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

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

The no. of common roots of $15th$ roots of unity which are also $25th$ the roots of unity is

  1. $4$
  2. $3$
  3. $5$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A complex number is both an nth root and an mth root of unity if and only if it is an LCM(n, m)th root of unity, or more precisely, the common roots are the kth roots of unity where k is the greatest common divisor of n and m. Here, the common roots of the 15th and 25th roots of unity are the gcd(15, 25) = 5th roots of unity. Thus, there are 5 common roots.

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

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

lf $a=\displaystyle \cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}, \alpha=a+a^{2}+a^{4}$ and $\beta=a^{3}+a^{5}+a^{6}$, then $\alpha, \beta$ are the roots of the equation

  1. $x^{2}+x+1=0$
  2. $x^{2}+x+2=0$
  3. $x^{2}+2x+2=0$
  4. $x^{2}+2x+3=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$a={ e }^{ i2\pi /7 }\ a^ 7=1\ \alpha +\beta $

$=a+a^ 2+a^ 3+a^ 4+a^ 5+a^ 6\ =\dfrac{a(a^ 6-1)}{(a-1)}\ =\dfrac{(a^ 7-a)}{(a-1)}\ =\dfrac{(1-a)}{(a-1)}$
$=-1$
$\alpha \beta =(a+a^ 2+a^ 4)(a^ 3+a^ 5+a^ 6)\ =(a^ 4+a^ 6+a^ 7+a^ 5+a^ 7+a^ 8+a^ 7+a^ 9+a^ {10})\ =(a^ 4+a^ 6+1+a^ 5+1+a+1+a^ 2+a^ 3)\ =(3+a+a^ 2+a^ 3+a^ 4+a^ 5+a^ 6)\ =(3-1)$
$=2 $
The equation can be written as
 $x^ 2-(\alpha +\beta )x +\alpha \beta  =x^ 2+x+2$
Hence, option B is correct.

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, $z = \cos  \displaystyle \frac{\pi}{8} + i  \sin  \frac{\pi}{8}$