Tag: algebra of complex numbers

Questions Related to algebra of complex numbers

If $\dfrac { z+2i }{ z-2i } $ is purely imaginary then $\left| z \right| $ is 

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

  1. $1$

  2. $2$

  3. $\dfrac { 1 }{ 2 } $

  4. $\dfrac { 1 }{ 4 } $

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Correct Option: A
Explanation:
$\dfrac{z+2i}{z-2i}$

$=\dfrac{z+2i}{z-2i}\times \dfrac{z+2i}{z+2i}$

$=\dfrac{z^2+4z-4}{z^2-4}$

Let $z=x+iy$

$=\dfrac{(x+iy)^2+4(x+iy)-4}{(x+iy)^2-4}$

$=\dfrac{x^2-y^2+2ixy-4+4x+4iy}{(x+iy)^2-4}$

$=\dfrac{(x^2-y^2+4x-4)+i(2xy+4y)}{(x+iy)^2-4}$

z is purely imaginary. So, Re(z)=0

$\dfrac{x^2-y^2+4x-4}{(x+iy)^2-4}=0$

$\Rightarrow x^2-y^2+4x-4=0$

$\therefore x^2-y^2+4x=4$

The above equation represents the hyperbola on x-axis, with $(1,0)$

$\therefore z=1$

Simplify the following :


$\left(\dfrac{1 \, + \, i}{1 \, - \, i}\right)^{4n \, + \, 1}$

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

  1. $1$

  2. $i$

  3. $0$

  4. None of these

Show answer
Correct Option: B
Explanation:

For $n$ is positive integer, $4n+1$ is an odd integer$(5,9,13,....)$


Now,


$\left(\dfrac{1+i}{1-i}\right)^{4n+1}$

$\implies \left(\dfrac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{4n+1}$

$\implies \left(\dfrac{(1-1+2i}{1-(i^2)}\right)^{4n+1}$

$\implies \left(\dfrac{2i}{2}\right)^{4n+1}\implies (i)^{4n+1}=i$.................[putting $n=4,9,13.....$]

$\left(\sqrt[3]{3}+\left(3^\cfrac{5}{6}\right)i\right)^3$ is an integer where $i=\sqrt{-1}$. The value of the integer is equal to.

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

  1. $24$

  2. $-24$

  3. $-22$

  4. $-21$

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Correct Option: B
Explanation:
$\rightarrow { \left( \sqrt [ 3 ]{ 3 } +{ 3 }^\cfrac{ 5}{6 }i \right)  }^{ 3 }=3{ \left( 1+\sqrt { 3 } i \right)  }^{ 3 }=3{ \left( 1+\sqrt { 3 } i \right)  }^{ 2 }\left( 1+\sqrt { 3 } i \right) $
$\rightarrow 3{ \left( 1+\sqrt { 3 } i \right)  }^{ 3 }=3\left( 1+3\sqrt { 3 } { i }^{ 3 }+3\sqrt { 3 } i\left( 1+\sqrt { 3 } i \right)  \right) $
$\rightarrow 3\left( 1-3\sqrt { 3 } i+3\sqrt { 3 } i-9 \right) $
$\rightarrow 3\left( -8 \right) =-24$

 The value of $\sqrt{i}$ is 

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

  1. $1-i$

  2. $1+i$

  3. $ \pm \left( {1 + i} \right)$

  4. $i-1$

  5. $\frac{{ \pm 1}}{{\sqrt 2 }}\left( {1 + i} \right)$

Show answer
Correct Option: E
Explanation:
We can write a complex number in the form $2=(a, b)=a+ib$
$z=\sqrt{i}$
Thus
$i=z^2=a^2-b^2+2abi=(a^2-b^2, 2ab)$
$a^2-b^2=0$
$2ab=1$
$2a^2=1$
$a^2=\dfrac{1}{2}$
$a=\pm \dfrac{1}{\sqrt{2}}$
$\sqrt{i}=\dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}=\pm \dfrac{1}{\sqrt{2}}(1+i)$.

If ${ \left( \sqrt { 3 } -i \right)  }^{ n }={ 2 }^{ n }, n\in Z$, then $n$ is multiple

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

  1. $6$

  2. $10$

  3. $9$

  4. $12$

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Correct Option: A
Explanation:
$(\sqrt {3}-i)^{n}=2^{n}$
$\Rightarrow \quad\left(\dfrac {\sqrt {3}-i}{2}\right) ^{n}=1$
$\Rightarrow \quad i \left(\dfrac {-1}{2}-\dfrac {i\sqrt {3}}{2}\right) ^{n}=1$
$\therefore \quad i n^{2n}=1$
$\therefore \quad i=1$ and $n^{2n}=1$
$’n ’$ is multiple of $’3 ’$ and $’4 ’$ 
$\Rightarrow \quad’n ’$ is multiple of $’12 ’$

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 real number if

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

  1. $n _1 = n _2 + 1$

  2. $n _1 = n _2 - 1$

  3. $n _1 = n _2$

  4. $n _1 > 0, n _2 > 0$

Show answer
Correct Option: D
Explanation:

$(1+i)^{n _{1}}+(1+i^{2})^{n _{2}}+(1+i^{5})^{n _{2}}+(1+i^{7})^{n _{2}}$
$=(1+i)^{n _{1}}+(1-i)^{n _{2}}+(1+i)^{n _{2}}+(1-i)^{n _{2}}$
$=2[1+:^{n _{1}}C _{2}i^{2}+:^{n _{1}}C _{4}i^{4}...]+2[1+:^{n _{2}}C _{2}i^{2}+:^{n _{2}}C _{4}i^{4}...]$
$=2[1+-:^{n _{1}}C _{2}+:^{n _{1}}C _{4}-...]+2[1-:^{n _{2}}C _{2}+:^{n _{2}}C _{4}-...]$
Hence
For all $n _{1}>0$ and $n _{2}>0$ the above expression yields real integral number.
Where $n _{1},n _{2}\epsilon N$.
Hence, option 'D' is correct.

What is the value of the sum
$\displaystyle \sum _{ n=2 }^{ 11 }{ \left( { i }^{ n }+{ i }^{ n+1 } \right)  } $ where $i=\sqrt { -1 } $?

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

  1. $i$

  2. $2i$

  3. $-2i$

  4. $1+i$

Show answer
Correct Option: C
Explanation:

$\displaystyle \sum _{n = 1}^{11} (i^{n} + i^{n + 1}) i = \sqrt {-1}$

Now $i^{2} = -1, i^{3} = -i, i^{4} = 1, i^{5} = i$ after this values repeats
$i^{2} + i^{3} + i^{4} + i^{5} = 0$
$\therefore \displaystyle \sum _{n = 2}^{11} i^{n} = i^{11} + i^{12} = -i + 1$
$\therefore \displaystyle \sum _{n = 2}^{11} i^{n} + i^{n + 1} = -2i$.