct Option: A
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: $A$

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

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

  1. $0$

  2. $-1$

  3. $\dfrac{-\sqrt{3}+i}{2}$

  4. $z$

Show answer
Correct Option: C
Explanation:

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


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

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

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

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

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 &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.

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