Tag: identities of complex numbers

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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 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 $D$.

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

Solve:

$\left ( \dfrac{2i}{1 \, + \, i} \right )^2$

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

We have,

$ {{\left( \dfrac{2i}{1+i} \right)}^{2}} $

$ \Rightarrow \dfrac{4{{i}^{2}}}{1+{{i}^{2}}+2i} $

$ \Rightarrow \dfrac{-4}{1-1+2i}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \because {{i}^{2}}=-1 \right) $

$ \Rightarrow \dfrac{-4}{2i} $

$ \Rightarrow \dfrac{-2}{i} $

$ \Rightarrow \dfrac{-2i}{{{i}^{2}}} $

$ \Rightarrow \dfrac{-2i}{-1} $

$ \Rightarrow 2i $


Hence, this is the answer.

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 $n = 4$.