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

If $\omega$ is a complex cube root of unity, then the equation $\left|z-\omega\right|^{2}+\left|z-\omega^{2}\right|^{2}=\lambda$ will represent a circle if

  1. $\lambda \epsilon\left(0,\dfrac{3}{2}\right)$
  2. $\lambda \epsilon\left[\dfrac{3}{2},\infty\right)$
  3. $\lambda \epsilon\left(0,3\right)$
  4. $\lambda \epsilon\left[1,\infty\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation |z - w|^2 + |z - w^2|^2 = lambda represents a circle if the constant term after simplification is positive. For w being the cube root of unity, this simplifies to a circle for certain values of lambda.

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

Let $\displaystyle z _{1}$ and $\displaystyle z _{2}$ be the $n^{th}$ roots of unity, which are ends of a line segment that subtends a right angle at the origin. Then, $n$ must be of the form

  1. $4k+1$
  2. $4k+2$
  3. $4k+3$
  4. $4k$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The $n^{th}$ roots of unity lie on a circle, where the angle between any 2 consecutive roots is $\dfrac { 2\pi  }{ n } $.
Hence, $ \dfrac{2r\pi}{n} = \dfrac{\pi}{2}$
$\Rightarrow 4r=n$
Hence, option D is correct.

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

Which one is not a root of the fourth root of unity

  1. $i$
  2. $1$
  3. $\dfrac { i } { \sqrt { 2 } }$
  4. $-i$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The fourth roots of unity are 1, -1, i, and -i. Evaluating option C gives (i / sqrt(2)), which does not equal any of these standard roots when raised to the fourth power, as (i / sqrt(2))^4 = (-1/4) != 1. Therefore, option C is not a root of unity.

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

If $z _{1},z _{2}$be two $nth$ roots of unity such that they represent two point $A,B$ in the Argand plane where $\angle AOB=60^{\circ}$ and $O$ is the orgin then the positive integer $n$ is of the form 

  1. $4k,k\:\epsilon\:N$
  2. $4k+3,k\:\epsilon\:N$
  3. $6k,k\:\epsilon\:N$
  4. $6k+5,k\:\epsilon\:N$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From above concept


Let consider $k=0$ for $z _{1}$

So second root will be in $60^{\circ}$ 

Thus $z _{2}=cos\dfrac{\pi}{3}+isin\dfrac{\pi}{3}$

Hence $\dfrac{\pi}{3}=\dfrac{2k\pi}{n}$

$\Rightarrow n=6k$

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

If ${ z } _{ 1 },{ z } _{ 2 }$ are two complex numbers and ${ \omega  }^{ k },k=0,1,...,n-1$ are the nth roots of unity, then $\displaystyle \sum _{ k=0 }^{ n-1 }{ { \left| { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right|  }^{ 2 } } $

  1. $<n\left( { \left| { z } _{ 1 } \right| }^{ 2 }+{ \left| { z } _{ 2 } \right| }^{ 2 } \right) $
  2. $=n\left( { \left| { z } _{ 1 } \right| }^{ 2 }+{ \left| { z } _{ 2 } \right| }^{ 2 } \right) $
  3. $>n\left( { \left| { z } _{ 1 } \right| }^{ 2 }+{ \left| { z } _{ 2 } \right| }^{ 2 } \right) $

  4. can't say
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have, $\displaystyle { \left| { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right|  }^{ 2 }=\left( { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right) \left( \overline { { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } }  \right) $


$\displaystyle =\left( { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right) \left( \overline { { z } _{ 1 } } +\overline { { z } _{ 2 } } { \omega  }^{ -k } \right) \quad \quad \quad \left[ { \omega  }^{ k }={ e }^{ i(2\pi k/n) }\Rightarrow { \omega  }^{ \overline { k }  }={ e }^{ -i(2\pi k/n) }={ \omega  }^{ -k } \right] $


$={ \left| { z } _{ 1 } \right|  }^{ 2 }+{ \left| { z } _{ 2 } \right|  }^{ 2 }+\overline { { z } _{ 1 } } { z } _{ 2 }{ \omega  }^{ k }+{ z } _{ 1 }\overline { { z } _{ 2 } } { \omega  }^{ -k }$

Therefore, we have

$\displaystyle \sum _{ k=0 }^{ n-1 }{ { \left| { z } _{ 1 }+{ z } _{ 2 }{ \omega  }^{ k } \right|  }^{ 2 } } =n\left( { \left| { z } _{ 1 } \right|  }^{ 2 }+{ \left| { z } _{ 2 } \right|  }^{ 2 } \right) +\overline { { z } _{ 1 } } { z } _{ 2 }\sum _{ k=0 }^{ n-1 }{ { \omega  }^{ k } } +{ z } _{ 1 }\overline { { z } _{ 2 } } \sum _{ k=0 }^{ n-1 }{ { \omega  }^{ -k } } $

$\displaystyle =n\left( { \left| { z } _{ 1 } \right|  }^{ 2 }+{ \left| { z } _{ 2 } \right|  }^{ 2 } \right) \left[ \sum _{ k=0 }^{ n-1 }{ { \omega  }^{ k } } =\sum _{ k=0 }^{ n-1 }{ { \omega  }^{ -k } }  \right] $

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

If  $z _ { 1 }$  and  $z  _ { 2 }$  be the  $n ^ { th }$  roots of unity which subtend right angle at the origin. Then  $n$  must be of the form

  1. $4 k + 1$
  2. $4 k + 2$
  3. $4 k + 3$
  4. $4 k$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The nth roots of unity are given by e^(i 2 pi k / n) for k = 0, 1, ..., n-1. These roots are equally spaced points on the unit circle, subtending an angle of 2 pi / n at the origin. If two roots subtend a right angle (pi / 2 radians) at the origin, then 2 pi / n must divide pi / 2 or be related such that n is a multiple of 4, making n of the form 4k.

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

The value of the expression $\left( \omega -1 \right) \left( \omega -{ \omega  }^{ 2 } \right) \left( \omega -{ \omega  }^{ 3 } \right) ...\left( \omega -{ \omega  }^{ n-1 } \right) ,$ where $\omega$ is the nth root of unity, is 

  1. $n{ \omega }^{ n-1 }$
  2. $n{ \omega }^{ n }$
  3. $\left( n-1 \right) { \omega }^{ n }$
  4. $\left( n-1 \right) { \omega }^{ n-1 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have, $\displaystyle { x }^{ n }-1=\left( x-1 \right) \left( x-\omega  \right) \left( x-{ \omega  }^{ 2 } \right) ...\left( x-{ \omega  }^{ n-1 } \right) $


$\displaystyle \Rightarrow \frac { { x }^{ n }-1 }{ x-\omega  } =\left( x-1 \right)  \left( x-{ \omega  }^{ 2 } \right) ...\left( x-{ \omega  }^{ n-1 } \right) $


Putting $x=\omega $ on both sides, we have

$\displaystyle \left( \omega -1 \right) \left( \omega -{ \omega  }^{ 2 } \right) ...\left( \omega -{ \omega  }^{ n-1 } \right) =\lim _{ x\rightarrow \omega  }{ \frac { { x }^{ n }-1 }{ x-\omega  }  } \left( \frac { 0 }{ 0 } form \right) $

$\displaystyle =\lim _{ x\rightarrow \omega  }{ \frac { n{ x }^{ n-1 } }{ 1 }  } =n{ \omega  }^{ n-1 }$

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

If $1,\ \alpha _{1},\ \alpha _{2},\ \alpha _{3},\ \alpha _{4},\ \alpha _{5},\ \alpha _{6}$ are sevan $7^{th}$ root of unity then $|(3-\alpha _{1})(3-\alpha _{3})(3-\alpha _{5})|$ is 

  1. $\sqrt {2186}$
  2. $\sqrt {1093}$
  3. $\sqrt {1023}$
  4. $\sqrt {511}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The product of (x - alpha_i) for all roots of unity is related to the polynomial x^n - 1. For n=7, the roots are 1, alpha_1, ..., alpha_6. The product (3 - alpha_1)(3 - alpha_2)...(3 - alpha_6) equals (3^7 - 1) / (3 - 1) = 2186 / 2 = 1093. The question asks for the product of three specific terms, which is the square root of the full product.

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

If $\alpha $ is a non-real root of $x^6=1$ then $\displaystyle \frac{\alpha ^5+\alpha ^3+\alpha +1}{\alpha ^2+1}=$

  1. -$\alpha ^2$

  2. 0
  3. $\alpha ^2$
  4. $\alpha $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\alpha$ is non real root of $x^6 = 1$

one possible complex value of 
$\alpha = \cos (2 \pi / 6) + i \sin (2\pi / 6)$
$\alpha = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i$
$\alpha^2 = \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i \right) \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i \right) = \dfrac{1}{4} + \dfrac{\sqrt{3}}{4} i + \dfrac{\sqrt{3}}{4} i - \dfrac{3}{4} = \dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i$
$\alpha^3 = \left(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i \right) \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i \right) = \dfrac{-1}{4} = \dfrac{\sqrt{3}}{4} i + \dfrac{\sqrt{3}}{4} - \dfrac{3}{4}= -1$
$\dfrac{\alpha^5 + \alpha^3 + \alpha + 1}{\alpha^2 + 1} = \alpha^3 + \dfrac{\alpha + 1}{\alpha^2 + 1}$
$= -1 + \dfrac{\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i + 1 \right)}{\left(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i + 1 \right)}$
$= \dfrac{\dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i - 1 + \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i + 1}{(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i + 1)}$
$= \dfrac{1}{\alpha^2 + 1} = \dfrac{\alpha^2}{\alpha^4 + \alpha^2}$
$= \alpha^4 = \alpha^2 . \alpha^2 = \left(\dfrac{\sqrt{3}}{2} i - \dfrac{1}{2} \right) \left(\dfrac{\sqrt{3}}{2} i - \dfrac{1}{2} \right) = -\dfrac{\sqrt{3}}{2} i - \dfrac{1}{2}$
$= \dfrac{\alpha^2}{\alpha^4 + \alpha^2} = \dfrac{\alpha^2}{\left(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i - \dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i\right)} = - \alpha^2$
Considering option C as $- \alpha^2$ , it is correct

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

The roots of the equation  $z^{5}+z^{4}+z^{3}+z^{2}+z+1=0$   are given by

  1. $-1$
  2. $\displaystyle -\frac{1}{2}+\frac{i\sqrt{3}}{2}$
  3. $\displaystyle \frac{1}{2}+\frac{i\sqrt{3}}{2}$
  4. $\displaystyle \frac{-1-i\sqrt{3}}{2}$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

$\displaystyle { z }^{ 5 }+{ z }^{ 4 }+{ z }^{ 3 }+{ z }^{ 2 }+z+1=0\ \Rightarrow \frac { { z }^{ 6 }-1 }{ z-1 } =0\ \Rightarrow z\neq 1\quad &amp; \quad { z }^{ 6 }=1=\cos { 0 } +i\sin { 0 } \ \Rightarrow z={ \left( \cos { 0 } +i\sin { 0 }  \right)  }^{ \frac { 1 }{ 6 }  }=\cos { \frac { 2k\pi  }{ 6 }  } +i\sin { \frac { 2k\pi  }{ 6 }  } \ \Rightarrow z=\cos { \frac { k\pi  }{ 3 }  } +i\sin { \frac { k\pi  }{ 3 }  } $
where $k=0,1,2,3,4,5$.

For $k=0$,
$z=1$ but $z\neq 1$

For $k=1$,
$\displaystyle z=\cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  } =\frac { 1+i\sqrt { 3 }  }{ 2 } $

For  $k=2$,
$\displaystyle z=\cos { \frac { 2\pi  }{ 3 }  } +i\sin { \frac { 2\pi  }{ 3 }  } =\frac { -1+i\sqrt { 3 }  }{ 2 } $

For  $k=3$,
$\displaystyle z=\cos { \frac { 3\pi  }{ 3 }  } +i\sin { \frac { 3\pi  }{ 3 }  } =-1$

For  $k=4$,
$\displaystyle z=\cos { \frac { 4\pi  }{ 3 }  } +i\sin { \frac { 4\pi  }{ 3 }  } =\frac { -1-i\sqrt { 3 }  }{ 2 } $

For $k=4$,
$\displaystyle z=\cos { \frac { 5\pi  }{ 3 }  } +i\sin { \frac { 5\pi  }{ 3 }  } =\frac { 1-i\sqrt { 3 }  }{ 2 } $

Hence, all the options A,B,C and D are correct.

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

If $1,\omega ,\omega ^{2},....\omega ^{n-1}$ are $n,n^{th}$ roots ofunity then the value of $\left ( 13-\omega  \right )\left ( 13-\omega ^{n-1} \right )$ equals

  1. $ \displaystyle \frac{13^{n}+1}{3}$
  2. $ \displaystyle\frac{13^{n}-1}{3}$
  3. $ \displaystyle 13^{n}-1$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(13-w)(13-w^{n-1})$
$=(13-w)(13-\dfrac{w^{n}}{w})$
$=(13-w)(13-\overline{w})$
$=169-13(w+\overline{w})+w\overline{w}$
$=169-13(Re(w))+1$
$=170-13(Re(w))$
Hence answer is none of these.

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

lf $z _{1},z _{2}$ are $n^{th}$ roots of unity which are ends of a line segment that subtends $\displaystyle \frac{\pi}{2}$ at the origin. 

then $\mathrm{n}$ is of the form.

  1. $4k +1$
  2. $4k + 2$
  3. $4k + 3$
  4. $4k$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$z _{1}= e^{i\dfrac{2k _{1}\pi}{n}}$         $z _2= e^{i\dfrac{2k _{2}\pi }{n}}$

Given, $z _{1}= z _{2}e^{i\ ^{\pi }/ _{2}}$

$\Rightarrow e^{i\left ( \dfrac{2k _{1}-2k _{2}}{n} \right ){\pi }} = e^{i\ ^{\pi }/ _{2}}$
$\Rightarrow \dfrac{2(k _{1}-k _{2})\pi }{n} = \dfrac{\pi }{2}$
$or\ 2= 4(k _{1}-k _{2})=4k$

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

If $\alpha,\ \beta,\ \gamma$ and $\Delta $ are the roots of the equation $x^{4}-1=0$, then the value of $\displaystyle \frac{a\alpha+b\beta+c\gamma+d\Delta}{a\gamma+b\Delta +c\alpha+d\beta}+\frac{a\gamma+b\Delta +c\alpha+d\beta}{a\alpha+b\beta+c\gamma+d\Delta }$ is

  1. $ 3\beta$
  2. $0$
  3. $ 2\gamma$
  4. $-2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Clearly,
$\alpha = e^{i0} = 1$
$\beta = e \frac{i2\pi}{4} = i$
$\gamma = e \frac{i4\pi}{4} = -1$
$\delta = e \frac{i6\pi}{4} = -i$
So, $\dfrac {a\alpha+b\beta+c\gamma+d\delta}{ a\gamma+b\delta+c\alpha+d\beta}=\dfrac {a+bi-c-di}{- a-bi+c+di}$
$=-1$
Similarly second expression is nothing but reciprocal of first
$=\frac{1}{-1}=-1$
Ans $ = -1-1 = -2$

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

The number of roots of the equation $z^{15}=1$ satisfying $|\arg(z)|<\pi/2$ is

  1. 6

  2. 7

  3. 8

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For the nth root of unity of complex number $z$ i.e. $z^n = 1$, there are 'n' total roots.
In the present case, $n=15$, thus, we have 15 roots.

$|arg(z)|<\cfrac {\pi}{2}$ 
$\Rightarrow -\cfrac {\pi}{2} < arg(z) < \cfrac {\pi}{2}$

Each root is at equal angular distance i.e. $\dfrac{2\pi}{15}$      ...(because $\dfrac{2\pi}{n}$).

$\therefore$ between $-\cfrac {\pi}{2}$ and $\cfrac {\pi}{2}$, their will be 7 roots (imcluding 1).
Hence, the correct option is B.