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 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

$\displaystyle \left ( \frac{1 + i}{1 - i} \right )^2 + \left(\frac{1 - i}{1 + i} \right )^2$ is equal to

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

$\left(\dfrac{1+i}{1-i}\right)^2+\left(\dfrac{1-i}{1+i}\right)^2=\left[\dfrac{(1+i)(1+i)}{(1-i)(1+i)}\right]^2+\left[\dfrac{(1-i)(1-i)}{(1+i)(1-i)}\right]^2$


$=\left[\dfrac{1+2i-1}{2}\right]^2+\left[\dfrac{1-2i-1}{2}\right]^2$

$=\dfrac{4i^2}{4}+\dfrac{4i^2}{4}=[-1]+[-1]=-2$

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

Find the value of $\displaystyle \left( 4+2i \right) \left( 4-2i \right) $ given that $\displaystyle { i }^{ 2 }=-1$. 

  1. $12$
  2. $20$
  3. $\displaystyle 16-4i$
  4. $\displaystyle 4+16i$
  5. $\displaystyle 12-16i$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

After expanding, we get $(4+2i)(4-2i)=16+8i-8i-4i^4$

According to the question $i^2=-1$
$\Rightarrow 16-4i^4$
$\Rightarrow 16-4 ( -1)$
$\Rightarrow 16+4=20$

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
$\Rightarrow 3\times (-1)+(-i)-(1)$
$\Rightarrow -3-i-1=-4-i$

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

The value of the sum $\displaystyle \sum _{ n=1 }^{ 13 }{ \left( { i }^{ n }+{ i }^{ n+1 } \right)  }$. where $i=\sqrt { -1 }$, equals 

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

Given:

$\displaystyle \sum _{n=1}^{13}(i^n+i^{n+1})$

So,
$\Rightarrow(i^1+i^{2})+(i^2+i^3)+(i^3+i^4)+........+(i^{13}+i^{14})$

We know that
$i^2=-1$
$i^3=-i$
$i^4=1$
$i^5=i$
$i^6=-1$
$i^7=-i$
$i^8=1$

Therefore,
$\Rightarrow(i-1)+(-1-i)+(-i+1)+........+(i-1)$

Same cycle upto $4^{th}$ term.

Therefore,

$\Rightarrow(i-1)+(-1-i)+(-i+1)+(1+i)+........+(i-1)$

So, all terms will cancel out with each other up to $12th$ term.

Therefore,
$\Rightarrow i-1$

Hence, this is the answer.