Mathematics

Complex Variables and Numbers

206 Questions

Complex numbers and variables form a crucial part of advanced mathematics syllabi. This topic covers roots of unity, Argand plane geometry, and Z-transforms. These concepts are frequently tested in engineering entrance exams and UPSC mathematics optional papers.

Roots of unityArgand plane geometryZ-transform sequencesExponential complex formsComplex number quadrantsRiemann hypothesis

Complex Variables and Numbers Questions

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

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).


  1. Both are correct

  2. Only first statement is true.

  3. Only second ststement is true

  4. None

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

De Moivre's theorem states (cos theta + i sin theta)^n = cos(n theta) + i sin(n theta) for integer n, and the values for fractional n are given by the corollary provided.

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

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

  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. $cis(3\alpha - 3\beta + 2\gamma )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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

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

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

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

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

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

  1. $0$
  2. $-\cfrac{1}{2}$
  3. $\cfrac{1}{2}$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let a = exp(i * 8pi/11). The sum is a + a^2 + a^3 + a^4 + a^5. This is a geometric series: a(1-a^5)/(1-a). The real part of this sum is -1/2.

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

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?

  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. $\\ \\ \\ { \left| { Z } _{ 1 } \right| }^{ 4 }+{ \left| { Z } _{ 2 } \right| }^{ 4 }={ \left| { Z } _{ 3 } \right| }^{ 8 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Calculate the modulus of Z1, Z2, Z3. |Z1| = |(1-i)/(1+i*sqrt(3))|^(1/6) = (sqrt(2)/2)^(1/6). Calculating the sum of squares leads to 3/2.

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

  1. $-\displaystyle\frac{\pi}{4}$
  2. $\displaystyle\frac{\pi}{4}$
  3. $-\displaystyle\frac{3\pi}{4}$
  4. $\displaystyle\frac{3\pi}{4}$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
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 $\dfrac{2n-1(\pi)}{4}$ $n\epsilon::Integers$

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 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

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 $z + z^{-1} = 1$, then $z^{100} + z^{-100}$ is equal to

  1. $i$
  2. $-i$
  3. $1$
  4. $-1$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $-1.$
Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

The modulus and amplitude of the complex number $[e^{3-i \tfrac{\pi}{4}}]^3$ are respectively.

  1. $e^9, \dfrac{\pi}{2}$
  2. $e^9, \dfrac{-\pi}{2}$
  3. $e^6, \dfrac{-3\pi}{4}$
  4. $e^9, \dfrac{-3\pi}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\arg(z)=\dfrac{-3\pi}{4}$.

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

If $x = \cos  \theta + i  \sin  \theta$ the value of $x^n + \dfrac{1}{x^n}$ is

  1. $2 \cos n \theta$
  2. $2 i \sin n \theta$
  3. $2 \sin n \theta$
  4. $2 i \cos n \theta$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 
$x^{n}+\dfrac{1}{x^{n}}=e^{in\theta}+e^{-in\theta}$
$=\cos n\theta+i\sin n\theta+\cos n\theta-i\sin n\theta$
$=2\cos n\theta$.
Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

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:

  1. $8$
  2. $12$
  3. $16$
  4. $24$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a regular polygon with n sides, the vertices are exp(i * 2pi * k / n). For adjacent vertices z1 and z2, the angle is 2pi/n. If z1 = r(cos theta + i sin theta), then Im z1 / Re z1 = tan theta = sqrt(2)-1. This corresponds to theta = pi/8. Since the angle between adjacent vertices is 2pi/n, and the symmetry relates to the origin, n=8.

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

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

  1. $-\cos 3x$
  2. $-\sin 3x$
  3. $\sin 3x$
  4. $\cos 3x$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 ${ \sin }^{ 3 }x-3\sin x\cos^{ 2 }x$
$=\sin x({ \sin }^{ 2 }x-3\cos^{ 2 }x)$
$=\sin x(-3+3{ \sin }^{ 2 }x+{ \sin }^{ 2 }x)$ 
$=\sin x(-3+4{ \sin }^{ 2 }x)$
$=-(3\sin x-4{ \sin }^{ 3 }x)$
$=-\sin3x$