Tag: algebra of complex numbers

Questions Related to algebra of complex numbers

Which  of the following is correct ?

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

  1. $2 + 3i > 1 + 4i$

  2. $2 + 2i > 3 + 3i$

  3. $5 + 8i > 5 + 7i$

  4. None of these

Show answer
Correct Option: C
Explanation:

Option $A$

$2+3i>1+4i$
$2^2+3^2>1^2+4^2$
$4+9>1+16$
$13>17$

It is not correct.

Option $B$
$2+2i>3+3i$

$2^2+2^2>3^2+3^2$
$4+4>9+9$
$8>18$

It is not correct.


Option $C$
$5+8i>5+7i$

$5^2+^82>5^2+7^2$
$25+64>25+49$
$89>74$

It is correct.

Hence, this is the answer.

Find the value of $\begin{vmatrix} 2+i & 2-i \ 1+i & 1-i \end{vmatrix}$ if $i^2=-1$.

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

  1. A complex quantity

  2. real quantity

  3. $0$

  4. cannot be determined

Show answer
Correct Option: A
Explanation:

$\left| \begin{matrix} 2+i & 2-i \ 1+i & 1-i \end{matrix} \right| \ =(2+i)(1-i)-(2+i)(1+i)\ =2-2i+i-{ i }^{ 2 }-2-2i-i-{ i }^{ 2 }\ =-4i+2$

So it is complex quantity

Find the value of $\dfrac{i^{4n+1}-i^{4n-1}}{2}$.

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

  1. $-1$

  2. $1$

  3. $-i$

  4. $i$

Show answer
Correct Option: D
Explanation:
We have,
$\dfrac{i^{4n+1}-i^{4n-1}}{2}$
$=\dfrac{(i^{4})^{n}.i-(i^{4})^{n}.i^{-1}}{2}$
$=\dfrac{i-i^{-1}}{2}$ [ Since $i^2=1\Rightarrow i^4=1$]
$=\dfrac{i+i}{2}$ [ As $i^2=-1\Rightarrow i=-\dfrac{1}{i}$]
$=i$.

$i^{242}=$

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

  1. $i$

  2. $-i$

  3. $1$

  4. $-1$

Show answer
Correct Option: D
Explanation:

$(i)^{242}$

$=(i^{2})^{121}$

$=(-1)^{121}$                              since $i=\sqrt{-1}$.
Hence
$(-1)^{121}=-1$
Thus
$i^{242}=-1$

$\displaystyle i+\frac{1}{i}=$

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

  1. $1$

  2. $-1$

  3. $0$

  4. $2i$

Show answer
Correct Option: C
Explanation:

Let $Z= i+\dfrac{1}{i}$


Mutiplying numerator and denominator by i. We get,

          $=i+\dfrac{i}{i^{2}}$

          $=i+\dfrac{i}{-1} \quad \dots (i^2=-1)$

          $=i-i$

      $Z=0$

Hence, 

$i+\dfrac{1}{i}=0$.

Evaluate :

 $(-\sqrt{-1})^{4n+3}, n \in N$

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

  1. -$i$

  2. $i$

  3. $1$

  4. -$1$

Show answer
Correct Option: B
Explanation:

$(-\sqrt{-1})^{4n+3} = (-i)^{4n+3}
= (-i)^{4n}(-i)^3
= {(-i)^4}^n(-i)^3
= 1 \times (-i)^3 = i$

Evaluate  $i^{135}$
maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

  1. $ -i$

  2. $ i$

  3. $ -1$

  4. $ 1$

Show answer
Correct Option: A

$\displaystyle \left ( i \right )^{457}$

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

  1. $\displaystyle -1 $

  2. $\displaystyle -i $

  3. $\displaystyle i $

  4. $\displaystyle 1 $

Show answer
Correct Option: C
Explanation:

The value of $(i)^{457} $

$=(i)^{456+1} $
$=(i)^{4\times 114}(i)$
$= 1\times i $
$= i \quad [\because i^{4n}=1]$

The smallest integer n such that $\displaystyle \left(\frac{1+i}{1-i}\right)^{n}= 1$ is

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

  1. 16

  2. 12

  3. 8

  4. 4

Show answer
Correct Option: D
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 $n =4$

$\displaystyle \left ( \frac{1 + i}{1 - i} \right )^2 + \left(\frac{1 - i}{1 + i} \right )^2$ is equal to

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

  1. $2i$

  2. $-2i$

  3. $-2$

  4. $2$

Show answer
Correct Option: C
Explanation:

$\left(\dfrac{1+i}{1-i}\right)^2+\left(\dfrac{1-i}{1+i}\right)^2=\left[\dfrac{(1+i)(1+i)}{(1-i)(1+i)}\right]^2+\left[\dfrac{(1-i)(1-i)}{(1+i)(1-i)}\right]^2$


$=\left[\dfrac{1+2i-1}{2}\right]^2+\left[\dfrac{1-2i-1}{2}\right]^2$

$=\dfrac{4i^2}{4}+\dfrac{4i^2}{4}=[-1]+[-1]=-2$