Tag: identities of complex numbers

Questions Related to identities of complex numbers

The value of $\displaystyle\sum _{ n=0 }^{ 100 }{ { i }^{ n! } } $ equals ( where $i=\sqrt { -1 } $  ):

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

  1. $-1$

  2. $i$

  3. $2i + 95$

  4. $96 + i$

Show answer
Correct Option: D
Explanation:
$\sum _{0}^{n} i^{n!}=i^0+i^1+i^2+i^{1\times2\times3}+.....+ i^{1\times 2.....\times 100}$
We know that , $i^{4n}=1$ , $i^{4n+1}=i$ , $i^{4n+2}=-1$ , $i^{4n+3}=-i$
So starting from n=4 every number will be 1
So our sum shortens to $1+i+(-1)+(-1)+97=96+i$

If $a ^ { 2 } + b ^ { 2 } = 1$, then $\dfrac { 1 + b + i a } { 1 + b - i a } = ?$

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

  1. 1

  2. 2

  3. $b + i a$

  4. $a + i b$

Show answer
Correct Option: C
Explanation:
$\dfrac{1+b+ia}{1+b-ia}\times\dfrac{b+ia}{b+ia}$

$=\dfrac{\left(1+b+ia\right)\times\left(b+ia\right)}{b+{b}^{2}-iab+ia+iab+{a}^{2}}$

$=\dfrac{\left(1+b+ia\right)\times\left(b+ia\right)}{1+b+ia}$ since ${a}^{2}+{b}^{2}=1$

$=b+ia$

If ${(1+i)}^{2n}+{(1-i)}^{2n}=-{2}^{n+1}$ where, $i=\sqrt{-1}$ for all those $n$, which are

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

  1. even

  2. odd

  3. multiple of $3$

  4. None of these

Show answer
Correct Option: A
Explanation:
In $(1+i)^{2n}+(1-i)^{2n}$
$=\left\{(1+i)^{2}\right\}^{n}+\left\{(1-i)^{2}\right\}^n$
$=(1+i^{2}+2i) ^{n}+(1+i^{2}-2i) ^{n}$
$=(1-1+2\ i)^{n}+(1-1-2\ i) ^{n}$
$=2^{n}i^{2}+i^{2}(-2) ^n$
$=i^{2}(2^{2}+(-2) ^n)$
When $n=2$
$= i^{2}(2^{2}+2^{2})=-1 \cdot 2^{2+1}$
Hence, $’n’$ must be even

If $z + \frac{1}{z} = 2\cos {6^0}$, then ${z^{1000}} + \frac{1}{{{z^{1000}}}} + 1$ is equal to 

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

  1. 0

  2. 1

  3. -1

  4. 2

Show answer
Correct Option: A
Explanation:

$z+\cfrac{1}{z}=2\cos {6}^{o}$

$z=\cos ]theta+i\sin ]theta$
$z+\cfrac{1}{z}=\cos \theta +i\sin\theta +\cos \theta -i\sin\theta $
$2\cos{6}^{o}=2\cos \theta $
$\theta ={6}^{o}$
${z}^{1000}+\cfrac{1}{{z}^{1000}}=\cos(1000\theta )+i\sin(1000\theta )+\cos (1000\theta )-i\sin (1000\theta )+1$
$=2\cos(1000\theta )+1$
$=2\cos({6000}^{o} )+1$
$=2\cos(5760+240)+1$
$2\cos({240}^{o})+1=2\cos({120}^{o})+1$
$=-2\cos 60+1-1+1=0$
$\therefore$ ${z}^{1000}+\cfrac{1}{{z}^{1000}}+1=0$

The value of $( 1 + i ) ^ { 4 } + ( 1 - i ) ^ { 4 }$ is

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

  1. $8$

  2. $8 i$

  3. $-8$

  4. $32$

Show answer
Correct Option: C
Explanation:

$(1+i)^4+(1-i)^4$


$\Rightarrow$  $[(1+i)^2]^2+[(1-i)^2]^2$

We know, $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=^2-2ab+b^2$

$\Rightarrow$  $[1+2i+i^2]^2+[1-2i+i^2]^2$      

$\Rightarrow$  $[1+2i-1]^2+[1-2i-1]^2$                       [ $i^2=-1$ ]

$\Rightarrow$  $(2i)^2+(-2i)^2$

$\Rightarrow$  $4i^2+4i^2$

$\Rightarrow$  $-4-4$

$\Rightarrow$  $-8$

$\therefore$   $(1+i)^4+(1-i)^4=-8$

For positive integers $n _1, n _2, $ the value of the expression $(1 + i)^{n _1} + (1 + i^3)^{n _1} + (1 + i^5)^{n _2} + (1 + i^7)^{n _2}$, where $i = \sqrt{-1}$ is a 

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

  1. real

  2. complex number

  3. $0$

  4. $i$

Show answer
Correct Option: A

If $\begin{vmatrix}6i & -3i & 1\4 & 3i & -1\20 & 3 & i\end{vmatrix} = x+ iy$, then 

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

  1. $x =3, y = 0$

  2. $x =1, y = 3$

  3. $x =0, y = 3$

  4. $x =0, y = 0$

Show answer
Correct Option: D
Explanation:

Given:-

       $ \begin{vmatrix} 6i & -3i & 1 \ 4 & 3i & -1 \ 20 & 3 & i \end{vmatrix}=x+iy$
To find value of $x$ and $ y$.
By solving the given diterment.
$ \begin{vmatrix} 6i & -3i & 1 \ 4 & 3i & -1 \ 20 & 3 & i \end{vmatrix}=6i\left[ 3i\left( i \right) -\left( 3 \right) \left( -1 \right)  \right] -\left( -3i \right) \left[ 4\left( i \right) -\left( 20 \right) \left( -1 \right)  \right] +1\left[ (4)\left( 3 \right) -\left( 20 \right) \left( 3i \right)  \right] $
$ 6i\left[ 3{ i }^{ 2 }+3 \right] +3i\left[ 4i+20 \right] +1\left[ 12-60i \right] $
 We know that $i=\sqrt { -1 }$ hence,$ { i }^{ 2 }=-1$
By substituting the value of ${ i }^{ 2 }$ we get
$ \begin{vmatrix} 6i & -3i & 1 \ 4 & 3i & -1 \ 20 & 3 & i \end{vmatrix}=6i\left[ -3+3 \right] +12{ i }^{ 2 }+60i+12-60i$
$=0+12(-1)+60i+12-60i$
$ =0$
 By comparing with given we get.
$ x+iy=0$
 If $x& y$ are real no. then the only possible solution
 for$ x+iy=0$ is
$ x=0$ and$ y=0$
 Hence the answer is $x=0\quad & \quad y=0$

Let $\displaystyle \Delta =\left | \begin{matrix}a _{11} & a _{12} & a _{13}\a _{21}  &a _{22}  &a _{23} \a _{31}  &a _{32}  &a _{33} \end{matrix} \right |$ and $\displaystyle a _{pq}= i^{p+q}$ where $\displaystyle i= \sqrt{-1}.$ The value of $\displaystyle \Delta $ is 

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

  1. real and positive

  2. real and negative

  3. $0$

  4. imaginary

Show answer
Correct Option: C
Explanation:

$\triangle =\left| \begin{matrix} { a } _{ 11 }\quad \quad  & { a } _{ 12 }\quad \quad  & { a } _{ 13 } \ { a } _{ 21 }\quad \quad  & { a } _{ 22 }\quad \quad  & { a } _{ 23 } \ { a } _{ 31 }\quad \quad  & { a } _{ 32 }\quad \quad  & { a } _{ 33 } \end{matrix} \right| \quad & \quad { a } _{ pq }={ i }^{ p+q }$

$\Rightarrow \quad \triangle =\left| \begin{matrix} { i }^{ 2 }\quad \quad  & { i }^{ 3 }\quad \quad  & { i }^{ 4 } \ { i }^{ 3 }\quad \quad  & { i }^{ 4 }\quad \quad  & { i }^{ 5 } \ { i }^{ 4 }\quad \quad  & { i }^{ 5 }\quad \quad  & { i }^{ 6 } \end{matrix} \right| ={ i }^{ 2+3+4 }\left| \begin{matrix} { 1 }\quad \quad  & 1\quad \quad  & 1 \ { i }\quad \quad  & { i }\quad \quad  & { i } \ { i }^{ 2 }\quad \quad  & { i }^{ 2 }\quad \quad  & { i }^{ 2 } \end{matrix} \right| $


$=i\left| \begin{matrix} 1 & 1 & 1 \ i & i & i \ -1 & -1 & -1 \end{matrix} \right| =-i\left| \begin{matrix} 1 & 1 & 1 \ i & i & i \ 1 & 1 & 1 \end{matrix} \right| $

$\therefore \quad \triangle =0$
Hence, option 'C' is correct.

The sequence $S=i+2{ i }^{ 2 }+3{ i }^{ 3 }+.......$ upto 100 times simplifies to where $i=\sqrt { -1 } $.

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

  1. $50(1-i)$

  2. $25i$

  3. $25(1+i)$

  4. $100(1-i)$

Show answer
Correct Option: A
Explanation:

$S=i+2i^2+3i^3\ldots+100i^{100}\S=i-2-3i+4+\ldots +100\S=i(1-3+5\ldots)+(-2+4\ldots)\S=50-50i=50(1-i) $

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

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

  1. $ 0
    $

  2. $ 1
    $

  3. $ -1
    $

  4. $ None.
    $

Show answer
Correct Option: A
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     $i^4=1$

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

$Z=0$


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