Tag: de moivre’s theorem and its applications

Questions Related to de moivre’s theorem and its applications

If $iz^4 + 1 = 0$, then z can take the value

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

  1. $\displaystyle \frac{1 + i}{\sqrt 2}$

  2. $\cos \displaystyle \frac{\pi}{8} + i \sin \frac{\pi}{8}$

  3. $\displaystyle \frac{1}{4 i}$

  4. $i$

Show answer
Correct Option: B
Explanation:
$iz^4 + 1 = 0 \Rightarrow z^4 = - \displaystyle \frac{1}{i} = \frac{i^2}{i} = i$
Let $z^4 = \cos  \displaystyle \frac{\pi}{2} + i  \sin  \frac{\pi}{2}$
$\therefore z = \displaystyle \left [ \cos  \frac{\pi}{2} + i  \sin  \frac{\pi}{2} \right ]^{1/4}$
U\sin g De-Moivre's theorem,
$(\cos   \theta + i  \sin  \theta)^n = \cos   n \theta + i  \sin   n \theta, n  \varepsilon I$
Hence, $z = \cos  \displaystyle \frac{\pi}{8} + i  \sin  \frac{\pi}{8}$

The product of the values of $\displaystyle{\left[ {\cos {\pi  \over 3} + i\sin {\pi  \over 3}} \right]^{{3 \over 4}}}$ is

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

  1. $-1$

  2. $1$

  3. $i$

  4. $-i$

Show answer
Correct Option: B
Explanation:

Given: $\displaystyle{\left[ {\cos {\pi  \over 3} + i\sin {\pi  \over 3}} \right]^{{3 \over 4}}}$


$=[e^{i(\pi/3)}]^{(3/4)}=e^{i\pi(1/3)(3/4)}=e^{4\pi i}=cos4\pi+isin{4\pi}=1-0i=1$ 

Number of integral values of n for which the quantity ${n+i}^{4}$ where ${i}^{2}=-1$, is an integer is

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

  1. $1$

  2. $2$

  3. $3$

  4. Infinite

Show answer
Correct Option: D
Explanation:

Simply $n+i^{4}$ = $n+1$ which is integer for every integral value of n .

So there are infinite values of n .

De Moivre's theorem

$(\cos\theta +i\sin \theta )=\cos n\theta $ if n is an integer and $\cos n\theta +i \sin n\theta $ is one of the values of $(\cos\theta +i\sin\theta )^{n}$, if n is a fraction.

Corollary : The q values of ($(\cos\theta +i\sin\theta )^{\frac{1}{q}}$ are obtained from

cos $\frac{2n\pi +\theta }{q}+i\sin\frac{2n\pi +\theta }{q}$ by putting n = 0, 1, 2, ..., (q - 1).


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

  1. Both are correct

  2. Only first statement is true.

  3. Only second ststement is true

  4. None 

Show answer
Correct Option: A

If $a = {\mathop{\rm cis}\nolimits} \alpha ,b = cis\beta ,c = cis\gamma $ then $\dfrac{{{a^3}{b^3}}}{{{c^2}}} = $

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

  1. $cis(3\alpha + 3\beta + 2\gamma )$

  2. $cis(3\alpha + 3\beta - 2\gamma )$

  3. $cis( - 3\alpha - 3\beta + 2\gamma )$

  4. $cis(3\alpha - 3\beta + 2\gamma )$

Show answer
Correct Option: B
Explanation:

$a=cis\alpha=Cos\alpha+iSin\alpha= \ e^{i\alpha}$

$b=cis\beta=Cos\beta+iSin\beta= \ e^{i\beta}$      

$a=cis\gamma=Cos\gamma+iSin\gamma= \ e^{i\gamma}$

$\therefore \dfrac{a^3b^3}{c^2}=\dfrac{({e^{i\alpha}})^3({e^{i\beta}})^3}{({e^{i\gamma}})^2}$

$=\dfrac{e^{3i\alpha}e^{3i\beta}} {e^{2i\gamma}}=\ e^{i(3\alpha+3\beta-2\gamma)}$

$=Cos(3\alpha+3\beta-2\gamma)+iSin(3\alpha+3\beta-2\gamma)$

$=cis(3\alpha+3\beta-2\gamma)$

If $a=\cos { \left( \cfrac { 8\pi  }{ 11 }  \right)  } +i\sin { \left( \cfrac { 8\pi  }{ 11 }  \right)  } $, then $Re(a+{a}^{2}+{a}^{3}+{a}^{4}+{a}^{5})=$

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

  1. $0$

  2. $-\cfrac{1}{2}$

  3. $\cfrac{1}{2}$

  4. $1$

Show answer
Correct Option: B

For ${ Z } _{ 1 }=\sqrt [ 6 ]{ \dfrac { 1-i }{ 1+i\sqrt { 3 }  }  } $, ${ Z } _{ 2 }=\sqrt [ 6 ]{ \dfrac { 1-i }{ \sqrt { 3 } +i }  } $, ${ Z } _{ 3 }=\sqrt [ 6 ]{ \dfrac { 1+i }{ \sqrt { 3 } -i }  } $ which of the following holds goods?

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

  1. $\sum { { \left| { Z } _{ 1 } \right| }^{ 2 } } =\dfrac { 3 }{ 2 } $

  2. ${ \left| { Z } _{ 1 } \right| }^{ 4 }+{ \left| { Z } _{ 2 } \right| }^{ 4 }={ \left| { Z } _{ 3 } \right| }^{ -8 }$

  3. $\sum { { \left| { Z } _{ 1 } \right| }^{ 3 }+{ \left| { Z } _{ 2 } \right| }^{ 3 }={ \left| { Z } _{ 3 } \right| }^{ -6 } } $

  4. $\ \ \ { \left| { Z } _{ 1 } \right| }^{ 4 }+{ \left| { Z } _{ 2 } \right| }^{ 4 }={ \left| { Z } _{ 3 } \right| }^{ 8 }$

Show answer
Correct Option: A

Given z is a complex number with modulus 1. Then the equation $\left[\dfrac{(1+ia)}{(1-ia)}\right]^4$ = z has

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

  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

Show answer
Correct Option: A
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

If $\sqrt{5 - 12i} + \sqrt{-5 - 12i} = z$, then principal value of arg z can be 

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

  1. $-\displaystyle\frac{\pi}{4}$

  2. $\displaystyle\frac{\pi}{4}$

  3. $-\displaystyle\frac{3\pi}{4}$

  4. $\displaystyle\frac{3\pi}{4}$

Show answer
Correct Option: A,B,C,D
Explanation:

Dividing and multiplying by $\sqrt{13}$
$=\sqrt{13}[(\dfrac{5}{13}-\dfrac{12i}{13})^{\dfrac{1}{2}}+(-\dfrac{5}{13}-\dfrac{12i}{13})^{\dfrac{1}{2}}]$
$=\sqrt{13}[e^{i\dfrac{-\theta}{2}}+e^{i\dfrac{\theta-\pi}{2}}]$
$=\sqrt{13}[cos\dfrac{\theta}{2}-isin\dfrac{\theta}{2}+sin\dfrac{\theta}{2}-icos\frac{\theta}{2}]$
$=\sqrt{13}[cos\dfrac{\theta}{2}+sin\dfrac{\theta}{2}-i(sin\dfrac{\theta}{2}+cos\dfrac{\theta}{2})]$
$=z$
Here $\theta=sin^{-1}(\dfrac{12}{13})$
Hence
$|Re(z)|=|Im(z)|$
Hence argument of $Z$ is in the form of $\dfrac{2n-1(\pi)}{4}$ $n\epsilon::Integers$

The value of $\displaystyle { \left( \frac { 1+i }{ \sqrt { 2 }  }  \right)  }^{ 8 }+{ \left( \frac { 1-i }{ \sqrt { 2 }  }  \right)  }^{ 8 }$ is equal to

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

  1. $4$

  2. $6$

  3. $8$

  4. $2$

Show answer
Correct Option: D
Explanation:

We have $\displaystyle { \left( \frac { 1+i }{ \sqrt { 2 }  }  \right)  }^{ 8 }+{ \left( \frac { 1-i }{ \sqrt { 2 }  }  \right)  }^{ 8 }$


$\displaystyle={ \left[ \cos { \frac { \pi  }{ 4 }  } +i\sin { \frac { \pi  }{ 4 }  }  \right]  }^{ 8 }+{ \left[ \cos { \frac { \pi  }{ 4 }  } -i\sin { \frac { \pi  }{ 4 }  }  \right]  }^{ 8 }$


$=\cos { 2\pi  } +i\sin { 2\pi  } +\cos { 2\pi  } -i\sin { 2\pi  } $      [by de-moivre's theorem]

$=2\cos { 2\pi  } =2\left( 1 \right) =2$