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 ${ z } _{ 1 },{ z } _{ 2 }$ 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

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

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 $w \neq 1$ is $n^{th}$ root of unity, then value of $ \displaystyle \sum _{k=0}^{n-1} \left| z _{1} w^{k} z _{2} \right| ^{2}$ is

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

The expression |z1 w^k z2|^2 simplifies to |z1|^2 * |w^k|^2 * |z2|^2. Since |w^k| = 1 for any root of unity, this is |z1|^2 * |z2|^2. Summing this n times yields n * |z1 z2|^2.

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

Let $z _1$ and $z _2$ be ${ n }^{ th }$ roots of unity which subtend 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

$Z _1 = e^{i\dfrac{2k _1\pi}{n}}$
$Z _2 = e^i{\frac{2k _2\pi}{n}}$
Now, $Z _1, Z _2$ subtend a right angle at origin
$\Rightarrow \frac {Z _1}{|Z _1|}=\frac {Z _2}{|Z _2|}e^{i(\frac{\pi}{2})}$

$\Rightarrow e^{i(k _1-k _2) \frac{2\pi}{n}} = e^{i(\frac{\pi}{2})} $

Hence, 

$ (k _1-k _2)\frac{2\pi}{n} = \frac{\pi}{2} $

$ \Rightarrow k _1 -k _2 = \frac{n}{4} $

As $k _1, k _2$ are integers, $n$ must be of the form $4k$

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

If 1, ${ a } _{ 1 },{ a } _{ 2 },....{ a } _{ n-1 }$ are the nth roots of unity then 
i) $\left( 1-{ a } _{ 1 } \right) \left( 1-{ a } _{ 2 } \right) \left( 1-{ a } _{ 3 } \right) ......\left( 1-{ a } _{ n-1 } \right) =n$
ii) $1+{ a } _{ 1 }+{ a } _{ 2 }+....+{ a } _{ n-1 }=0$
iii) $\dfrac { 1 }{ 2-{ a } _{ 1 } } +\dfrac { 1 }{ 2-{ a } _{ 2 } } +....+\dfrac { 1 }{ 2-{ a } _{ n-1 } } =\dfrac { \left( n-2 \right) { 2 }^{ n-1 }+1 }{ { 2 }^{ n }-1 } $

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

All three identities are standard properties of n-th roots of unity. (i) Product of (1 - a_i) is n. (ii) Sum of roots is 0. (iii) The partial fraction sum is a known identity derived from the derivative of the polynomial x^n - 1.

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

If $1, z _1, z _2, z _3, ...., z _{n-1}$ be the nth roots of unity and $\omega$ be a non-real complex cube root of unity, then the product
$\Pi _{r=1}^{n-1}(\omega-z _r)$ can be equal to

  1. $0$
  2. $1$
  3. $-1$
  4. $1+\omega$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

$x^n-1=(x-1)(x-z _1)(x-z _2)....(x-z _{n-1})$
$\Rightarrow \dfrac {x^n-1}{x-1}=(x-z _1)(x-z _2)....(x-z _{n-1})$
Putting $x=\omega$, we have
$\Pi _{r=1}^{n-1}(\omega-z _r)=\dfrac {\omega^n-1}{\omega-1}=\left{\begin{matrix}0 & if \ n=3k, k\epsilon Z \ 1, & if\  n=3k+1, k\epsilon Z \ 1+\omega, & if\   n=3k+2, k\epsilon Z\end{matrix}\right.$

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 $n$th root of unity, then $\displaystyle \sum _{r=1}^{n} (ar + b)\omega^{r-1}$ is equal to

  1. $\displaystyle \frac{n(n+1)a}{2}$
  2. $\displaystyle \frac{nb}{1-n}$
  3. $\displaystyle \frac{na}{\omega - 1}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Upon expanding, we get
$(a+b)+(2a+b)w+(3a+b)w^{2}+...(na+b)w^{n-1}$
$=a(1+2w+3w^{2}+...nw^{n-1})+b(1+w+w^{2}+...w^{n-1})$
$=a(1+2w+3w^{2}+...nw^{n-1})+b(\cfrac{1-w^{n}}{1-w})$
$=a(1+2w+3w^{2}+...nw^{n-1})+0$

Let
$S=a(1+2w+3w^{2}+...nw^{n-1})$
$Sw=a(w+2w^{2}+3w^{3}+...(n-1)w^{n-1}-nw^{n})$
$S(1-w)=a(1+w+w^{2}....w^{n-1})-anw^{n}$
$S(1-w)=a(\cfrac{1-w^{n}}{1-w})-anw^{n}$
$S(1-w)=a(0)-an$
$S=-\cfrac{an}{1-w}=\dfrac{an}{w-1}$
Hence, option 'C' is correct.

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

Find all those roots of the equation $z^{12} - 56z^6 - 512 = 0$ whose imaginary part is positive.

  1. $2, 2 \left ( cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2,$$ 2^{2/3} \left ( cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{2/3} \left ( cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{2/3} \left ( cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$
  2. $2, 2 \left ( cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2, $$2^{1/3} \left ( cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{1/3} \left ( cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{1/3} \left ( cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$
  3. $2, 2 \left ( cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2,$$ 2^{1/3} \left ( -cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{1/3} \left ( -cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{1/3} \left ( -cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$
  4. $2, 2 \left ( -cos \frac{\pi}{3} + i sin \frac{\pi}{3} \right ), 2 \left ( -cos \frac{2\pi}{3} + i sin \frac{2\pi}{3} \right ), - 2,$$ 2^{2/3} \left ( cos \frac{\pi}{6} + i sin \frac{\pi}{6} \right ), 2^{2/3} \left ( cos \frac{\pi}{2} + i sin \frac{\pi}{2} \right ), 2^{2/3} \left ( cos \frac{5\pi}{6} + i sin \frac{5\pi}{6} \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$(z^{6}-28)^{2}-784-512=0$
$(z^{6}-28)^{2}=1296$
$z^{6}-28=\pm36$
$z^{6}=64$ and $z^{6}=-8$
$z^{3}=\pm8$
$z=2$ and $z=-2$ ...(i)
$z^{6}=2^{3}.e^{i(2k-1)\pi}$
$z=2^{\frac{1}{2}}(e^{i\frac{(2k-1)\pi}{6}})$ where $k=1,2,3..6$.

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

If $n\ge 3$ and $1,\alpha _1, \alpha _2, ... , \alpha _{n-1}$ are $nth$ roots of unity, then the value of $\displaystyle\sum _{1 \le i < j \le n-1}{\alpha _i\alpha _j}$ is

  1. $0$
  2. $1$
  3. $-1$
  4. $(-1)^n$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given that $ 1,{ \alpha  } _{ 1 },{ \alpha  } _{ 2 },....,{ \alpha  } _{ n }$ are $n$th roots of unity

$ \Rightarrow x^{n}=1$
So the sum of roots is $0$
$\Rightarrow 1+{ \alpha  } _{ 1 }+{ \alpha  } _{ 2 }+....,{ +\alpha  } _{ n }=0$
Sum of product of roots taken two at a time is $0$
$\Rightarrow 1({ \alpha  } _{ 1 }+{ \alpha  } _{ 2 }+....,{ +\alpha  } _{ n })+\sum _{ 1\le i<j\le n-1 }^{  }{ { \alpha  } _{ i }{ \alpha  } _{ j } } =0$
$\Rightarrow \sum _{ 1\le i<j\le n-1 }^{  }{ { \alpha  } _{ i }{ \alpha  } _{ j } } =-({ \alpha  } _{ 1 }+{ \alpha  } _{ 2 }+....,{ +\alpha  } _{ n })=-(-1)=1$

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

$\alpha _{1},\alpha _{2},\alpha _{3},\alpha _{4},.........\alpha _{100},$ are all the $100^{th}$ roots of unity. Then the numerical value of $\sum _{1 \leq i}^{ }  \sum _{j \leq 100}^{ } (\alpha _{i}\alpha _{j})^{5}$ is



  1. 20

  2. 0
  3. $(20)^{1/20}$

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

$\sum _{1 \leq i}^{ }  \sum _{j \leq 100}^{ } \alpha _{i}^{5} \alpha _{j}^{5}=(\alpha _{1}^{5}+\alpha _{2}^{5}.......+\alpha _{100}^{5})^{2}-(\alpha _{1}^{10}+\alpha _{2}^{10}.......+\alpha _{100}^{10})$
=0-0=0