Tag: de moivre’s theorem and its applications

Questions Related to de moivre’s theorem and its applications

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

What is the real part of $(\sin x + i \cos x)^{3}$ where $i = \sqrt {-1}$?

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

  1. $-\cos 3x$

  2. $-\sin 3x$

  3. $\sin 3x$

  4. $\cos 3x$

Show answer
Correct Option: B
Explanation:
${ (\sin x+i\cos x) }^{ 3 }={ \sin }^{ 3 }x-i{ \cos }^{ 3 }x+3i\sin x\cos x(\sin x+i\cos x)$ 
$={ \sin }^{ 3 }x-i{ \cos }^{ 3 }x+3i{ \sin }^{ 2 }x\cos x-3\sin x\cos^{ 2 }x$
$={ \sin }^{ 3 }x-3\sin x\cos^{ 2 }x+i(3{ \sin }^{ 2 }x\cos x-{ \cos }^{ 3 }x)$
Real part is ${ \sin }^{ 3 }x-3\sin x\cos^{ 2 }x$
$=\sin x({ \sin }^{ 2 }x-3\cos^{ 2 }x)$
$=\sin x(-3+3{ \sin }^{ 2 }x+{ \sin }^{ 2 }x)$ 
$=\sin x(-3+4{ \sin }^{ 2 }x)$
$=-(3\sin x-4{ \sin }^{ 3 }x)$
$=-\sin3x$

If $(\cos  \theta  + i  \sin  \theta)(\cos  2 \theta 
+ i  \sin  2  \theta) ... (\cos  n  \theta + i  \sin  n  \theta) = 1$, then the value of $\theta$ is , $m\in N$  

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

  1. $4m\pi$

  2. $\displaystyle \frac{2m\pi}{n(n+1)}$

  3. $\displaystyle \frac{4m\pi}{n(n+1)}$

  4. $\displaystyle \frac{m\pi}{n(n+1)}$

Show answer
Correct Option: C
Explanation:

Changing the above expression to Eular's form, we get
$e^{i\theta}e^{2i\theta}e^{3i\theta}...e^{in\theta})=1$
$e^{i(\theta+2\theta+3\theta+...n\theta}=1$
$e^{i\cfrac{n(n+1)}{2}\theta}=e^{2m\pi}$
Therefore, simplifying we get
$\dfrac{n(n+1)}{2}\theta=2m\pi$
$\theta=\dfrac{4m\pi}{n(n+1)}$

Statement 1: The product of all values of $(cos\alpha+i sin \alpha)^{\frac {3}{5}}$ is $cosn 3\alpha+i sin 3\alpha$.
Statement 2: The product of fifth roots of unity is 1.

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

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.

Show answer
Correct Option: B
Explanation:

Let
$z=(cos\theta+isin\theta)^{3}$
Taking the fifth root, we get
$z^{\dfrac{1}{5}}=(cos\theta+isin\theta)^{\dfrac{3}{5}}=x$
Now there will be $5$, corresponding values of $x$.
Product if all the $5$ values will be
$x^{5}$
$=(cos\theta+isin\theta)^{3}$
$=cos3\theta+isin3\theta$ ... using De-Moivre's rule.
Now consider $x^{n}=1$
Hence if $n$ is odd, the nth roots of unity will be
$1,a _{1},\overline{a _{1}},a _{2},\overline{a _{2}}....$
Now $a _{1}.\overline{a _{1}}=|a _{1}|^{2}=1$
$a _{2}.\overline{a _{2}}=|a _{2}|^{2}=1$
:
:
Hence one root will be one, and the rest $n-1$ roots will occur in pair with its conjugate.
Hence product will be $1$.
Substituting, $n=5$, we get the roots as
$1,a _{1},\overline{a _{1}},a _{2},\overline{a _{2}}$
Hence product if all the roots will be $1$. And sum of all the roots will be $0$.
Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

If $z _1$ and $z _2$ are the complex roots of the equation $(x-3)^3+1 = 0$, then $z _1 + z _2$ equals to 

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

  1. 1

  2. 3

  3. 5

  4. 7

Show answer
Correct Option: D
Explanation:

$cis\left( \theta  \right) =\cos { \theta  } +i\sin { \theta  } $
De Moivre's Theorem for fractional power:
${ \left( cis\theta  \right)  }^{ \frac { 1 }{ n }  }=cis\left( \frac { 2k\Pi +\theta  }{ n }  \right) $

${ \left( x-3 \right)  }^{ 3 }+1=0$
$\Longrightarrow x=3+{ \left( cis\left( \Pi  \right)  \right)  }^{ \frac { 1 }{ 3 }  }$
$x=3+{ \left( cis\left( \frac { 2k\Pi +\Pi  }{ 3 }  \right)  \right)  }$      ...{De Moivre's Theorem}
Where, $k=0,1,2$
for  $k=0$,
$x _{ 1 }=3+cis\left( \frac { \Pi  }{ 3 }  \right)$ 

for $k=1$,
$x _{ 2 }=3+cis\left( \Pi  \right) $

for $k=2,$
$x _{ 3 }=3+cis\left( \frac { 5\Pi  }{ 3 }  \right) $

$\Longrightarrow { x } _{ 1 }+{ x } _{ 3 }=6+cis\left( \frac { \Pi  }{ 3 }  \right) +cis\left( \frac { 5\Pi  }{ 3 }  \right) \ \Longrightarrow { x } _{ 1 }+{ x } _{ 3 }=7$
 
Ans: D