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 the nth roots of unity complex numbers maths

Let the four roots of unity be $z _1, z _2, z _3$, and $z _4$, respectively.
Statement 1: $z _1^2+z _2^2+z _3^2+z _4^2=0$
Statement 2: $z _1+z _2+z _3+z _4=0$.

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$x^{4}=1$
$x^{2}=\pm1$
$x^{2}=1$ and $x^{2}=-1$
$x=\pm1$ and $x=\pm i$
Hence the four roots are
$1,-1,i,-i$.
Now
$z _{1}=1=-z _{2}$
$z _{3}=i=-z _{4}$
Hence
$z _{1}^{2}+z _{2}^{2}+z _{3}^{2}+z _{4}^{2}$
$=1+1+(i)^{2}+(-i)^{2}$
$=2-2$
$=0$ ...(i)
And also
$z _{1}+z _{2}+z _{3}+z _{4}$
$=1-1+i-i$
$=0$ ...(ii)
Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

Multiple choice the nth roots of unity complex numbers maths

If $\alpha _1, \alpha _2, \alpha _3, \alpha _4$ be the roots of $x^5 - 1 = 0$ then find $\displaystyle \frac{\omega - \alpha _1}{\omega^2 - \alpha _1} \cdot \frac{\omega - \alpha _2}{\omega^2 - \alpha _2} \cdot \frac{\omega - \alpha _3}{\omega^2 - \alpha _3} \cdot \frac{\omega - \alpha _4}{\omega^2 - \alpha _4} $

  1. $\omega^2$
  2. $1$
  3. $\omega$
  4. $(\omega-\alpha _1)(\omega-\alpha _2)(\omega-\alpha _3)(\omega-\alpha _4)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^5 - 1 = 0$ has roots $1, \alpha _1, \alpha _2, \alpha _3, \alpha _4$
$\therefore (x^5 - 1) = (x- 1) (x - \alpha _1) (x - \alpha _2) (x - \alpha _3) (x- \alpha _4)$
$\Rightarrow \displaystyle \frac{x^5 -1}{x - 1} = (x - \alpha _1) (x- \alpha _2) (x - \alpha _3) (x - \alpha _4)$           ........   (1)
Putting $x = \omega$ (1) we have
$\displaystyle \frac{\omega^5 - 1}{\omega - 1} = (\omega - \alpha _1) (\omega - \alpha _2) (\omega - \alpha _3) (\omega - \alpha _4)$
$\displaystyle \frac{\omega^2 - 1}{\omega - 1} = (\omega - \alpha _1) (\omega - \alpha _2) (\omega - \alpha _3) (\omega - \alpha _4)$         ....... (2)
and putting $x = \omega^2$ in (1) we have
$\displaystyle \frac{\omega^{10} - 1}{\omega^2 - 1} = (\omega^2- \alpha _1) (\omega^2 - \alpha _2) (\omega^2 - \alpha _3) (\omega^2 - \alpha _4)$
$\Rightarrow \displaystyle \frac{\omega - 1}{\omega^2 - 1} = (\omega^2- \alpha _1) (\omega^2 - \alpha _2) (\omega^2 - \alpha _3) (\omega^2 - \alpha _4)$           ....... (3)
Dividing (2) by (3)
then $\displaystyle \frac{\omega - \alpha _1}{\omega^2 - \alpha _1} \cdot \frac{\omega - \alpha _2}{\omega^2 - \alpha _2} \cdot \frac{\omega - \alpha _3}{\omega^2 - \alpha _3} \cdot \frac{\omega - \alpha _4}{\omega^2 - \alpha _4} \cdot = \frac{(\omega^2 - 1)^2}{(\omega - 1)^2}$
                                                                                  $= \displaystyle \frac{\omega^4 + 1 - 2 \omega^2}{\omega^2 + 1 - 2 \omega}$
                                                                                   $= \displaystyle \frac{\omega + 1 - 2 \omega^2}{\omega^2 + 1 - 2 \omega}$
                                                                                   $= \displaystyle \frac{- \omega^2 - 2 \omega^2}{- \omega - 2 \omega}$
                                                                                    $= \displaystyle \frac{- 3 \omega^2}{- 3 \omega}$
                                                                                    $= \omega$

Ans: C

Multiple choice the nth roots of unity complex numbers maths

If $\alpha$ is the n$^{th}$ root of unity, then $1+2\alpha+3\alpha^2+.... $ to $n$ terms equal to

  1. $\displaystyle \frac {-n}{(1-\alpha)^2}$
  2. $\displaystyle \frac {-n}{1-\alpha}$
  3. $\displaystyle \frac {-2n}{1-\alpha}$
  4. $\displaystyle \frac {-2n}{(1-\alpha)^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$S=1+2\alpha+3\alpha^{2}+...n\alpha^{n-1}$
$\alpha S=\:\:\alpha+2\alpha^{2}+3\alpha^{3}+...(n-1)\alpha^{n-1}+n\alpha^{n}$
$S(1-\alpha)=1+\alpha+\alpha^{2}+\alpha^{3}+...\alpha^{n-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{1-\alpha^{n}}{1-\alpha}-n\alpha^{n}$
Now $\alpha^{n}=1$ since it is the $n^{th}$ root of unity.
Therefore,
$S(1-\alpha)=-n$
$S=\dfrac{-n}{1-\alpha}$

Multiple choice the nth roots of unity complex numbers maths

If n is an odd positive integer and $ I,\alpha _{1},\alpha _{2},....\alpha _{n-1}$ are the $n,n^{th}$ roots of unity, then $\left ( 3+\alpha ^{1} \right )\left ( 3+\alpha ^{2} \right )....\left ( 3+\alpha ^{n-1} \right )$ equals

  1. $\displaystyle \frac{3^{n}+1}{4}$
  2. $\displaystyle \frac{3^{n}-1}{2}$
  3. $\displaystyle \frac{3^{n}-1}{4}$

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

$x^{n}-1=(x-1)(x-\alpha _{1})(x-\alpha _{2})(x-\alpha _{3})...(x-\alpha _{n-1})$
$\dfrac{x^{n}-1}{x-1}=(x-\alpha _{1})(x-\alpha _{2})(x-\alpha _{3})...(x-\alpha _{n-1})$
$(x-\alpha _{1})(x-\alpha _{2})(x-\alpha _{3})...(x-\alpha _{n-1})=1+x+x^{2}+...x^{n-1}$
Substituting  $x=-3$.
$(3+\alpha _{1})(3+\alpha _{2})(3+\alpha _{3})...(3+\alpha _{n-1})=1-3+3^{2}+...(-1)^{n-1}3^{n-1}$
Therefore $(3+\alpha _{1})(3+\alpha _{2})(3+\alpha _{3})...(3+\alpha _{n-1})$
$=\dfrac{1-(-3)^{n}}{1-(-3)}$
Now, $n$ is odd, therefore
$=\dfrac{3^{n}+1}{4}$

Multiple choice the nth roots of unity complex numbers maths

$(1-\omega +\omega^{2})(1-\omega^{2}+\omega^{4})(1-\omega^{4}+\omega^{8})......$to 2n factors =

  1. 2
  2. $2^{2n}$

  3. 2n

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

$(1−ω+ω^2)(1−ω^2+ω^4)(1−ω^4+ω^8)$...... to $2n$

we have,

$1+w+w^2=0$

and $w^3=1$

$\Rightarrow (-w-w)(-w^2-w^2)(-w-w)....................2n$

$=(-2w)(-2w^2)(-2w).............2n$

here, we can see, 2 consecutive terms are same and are even, so we get,

$=(4w^3)(4w^3)(4w^3).........2n$

$=4 \times 4 \times .........2n$

$=2^2.........2n$

for $2n$ terms, we get,

$=2^{2n}$
Multiple choice the nth roots of unity complex numbers maths

If $\alpha$ is the $n^{th}$ root of unity, then $1+2\alpha+3\alpha^{2}+...$ to $n$ terms is equal to

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

$S=1+2\alpha+3\alpha^{2}+....n\alpha^{n-1}$
$S\alpha=\alpha+2\alpha^{2}+3\alpha^{3}...(n-1)\alpha^{n-1}+n\alpha^{n}$
$S(1-\alpha)=1+\alpha+\alpha^{2}+...\alpha^{n-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{\alpha^{n}-1}{\alpha-1}-n\alpha^{n}$
Since $\alpha$ is the nth root of unity, hence $\alpha^{n}=1$
Thus
$S(1-\alpha)=\dfrac{\alpha^{n}-1}{\alpha-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{1-1}{\alpha-1}-n$
$S(1-\alpha)=-n$
$S=-\dfrac{n}{1-\alpha}$

Multiple choice the nth roots of unity complex numbers maths

If the fourth roots of unity are $\displaystyle\ z _{1},z _{2},z _{3},z _{4}$ then $\displaystyle\ z _{1}^{2}+z _{2}^{2}+z _{3}^{2}+z _{4}^{2}$ is equal to

  1. $\displaystyle\ 1$
  2. $\displaystyle\ 0$
  3. $\displaystyle\ i$

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

If $z$ be the fourth roots of unity then, $z^4=1$
$\Rightarrow (z^4-1)=0\Rightarrow (z^2-1)(z^2+1)=0$
$\Rightarrow z=\pm 1, \pm i,$ where $i^2=-1$
$\therefore z _1^2+z _2^2+z _3^2+z _4^2=1+1+i^2+i^2=2-2=0$

Multiple choice the nth roots of unity complex numbers maths

If $a = cos \dfrac{2\pi}{7}+i  sin\dfrac{2\pi}{7}$, then find the quadratic equation whose roots are $a = a + a^2 + a^4$ and $\beta = a^3 + a^5 + a^6$.

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

$a = cos (2\pi/7)+i  sin(2\pi/7)$
$\Longrightarrow a^7 = [cos(2\pi/7)+i  sin(2\pi/7)]^7$
$= cos 2\pi + i sin 2\pi = 1$          (1)
$S = \alpha + \beta = (a + a^2 + a^4) + (a^3 + a^5 + a^6)$
$= a +a^2 +a^3 +a^4 + a^5 +a^6 = \frac{a(1-a^6)}{1-a}$
$= \frac{a-a^7}{1-a} = \frac{a-1}{1-a} = -1$                           (2)
$P= \alpha \beta = (a+a^2+a^4)(a^3+a^5+a^6)$
$= a^4+a^6 +a^7+a^5+a^7+a^8+a^7+a^9+a^{10}$
$= a^4+a^6+1+a^5+1+a+1+a^2+a^3$         [From Eq. (1)] 
$= 3+(a+ a^2+ a^3+ a^4+ a^5 + a^6)$ 
$= 3+S = 3-1=2$              [From Eq. (2)]
Therefore, the required equation is 
$x^2 -Sx + P = 0$
$\Longrightarrow x^2 + x + 2=0$


Ans: D

Multiple choice the nth roots of unity complex numbers maths

If $\displaystyle \omega $ is fifth root of unity, then $\displaystyle \log _2 \mid 1+\omega +\omega ^{2}+\omega ^{3}-\omega ^{-1}\mid $ is equal to

  1. $1$
  2. $0$
  3. $-1$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$1+w+w^{2}+...w^{3}-\dfrac{1}{w}$
$=\dfrac{w+w^{2}+w^{3}+...w^{4}-1}{w}$


$=\dfrac{1}{w}[\dfrac{w(1-w^{4})}{1-w}-1]$

$=\dfrac{1}{w}[\dfrac{w-w^{5}}{1-w}-1]$

$=\dfrac{1}{w}[\dfrac{w-1}{1-w}-1]$

$=\dfrac{1}{w}[-2]$

Now
$|\dfrac{-2}{w}|$

$=\dfrac{|-2|}{|w|}$

$=\dfrac{2}{1}$

$=2$

$=log _{2}(2)$

$=1$
Hence, option 'A' is correct.

Multiple choice the nth roots of unity complex numbers maths

If $w$ be complex $n^{th}$ root of unity and $r$ is an integer not divisible by $n$, then the sum of the $r$th powers of the nth roots of unity is

  1. $0$
  2. $1$
  3. $w$
  4. $n$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We have, for $n^{th}$ root of unity denoted by $\alpha i$
$1 + \alpha _{1}+\alpha _{2}+\alpha _{3}---------\alpha _{n-1}=\dfrac{1(1-(\alpha)^{2})}{1-\alpha}$                    [where $\alpha _{i}= e^{i \dfrac{2 \pi k}{n}}$ Thus, $\alpha _{1}=e^{i\dfrac{2 \pi}{n}}$ $\alpha _{1}=e^{i\dfrac{4 \pi}{n}}$---------------form a G.P. with ratio $\alpha = e^{i 2 \pi /2}$]
for power r, expression changes to,
$1+\alpha _{1}^{r}+\alpha _{2}^{r}+\alpha _{3}^{r}---------\alpha _{n-1}^{r}=\dfrac{1-(\alpha^{r})^{2}}{}$ [Given $r \neq kn$ $\Rightarrow \alpha^{r} \neq 1$]
$=\dfrac{1-(\alpha^{n})^{r}}{1-\alpha^{r}}$
But $\alpha^{2}=1$
Thus $\dfrac{1-1}{1-\alpha^{r}}=0$
Multiple choice the nth roots of unity complex numbers maths

If $1,\alpha,\alpha^ 2......\alpha^{n}$ are the $n^{th}$ roots of unity then $^nC _1+ ^nC _2.\alpha + ^nC _3.\alpha^2 ........+^nC _n.\alpha^{n}$ is equal to

  1. $\displaystyle \frac{1}{\alpha}$
  2. $\displaystyle \frac{1}{\alpha} (2^n -1)$
  3. $\alpha$
  4. $\displaystyle \frac{1}{\alpha} \left[ (1+\alpha)^n - 1\right]$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$^nC _1 + ^nC _2.\alpha + ^nC _3.\alpha^2 + ......... + ^nC _n.\alpha^{n1}$

$\Rightarrow \displaystyle \frac{1}{\alpha} [^n C _1. \alpha + ^n C _2 . \alpha^2 + ....... + ^n C _n. \alpha^n]$

$\Rightarrow \displaystyle \frac{1}{\alpha} [^nC _0.1+^n C _1. \alpha + ^n C _2 . \alpha^2 + ....... + ^n C _n. \alpha^n -1] \quad \dots (^nC _0=1)$

$\Rightarrow \displaystyle \frac{1}{\alpha} [(1 + \alpha)^n - 1]$

$^nC _1 + ^nC _2.\alpha + ^nC _3.\alpha^2 + ......... + ^nC _n.\alpha^{n1}=\dfrac{1}{\alpha} [(1 + \alpha)^n - 1]$

Multiple choice the nth roots of unity 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
C Correct answer
Explanation


$1+\alpha+...+\alpha^{5}=0$ [sum of n roots of unity]
$\Rightarrow 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 }=-\left( { \alpha  }^{ 2 }+{ \alpha  }^{ 4 } \right) $
$\displaystyle \Rightarrow 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 }=-{ \alpha  }^{ 2 }\left( { 1+\alpha  }^{ 2 } \right) $
$\displaystyle \Rightarrow \frac { 1+\alpha +{ \alpha  }^{ 3 }+{ \alpha  }^{ 5 } }{ { 1+\alpha  }^{ 2 } } =-{ \alpha  }^{ 2 }$

Multiple choice the nth roots of unity complex numbers maths

If $(2 + i \sqrt 3)$ is a root of the equation $x^2 + px + q = 0$, where p and q are real, then (p, q) equals to

  1. $(4, 7)$
  2. $(-4, -7)$
  3. $(-4, 7)$
  4. $(4, -7)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since $2 + i \sqrt 3$ is one root, then other root will be $2 - i \sqrt 3$.
$\therefore x^2 + px + q = 0$ is given equatiion
$\therefore$ Sum of roots $= 2 + i \sqrt 3 + 2 - i \sqrt 3 = p$
$\therefore p = - 4$
Product of roots $q = 4 + 3 = 7$

Multiple choice the nth roots of unity complex numbers maths

In the multiplicative group of $n^{th}$ roots of unity the inverse of ${ \omega  }^{ k },\left( k<n \right) $ is

  1. ${ \omega }^{ { 1 }/{ k } }$
  2. ${ \omega }^{ -1 }$
  3. ${ \omega }^{ n-k }$
  4. ${ \omega }^{ { n }/{ k } }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since $\omega $ is the $n^{th}$ root of unity, hence $\omega^{n}=1$.
Now inverse of $\omega ^{k}$ where $k<n$ is
$=\dfrac{1}{\omega ^{k}}$
$=\dfrac{\omega^{n}}{\omega^{k}}$
$=\omega^{n-k}$.
Hence inverse of $\omega^{k}$
$={\omega^{-k}}$
$=\omega^{n-k}$. 

Multiple choice the nth roots of unity complex numbers maths

The 4th roots of unity in the argand plane form a

  1. Square

  2. Rectangle

  3. Parallelogram

  4. Rhombus
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The fourth roots of unity are $1,i,-1, -i $. Hence, they form a square in the Argand plane. 
We can see that by plotting the points on the graph with coordinate system according to that of complex numbers.