Tag: powers of imaginary unit i

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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 $\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.

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.