Mathematics

Complex Variables and Numbers

157 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 finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $1,$$\alpha _{1},\alpha _{2,} \alpha _{3},\alpha _{4}$ be the  roots of $z^{5}-1=0$ and $\omega $ be an imaginary cube root of unity, 


then  $ \displaystyle \left ( \frac{\omega -\alpha _{1}}{\omega ^{2}-\alpha _{1}} \right )\left ( \frac{\omega -\alpha _{2}}{\omega ^{2}-\alpha _{2}} \right )\left ( \frac{\omega -\alpha _{3}}{\omega ^{2}-\alpha _{3}} \right )\left ( \frac{\omega -\alpha _{4}}{\omega ^{2}-\alpha _{4}} \right )$ is ?

  1. $\omega $
  2. $\omega^{2}$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have,
$z^{5}-1=\left ( z-1 \right )\left ( z-\alpha _{1} \right )\left ( z-\alpha _{2} \right )\left ( z-\alpha 3 \right )\left ( z-\alpha _{4} \right )$
putting z=w,
$\Rightarrow w^{5}-1=\left ( w-1 \right )\left ( w-\alpha _{1} \right )\left ( w-\alpha _{2} \right )\left ( w-\alpha _{3} \right )\left ( w-\alpha _{4} \right )..............(1)$
Similarly putting $z=w^{2}$
$\left ( w^{2} \right )^{5}-1=w^{10}-1=\left ( w^{2}-1 \right )\left ( w^{2}-\alpha _{1} \right )\left ( w^{2}-\alpha _{2}\right )\left ( w^{2} -\alpha _{3}\right )\left ( w^{2}-\alpha _{4} \right ).........(2)$
So the required expression reduces to 
$ \displaystyle =\frac{\left ( w^{5} -1\right )/\left ( w-1 \right )}{\left ( w^{10}-1 \right )/\left ( w^{2} -1\right )}$
$=\frac{\left ( w^{2} -1\right )/\left ( w-1 \right )}{\left ( w-1 \right )/\left ( w^{2} -1\right )}$
$\left [ \because w^{3}=1,w^{5} =w^{3}w^{2}=w^{2}\right ]$
$=\frac{\left ( w^{2}-1 \right )^{2}}{\left ( w-1 \right )^{2}}$
$=\frac{\left ( w-1 \right )^{2}\left ( w+1 \right )^{2}}{\left ( w-1 \right )^{2}}$
$=1+2w+w^{2}$
$=w\left [ \because 1+w^{2} =-w\right ]$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $\omega$ is the complex cube root of unity, then inverse of $\begin{bmatrix} \omega  & 0 & 0 \ 0 & { \omega  }^{ 2 } & 0 \ 0 & 0 & { \omega  }^{ 2 } \end{bmatrix}$ is

  1. $\begin{bmatrix} -\omega & 0 & 0 \\ 0 & { \omega }& 0 \\ 0 & 0 & { \omega }^{ 2 } \end{bmatrix}$
  2. $\begin{bmatrix} \omega^{2} & 0 & 0 \\ 0 & { \omega }& 0 \\ 0 & 0 & 1 \end{bmatrix}$
  3. $\begin{bmatrix} \omega^{3} & 0 & 0 \\ 0 & { \omega } & 0 \\ 0 & 0 & 1 \end{bmatrix}$
  4. $\begin{bmatrix} \omega & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & { \omega }^{ 2 } \end{bmatrix}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$M=\begin{bmatrix} w & 0 & 0 \\ 0 & { w }^{ 2 } & 0 \\ 0 & 0 & { w }^{ 2 } \end{bmatrix}$
$(M)=w3w^2.w^3$
$=w^3.w^2$
$=w^3(w^3=1)$
$=(1)$

$adjA\Rightarrow \begin{bmatrix} { w }^{ 5 } & 0 & 0 \\ 0 & { w }^{ 4 } & 0 \\ 0 & 0 & { w }^{ 3 } \end{bmatrix}$

${ A }^{ -1 }=\dfrac { 1 }{ 1 } \begin{bmatrix} { w }^{ 2 }.{ w }^{ 3 } & 0 & 0 \\ 0 & { w }.{ w }^{ 3 } & 0 \\ 0 & 0 & \left( { w }^{ 3 } \right)  \end{bmatrix}$

$=\dfrac { 1 }{ 1 } \begin{bmatrix} { w }^{ 2 }.{ w }^{ 3 } & 0 & 0 \\ 0 & { w }.{ w }^{ 3 } & 0 \\ 0 & 0 & \left( { w }^{ 3 } \right)  \end{bmatrix}$
option $B$ is correct

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

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

  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. $i$
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $z = \cos  \displaystyle \frac{\pi}{8} + i  \sin  \frac{\pi}{8}$
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 $z + z^{-1} = 1$, 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 $z _1$ and $z _2$ are the complex roots of the equation $(x-3)^3+1 = 0$, then $z _1 + z _2$ equals to 

  1. 1

  2. 3

  3. 5

  4. 7
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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 $\dfrac{(1+ia)}{(1-ia)}$ = $z$ 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)
$\displaystyle &amp; \quad \left| z \right| =1$
$\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