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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.