Tag: complex numbers and linear inequations

Questions Related to complex numbers and linear inequations

$z _1$ 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

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

  1. $-1$

  2. $i$

  3. $2i + 95$

  4. $96 + i$


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 $\dfrac { 1 + b + i a } { 1 + b - i a } = ?$

  1. 1

  2. 2

  3. $b + i a$

  4. $a + i b$


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

  1. even

  2. odd

  3. multiple of $3$

  4. None of these


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 

  1. 0

  2. 1

  3. -1

  4. 2


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

  1. $8$

  2. $8 i$

  3. $-8$

  4. $32$


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 

  1. real

  2. complex number

  3. $0$

  4. $i$


Correct Option: A