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

If $\omega, \omega^2, \omega^3, ........ \omega^{n - 1}$ are nth roots of unity then $(1- \omega) (1- \omega^2) ....... (1 - \omega^{n  -1})$ equals:

  1. $0$
  2. $1$
  3. $n$
  4. $n^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since $1, \omega, \omega^2, \omega^3, ......... \omega^{n - 1}$ are nth roots of unity, therefore, we have the identity
$(x - 1)(x - \omega)(x - \omega^2) ...... (x - \omega^{n - 1})$
$= x^n - 1$
or $(x - \omega)(x - \omega^2) ...... (x - \omega^{n - 1}) = \displaystyle \frac{x^n - 1}{x - 1}$
$= x^{n - 1} + x^{n - 2} + ...... + x + 1$
Putting x = 1 on both sides, we get $(1 - \omega) (1 - \omega^2) (1 - \omega^{n - 1}) = n$

Multiple choice the nth roots of unity complex numbers maths

If $2 + i$ and $\sqrt {5} - 2i$ are the roots of the equation $(x^{2} + ax + b)(x^{2} + cx + d) = 0$, where $a, b, c, d$ are real constants, then product of all roots of the equation is

  1. $40$
  2. $9\sqrt {5}$
  3. $45$
  4. $35$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2 - i$ and $\sqrt {5} + 2i$ are other roots.
So, Product is $(2 + i)(2 - i)(\sqrt {5} + 2i)(\sqrt {5} - 2i)$
$= 5\times 9 = 45$

Multiple choice the nth roots of unity complex numbers maths

$1 , z _1, z _2, z _3, ..., z _{n-1}$ are the $n$th roots of unity, then the value of $\displaystyle\frac{1}{(3-z _1)} +\displaystyle\frac{1}{(3-z _2)} + ... +\displaystyle\frac{1}{(3-z _{n-1})}$ is equal to  

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

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

If $\alpha _1, \alpha _2, ... , \alpha _m$ are the roots of polynomial equation
$\quad f(x) = a _0x^m + a _1x^{m-1} + ... + a _{m-1}x + a _m = 0$
Then,
$\quad f(x) = a _0(x-\alpha _1) ... (x-\alpha _m).$


and $\quad \displaystyle\frac{f'(x)}{f(x)} = \displaystyle\frac{1}{x-\alpha _1}+...+\displaystyle\frac{1}{x-\alpha _m}$

The equation in question is ${ x }^{ n }-1=0$

 $f(x)={ x }^{ n }-1=(x-1)(x-{ z } _{ 1 })...(x-{ z } _{ n-1 })$


Thus, $\dfrac { f'(x) }{ f(x) } =\dfrac { n{ x }^{ n-1 } }{ { x }^{ n }-1 } =\dfrac { 1 }{ x-1 } +\dfrac { 1 }{ x-{ z } _{ 1 }  } +...+\dfrac { 1 }{ x-{ z } _{ n-1 } } $

Substituting $x=3$:

$\dfrac { 1 }{ 3-1 } +\dfrac { 1 }{ 3-{ z } _{ 1 }  } +...+\dfrac { 1 }{ 3-{ z } _{ n-1 } } =\dfrac { n{ 3 }^{ n-1 } }{ { 3 }^{ n }-1 } $

Hence, $\dfrac { 1 }{ 3-{ z } _{ 1 }  } +...+\dfrac { 1 }{ 3-{ z } _{ n-1 } } =\dfrac { n{ 3 }^{ n-1 } }{ { 3 }^{ n }-1 } - \dfrac { 1 }{ 2 }$

Hence,  (A) is correct.

Multiple choice the nth roots of unity complex numbers maths

If $1,\omega,\omega^{2},...,\omega^{n-1}$ are $n^{th}$ roots of unity, then the value of $(5-\omega)(5-\omega^{2})...(5-\omega^{n-1})=$

  1. $\displaystyle \frac{5^{n}-2}{4}$
  2. $\displaystyle \frac{5^{n}+2}{4}$
  3. $\displaystyle \frac{5^{n}+1}{4}$
  4. $\displaystyle \frac{5^{n}-1}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${ x }^{ n }-1=0$ has n roots (of unity).
Thus, ${ x }^{ n }-1=(x-1)(x-\omega )(x-{ \omega  }^{ 2 })...(x-{ \omega  }^{ n-1 })$.
Substitute $x=5$: 
${ 5 }^{ n }-1=(5-1)(5-\omega )(5-{ \omega  }^{ 2 })...(5-{ \omega  }^{ n-1 })$
=> $(5-\omega )(5-{ \omega  }^{ 2 })...(5-{ \omega  }^{ n-1 })=\dfrac { { 5 }^{ n }-1 }{ 4 } $
Hence, option D is correct.

Multiple choice the nth roots of unity complex numbers maths

If $ 1,\alpha ,\alpha ^{2} .....\alpha ^{n-1}$ are n roots of unity then ,$1.\alpha .\alpha ^{2}....\alpha ^{n-1}$ equals

  1. $\left ( -1 \right )^{n-1}$

  2. 0

  3. 1

  4. -1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$1,\alpha ,{ \alpha  }^{ 2 }...{ \alpha  }^{ n-1 }$ are the nth roots of unity. Thus, they are solutions of the equation: ${ x }^{ n }-1=0$. 
Thus, product of roots $= { (-1) }^{ n }(\dfrac { -1 }{ 1 } )$
(i.e. ${ (-1) }^{ n }$constant term / coefficient of ${ x }^{ n }$)
Thus, the product = ${ (-1) }^{ n }(\dfrac { -1 }{ 1 } )={ (-1) }^{ n }(\dfrac { -1 }{ 1 } )={ (-1) }^{ n+1 }={ (-1) }^{ 2 }{ (-1) }^{ n-1 }=1{ (-1) }^{ n-1 }={ (-1) }^{ n-1 }$
Hence, (A) is correct.

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If $A =\begin{bmatrix} 1&9  & -7\ i & \omega^n & 8\ 1 & 6 &\omega^{2n} \end{bmatrix}$ where $i= \sqrt{-1} $ and $\omega$ is complex cube root of unity, then tr(A) will be 

  1. $1, \,if \,n = 3k,\, k \in\, N$
  2. $3, \,if \,n = 3k,\, k \in\, N$
  3. $0,\, if \,n\neq \,3k,\, k \epsilon \in N$
  4. $-1,\, if \,n\neq \,3k, \,k \epsilon \in N$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$tr(A)=1+\omega^n+\omega^{2n}$
if $n=3k$ i.e Mutiple of $3$.
$\Rightarrow tr(A)=1+\omega^{3k}+\omega^{6k}=1+1+1=3$
if $n\neq 3k$ i.e not a multiple of $3$.
then $n=3k+1$ or $n=3k+2$
$\Rightarrow tr(A)=1+\omega^{3k+1}+\omega^{2(3k+1)}=1+\omega+\omega^2=0$
Hence, options B and C.

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

The smallest integer n such that $\displaystyle \left(\frac{1+i}{1-i}\right)^{n}= 1$ is

  1. 16

  2. 12

  3. 8

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

$\displaystyle \left ( \frac{1 + i}{1 - i} \right )^n = 1$         ${ \because -i = \displaystyle \Rightarrow \frac{1}{i}}$
$\displaystyle \Rightarrow\left ( \frac{1 + i}{\displaystyle 1 + \frac{1}{i}} \right )^n = 1$
$i^n = 1$
so min value of $n =4$

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If $i^{2} = -1$, calculate the value of $3i^{2} + i^{3} - i^{4}$.

  1. $-4 - i$
  2. $-2 - i$
  3. $2 + i$
  4. $4 + i$
  5. $6 + 2i$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$i$ is an imaginary number whose value is $\sqrt { -1 } $

So, $i^2=-1$
$i^3=i^2*i=-1*i=-i$
$i^4=(i^2)^2={(-1)}^2=1$
So the value of $3i^2+i^3-i^4$ is
$\Rightarrow 3\times (-1)+(-i)-(1)$
$\Rightarrow -3-i-1=-4-i$