Tag: complex numbers and linear inequations

Questions Related to complex numbers and linear inequations

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     $i^4=1$

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

$Z=0$


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

The value of the sum $\displaystyle \sum _{n=1}^{13}(i^n+i^{n+1})$, where $i=\sqrt {-1}$, equals

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

  1. i

  2. i-1

  3. -i

  4. 0

Show answer
Correct Option: B
Explanation:

We have $i^2 = -1$
Thus, $\displaystyle \sum _{n=1}^{4}(i^n+i^{n+1}) = (i^1 + i^2) + (i^2 + i^3) + (i^3 + i^4) + (i^4 + i^5) = (i - 1) + (-1 - i) + (-i + 1) + (1 + i) = 0$
\Rightarrow $\displaystyle \sum _{n=1}^{12}(i^n+i^{n+1}) = 0$
Now, only remains is $i^{13} + i^{14} = i - 1$

The value of $5\sqrt {-8}$ is 

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

  1. $10i\sqrt {4}$

  2. $20i\sqrt {2}$

  3. $10i\sqrt {2}$

  4. None of these

Show answer
Correct Option: C
Explanation:

$5\sqrt {-8}$ $= 5\sqrt {8}\times \sqrt {-1}$

$=5\ i \sqrt {8}$ $= 5\ i \sqrt {4\times 2}$

$= 10\ i \sqrt {2}$
So, option C is correct.

The value of $2\sqrt {-49}$ is equal to

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

  1. $-14$

  2. None of these

  3. $14$

  4. $14i$

Show answer
Correct Option: D
Explanation:

$2\sqrt {-49}$ $= 2\sqrt {-1 \times 7\times 7}$

$= 2\times 7\sqrt {-1}$ $= 14i$
So, option D is correct.

The value of $\sqrt {-36} $ is

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

  1. $6$

  2. $-6$

  3. $6i$

  4. None of these

Show answer
Correct Option: C
Explanation:

$\sqrt {-36}$ $=\sqrt {-1 \times 6 \times 6}$

$= 6\sqrt {-1}$   ...............$(\because \sqrt {-1} = i)$
$= 6i$
So, option $C$ is correct.

If $(i^{413})(i^x)=1$, then determine the one possible value of x.

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

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: D
Explanation:

${ i }^{ 413 }{ i }^{ x }=1$

$\Rightarrow \quad { i }^{ 413 }=1$
now, $\left( 413+x \right) $ must be a multiple of 4 becouse ${ i }^{ 4 }=1$
$\therefore \quad \left( 413+3 \right) $ is divisible by $4$
                                     hence $x=3$

Evaluate and write in standard form $(4-2i)(-3+3i)$, where ${i}^{2}=-1$.

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

  1. $6+18i$

  2. $-6+18i$

  3. $12+18i$

  4. $6-18i$

Show answer
Correct Option: B
Explanation:

Consider $(4-2i)(-3+3i)$

$\Rightarrow (4-2i)(-3+3i)=4(-3)+4(3i)-2i(-3)-2i(3i)$
$=-12+12i+6i-6i^2$
$=-12+18i+6$       ..... (as $i^2=-1$)
$=-6+18i$