Tag: complex numbers

Questions Related to complex numbers

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]

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

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:

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


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

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

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

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