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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

 Find the value of $\dfrac{i^6 + i^7 + i^8 + i^9}{i^2 + i^3}$

  1. $ 0
    $
  2. $ 1
    $
  3. $ -1
    $
  4. $ None.
    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $Z=\dfrac{i^6+i^7+i^8+i^9}{i^2+i^3}$


$=\dfrac{(i^2)^3-(i^2)^3i+(i^4)^2+(i^4)^2i}{-1-i}$


We know that, $i^2=-1$    and     $i^4=1$

$=\dfrac{-1-i+1+i}{-1-i}$

$Z=0$


$\therefore \dfrac{i^6+i^7+i^8+i^9}{i^2+i^3}=0$

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

The value of the sum $\displaystyle \sum _{n=1}^{13}(i^n+i^{n+1})$, 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

We have $i^2 = -1$
Thus, $\displaystyle \sum _{n=1}^{4}(i^n+i^{n+1}) = (i^1 + i^2) + (i^2 + i^3) + (i^3 + i^4) + (i^4 + i^5) = (i - 1) + (-1 - i) + (-i + 1) + (1 + i) = 0$
\Rightarrow $\displaystyle \sum _{n=1}^{12}(i^n+i^{n+1}) = 0$
Now, only remains is $i^{13} + i^{14} = i - 1$

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

If $(i^{413})(i^x)=1$, then determine the one possible value of x.

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

${ i }^{ 413 }{ i }^{ x }=1$

$\Rightarrow \quad { i }^{ 413 }=1$
now, $\left( 413+x \right) $ must be a multiple of 4 becouse ${ i }^{ 4 }=1$
$\therefore \quad \left( 413+3 \right) $ is divisible by $4$
                                     hence $x=3$

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 $D$.