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

Solve:

$\left ( \dfrac{2i}{1 \, + \, i} \right )^2$

  1. $-i$
  2. $i$
  3. $2i$
  4. $1-i$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have,

$ {{\left( \dfrac{2i}{1+i} \right)}^{2}} $

$ \Rightarrow \dfrac{4{{i}^{2}}}{1+{{i}^{2}}+2i} $

$ \Rightarrow \dfrac{-4}{1-1+2i}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \because {{i}^{2}}=-1 \right) $

$ \Rightarrow \dfrac{-4}{2i} $

$ \Rightarrow \dfrac{-2}{i} $

$ \Rightarrow \dfrac{-2i}{{{i}^{2}}} $

$ \Rightarrow \dfrac{-2i}{-1} $

$ \Rightarrow 2i $


Hence, this is the answer.

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

Find the least value of $n$ for which $\left (\dfrac {1 + i}{1 - i}\right )^{n} = 1$.

  1. $4$
  2. $3$
  3. $-4$
  4. $1$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$\because \dfrac {1 + i}{1 - i} = \dfrac {1 + i}{1 - i}\times \dfrac {1 + i}{1 + i}$
$= \dfrac {(1 + i)^{2}}{1 - i^{2}} = \dfrac {1 + 2i + i^{2}}{1 - i^{2}}$
$= \dfrac {1 + 2i - 1}{1 + 1} = i$
$\therefore \left (\dfrac {1 + i}{1 - i}\right )^{n} = 1$
$\Rightarrow i^{n} = 1$
Thus, $i^{n}$ will be positive integer, if $n = 4$.

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

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

  1. $1$
  2. $2$
  3. $\dfrac { 1 }{ 2 } $
  4. $\dfrac { 1 }{ 4 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $(1,0)$

$\therefore z=1$
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

Simplify the following :


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

  1. $1$
  2. $i$
  3. $0$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $n=4,9,13.....$]

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

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

  1. $24$
  2. $-24$
  3. $-22$
  4. $-21$
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $
$\rightarrow 3\left( -8 \right) =-24$
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

 The value of $\sqrt{i}$ is 

  1. $1-i$
  2. $1+i$
  3. $ \pm \left( {1 + i} \right)$
  4. $i-1$
  5. $\frac{{ \pm 1}}{{\sqrt 2 }}\left( {1 + i} \right)$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation
We can write a complex number in the form $2=(a, b)=a+ib$
$z=\sqrt{i}$
Thus
$i=z^2=a^2-b^2+2abi=(a^2-b^2, 2ab)$
$a^2-b^2=0$
$2ab=1$
$2a^2=1$
$a^2=\dfrac{1}{2}$
$a=\pm \dfrac{1}{\sqrt{2}}$
$\sqrt{i}=\dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}=\pm \dfrac{1}{\sqrt{2}}(1+i)$.
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

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

  1. $6$
  2. $10$
  3. $9$
  4. $12$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $’12 ’$
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

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

  1. $n _1 = n _2 + 1$
  2. $n _1 = n _2 - 1$
  3. $n _1 = n _2$
  4. $n _1 > 0, n _2 > 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. $i$
  2. $2i$
  3. $-2i$
  4. $1+i$
Reveal answer Fill a bubble to check yourself
C Correct answer
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
$i^{2} + i^{3} + i^{4} + i^{5} = 0$
$\therefore \displaystyle \sum _{n = 2}^{11} i^{n} = i^{11} + i^{12} = -i + 1$
$\therefore \displaystyle \sum _{n = 2}^{11} i^{n} + i^{n + 1} = -2i$.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

If the measures of the sides of a triangle are ________, then it is not a right angled triangle.

  1. $3,4,5$
  2. $5,12,13$
  3. $8,24,26$
  4. $7,24,25$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} 8,24,26 \ \sin  ce, \ { \left( 8 \right) ^{ 2 } }+{ \left( { 24 } \right) ^{ 2 } }\ne { \left( { 26 } \right) ^{ 2 } } \end{array}$