Tag: algebra of complex numbers

Questions Related to algebra of complex numbers

The value of $\sqrt {-1} $ is

  1. $1$

  2. $-1$

  3. $i$ $(iota)$

  4. none of these


Correct Option: C
Explanation:

$\sqrt {-1} = i$ $(iota)$

$i$ is used to represent an imaginary number.
So, option $C$ is correct.

The value of $-3\sqrt {-10}$ is equal to

  1. $-3\sqrt {10}$

  2. $3\sqrt {10}$

  3. $-3i\sqrt {10}$

  4. None of these


Correct Option: C
Explanation:

$-3\sqrt {-10}$ $= -3\sqrt {-1} \times \sqrt {10}$


$= -3 i \sqrt {10}$
So, option C is correct.

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. $\displaystyle 12-16i$


Correct Option: B
Explanation:

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

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

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. $6 + 2i$


Correct Option: A
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
$\Rightarrow 3\times (-1)+(-i)-(1)$
$\Rightarrow -3-i-1=-4-i$

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. $0$


Correct Option: B
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.

Evaluate: $i^{24} + \left(\dfrac{1}{i}\right)^{26}$

  1. $0$

  2. $1$

  3. $-1$

  4. $2$


Correct Option: A
Explanation:

$ i^{24} + (\cfrac{1}{i})^{26}$

$ i^4 = 1$
$= 1 + \cfrac{1}{i^2}$
$ = \cfrac{i^2 +1}{i^2}  = 0$  (As $i^2 = -1$)

$z _1$ and $z _2$ are two complex numbers such that $|z _1|= |z _2|$ and $arg (z _1)+arg(z _2)=\pi$, then $z _1$ is equal to

  1. $2\overline{z} _2$

  2. $\overline{z} _2$

  3. $-\overline{z} _2$

  4. None of these


Correct Option: A

When simplified the value of $[i^{57}-(1/i^{25})]$ is?

  1. $0$

  2. $2i$

  3. $-2i$

  4. $2$


Correct Option: B
Explanation:

${i}^{57}-(\cfrac{1}{{i}^{25}})$

${({i}^{4})}^{14}.i-\cfrac { 1 }{ { \left( { i }^{ 4 } \right)  }^{ 8 }.i } $
$\Rightarrow$ $i-\cfrac{1}{i}$
$\Rightarrow$ $i-\cfrac{i}{{i}^{2}}$
$\Rightarrow$ $i+i$
$=2i$

The value of $i^{n}+i^{n+1}+i^{n+3}, n \epsilon N$ is 

  1. $0$

  2. $1$

  3. $2$

  4. $none\ of\ these$


Correct Option: B

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. $\frac { 1 - i\sqrt { 3 } }{ 2 }$


Correct Option: B