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

If $z + \frac{1}{z} = 2\cos {6^0}$, then ${z^{1000}} + \frac{1}{{{z^{1000}}}} + 1$ is equal to 

  1. 0

  2. 1

  3. -1

  4. 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$z+\cfrac{1}{z}=2\cos {6}^{o}$

$z=\cos ]theta+i\sin ]theta$
$z+\cfrac{1}{z}=\cos \theta +i\sin\theta +\cos \theta -i\sin\theta $
$2\cos{6}^{o}=2\cos \theta $
$\theta ={6}^{o}$
${z}^{1000}+\cfrac{1}{{z}^{1000}}=\cos(1000\theta )+i\sin(1000\theta )+\cos (1000\theta )-i\sin (1000\theta )+1$
$=2\cos(1000\theta )+1$
$=2\cos({6000}^{o} )+1$
$=2\cos(5760+240)+1$
$2\cos({240}^{o})+1=2\cos({120}^{o})+1$
$=-2\cos 60+1-1+1=0$
$\therefore$ ${z}^{1000}+\cfrac{1}{{z}^{1000}}+1=0$

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

The value of $( 1 + i ) ^ { 4 } + ( 1 - i ) ^ { 4 }$ is

  1. $8$
  2. $8 i$
  3. $-8$
  4. $32$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$(1+i)^4+(1-i)^4$


$\Rightarrow$  $[(1+i)^2]^2+[(1-i)^2]^2$

We know, $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=^2-2ab+b^2$

$\Rightarrow$  $[1+2i+i^2]^2+[1-2i+i^2]^2$      

$\Rightarrow$  $[1+2i-1]^2+[1-2i-1]^2$                       [ $i^2=-1$ ]

$\Rightarrow$  $(2i)^2+(-2i)^2$

$\Rightarrow$  $4i^2+4i^2$

$\Rightarrow$  $-4-4$

$\Rightarrow$  $-8$

$\therefore$   $(1+i)^4+(1-i)^4=-8$

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

For positive integers $n _1, n _2, $ the value of the expression $(1 + i)^{n _1} + (1 + i^3)^{n _1} + (1 + i^5)^{n _2} + (1 + i^7)^{n _2}$, where $i = \sqrt{-1}$ is a 

  1. real

  2. complex number
  3. $0$
  4. $i$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Simplify the terms: (1+i)^n1 + (1-i)^n1 + (1+i)^n2 + (1-i)^n2. Since (1+i) and (1-i) are conjugates, their sum to any power n is real. Thus, the entire expression is real.

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

If $i=\sqrt{-1}$, then select from the following having the greatest value.

  1. $i^4+i^3+i^2+i$
  2. $i^8+i^6+i^4+i^2$
  3. $i^{12}+i^9+i^6+i^3$
  4. $i^{16}+i^{12}+i^8+i^4$
  5. $i^{20}+i^{15}+i^{10}+i^5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given, $i=\sqrt {-1}$

The value of option $A$ is $1-i-1+i = 0$
The value in option $B$ is $1-1+1-1 = 0$
The value in option $C$ is $1+i-1-i = 0$
The value in option $D$ is $1+1+1+1 = 4$
The value in option $E$ is $1-i-1+i = 0$
So, the correct answer is option $D$.

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

Find the least value of $n$ for which $\left (\dfrac {1 + i}{1 - i}\right )^{n} = 1$.

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

$\because \dfrac {1 + i}{1 - i} = \dfrac {1 + i}{1 - i}\times \dfrac {1 + i}{1 + i}$
$= \dfrac {(1 + i)^{2}}{1 - i^{2}} = \dfrac {1 + 2i + i^{2}}{1 - i^{2}}$
$= \dfrac {1 + 2i - 1}{1 + 1} = i$
$\therefore \left (\dfrac {1 + i}{1 - i}\right )^{n} = 1$
$\Rightarrow i^{n} = 1$
Thus, $i^{n}$ will be positive integer, if $n = 4$.

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

$\left(\sqrt[3]{3}+\left(3^\cfrac{5}{6}\right)i\right)^3$ is an integer where $i=\sqrt{-1}$. The value of the integer is equal to.

  1. $24$
  2. $-24$
  3. $-22$
  4. $-21$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\rightarrow { \left( \sqrt [ 3 ]{ 3 } +{ 3 }^\cfrac{ 5}{6 }i \right)  }^{ 3 }=3{ \left( 1+\sqrt { 3 } i \right)  }^{ 3 }=3{ \left( 1+\sqrt { 3 } i \right)  }^{ 2 }\left( 1+\sqrt { 3 } i \right) $
$\rightarrow 3{ \left( 1+\sqrt { 3 } i \right)  }^{ 3 }=3\left( 1+3\sqrt { 3 } { i }^{ 3 }+3\sqrt { 3 } i\left( 1+\sqrt { 3 } i \right)  \right) $
$\rightarrow 3\left( 1-3\sqrt { 3 } i+3\sqrt { 3 } i-9 \right) $
$\rightarrow 3\left( -8 \right) =-24$
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

 The value of $\sqrt{i}$ is 

  1. $1-i$
  2. $1+i$
  3. $ \pm \left( {1 + i} \right)$
  4. $i-1$
  5. $\frac{{ \pm 1}}{{\sqrt 2 }}\left( {1 + i} \right)$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation
We can write a complex number in the form $2=(a, b)=a+ib$
$z=\sqrt{i}$
Thus
$i=z^2=a^2-b^2+2abi=(a^2-b^2, 2ab)$
$a^2-b^2=0$
$2ab=1$
$2a^2=1$
$a^2=\dfrac{1}{2}$
$a=\pm \dfrac{1}{\sqrt{2}}$
$\sqrt{i}=\dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}=\pm \dfrac{1}{\sqrt{2}}(1+i)$.
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If ${ \left( \sqrt { 3 } -i \right)  }^{ n }={ 2 }^{ n }, n\in Z$, then $n$ is multiple

  1. $6$
  2. $10$
  3. $9$
  4. $12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$(\sqrt {3}-i)^{n}=2^{n}$
$\Rightarrow \quad\left(\dfrac {\sqrt {3}-i}{2}\right) ^{n}=1$
$\Rightarrow \quad i \left(\dfrac {-1}{2}-\dfrac {i\sqrt {3}}{2}\right) ^{n}=1$
$\therefore \quad i n^{2n}=1$
$\therefore \quad i=1$ and $n^{2n}=1$
$’n ’$ is multiple of $’3 ’$ and $’4 ’$ 
$\Rightarrow \quad’n ’$ is multiple of $’12 ’$
Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

Suppose z and $\omega$ are two complex numbers such that $|z| \leq 1, |\omega| \leq 1$, and $|z+i\omega|=|z-i\omega|=2$.

Which of the following is true about $|z|$ and $|\omega|$?

  1. $|z|=|\omega|=\frac {1}{2}$
  2. $|z|=\frac {1}{2}, |\omega|=\frac {3}{4}$
  3. $|z|=|\omega|=\frac {3}{4}$
  4. $|z|=|\omega|=1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,


$|z+i\omega|\le |z|+|i\omega|\le |z|+|i||\omega|\le 2$

$|z-i\omega|\le |z|+|-i\omega|\le |z|+|i||-\omega|=|z|+|i||\omega|\le 2$

Also,

$|z+i\omega|=|z-i\omega|$

and

$ |z+i\omega|=2$ 

Hence, 

$|z|  = |\omega| = 1 $

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

$\begin{array} { l } { \text { If } z _ { 1 } \text { and } z _ { 2 } \text { are complex numbers, then } \left| z _ { 1 } + z _ { 2 } \right| ^ { 2 } = \left| z _ { 1 } \right| ^ { 2 } + \left| z _ { 2 } \right| ^ { 2 } \text { if and only if } z _ { 1 } \overline { z } _ { 2 } \text { is } } \ { \text { purely imaginary. } } \end{array}$

  1. True

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

|z1 + z2|^2 = (z1 + z2)(conjugate(z1) + conjugate(z2)) = |z1|^2 + |z2|^2 + z1*conjugate(z2) + conjugate(z1)*z2. For this to equal |z1|^2 + |z2|^2, the cross terms must sum to zero: z1*conjugate(z2) + conjugate(z1*conjugate(z2)) = 0. This means 2*Re(z1*conjugate(z2)) = 0, so z1*conjugate(z2) must be purely imaginary.