Tag: complex numbers and linear inequations

Quizzes (4) Questions (127)

Questions Related to complex numbers and linear inequations

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

The value of ${ i }^{ \frac { 1 }{ 3 }  }$ is:

  1. $\frac { \sqrt { 3 } - i }{ 2 }$
  2. $\frac { \sqrt { 3 } + i }{ 2 }$
  3. $\frac { 1 + i\sqrt { 3 } }{ 2 }$
  4. $\frac { 1 - i\sqrt { 3 } }{ 2 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The cube roots of i can be found using De Moivre's Theorem. i = cos(pi/2) + i sin(pi/2). The roots are cos((pi/2 + 2k*pi)/3) + i sin((pi/2 + 2k*pi)/3) for k=0, 1, 2. For k=0, we get cos(pi/6) + i sin(pi/6) = sqrt(3)/2 + i/2.

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

The value of $\displaystyle\sum _{ n=0 }^{ 100 }{ { i }^{ n! } } $ equals ( where $i=\sqrt { -1 } $  ):

  1. $-1$
  2. $i$
  3. $2i + 95$
  4. $96 + i$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\sum _{0}^{n} i^{n!}=i^0+i^1+i^2+i^{1\times2\times3}+.....+ i^{1\times 2.....\times 100}$
We know that , $i^{4n}=1$ , $i^{4n+1}=i$ , $i^{4n+2}=-1$ , $i^{4n+3}=-i$
So starting from n=4 every number will be 1
So our sum shortens to $1+i+(-1)+(-1)+97=96+i$
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If $a ^ { 2 } + b ^ { 2 } = 1$, then $\dfrac { 1 + b + i a } { 1 + b - i a } = ?$

  1. 1

  2. 2
  3. $b + i a$
  4. $a + i b$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\dfrac{1+b+ia}{1+b-ia}\times\dfrac{b+ia}{b+ia}$

$=\dfrac{\left(1+b+ia\right)\times\left(b+ia\right)}{b+{b}^{2}-iab+ia+iab+{a}^{2}}$

$=\dfrac{\left(1+b+ia\right)\times\left(b+ia\right)}{1+b+ia}$ since ${a}^{2}+{b}^{2}=1$

$=b+ia$
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If ${(1+i)}^{2n}+{(1-i)}^{2n}=-{2}^{n+1}$ where, $i=\sqrt{-1}$ for all those $n$, which are

  1. even

  2. odd
  3. multiple of $3$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
In $(1+i)^{2n}+(1-i)^{2n}$
$=\left\{(1+i)^{2}\right\}^{n}+\left\{(1-i)^{2}\right\}^n$
$=(1+i^{2}+2i) ^{n}+(1+i^{2}-2i) ^{n}$
$=(1-1+2\ i)^{n}+(1-1-2\ i) ^{n}$
$=2^{n}i^{2}+i^{2}(-2) ^n$
$=i^{2}(2^{2}+(-2) ^n)$
When $n=2$
$= i^{2}(2^{2}+2^{2})=-1 \cdot 2^{2+1}$
Hence, $’n’$ must be even
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.