Tag: complex numbers and linear inequations

Quizzes (4) Questions (127)

Questions Related to complex numbers and linear inequations

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

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

  1. $x =3, y = 0$
  2. $x =1, y = 3$
  3. $x =0, y = 3$
  4. $x =0, y = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $x=0\quad & \quad y=0$

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

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 

  1. real and positive

  2. real and negative
  3. $0$

  4. imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $ 0
    $
  2. $ 1
    $
  3. $ -1
    $
  4. $ None.
    $
Reveal answer Fill a bubble to check yourself
A Correct answer
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     $i^4=1$

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

$Z=0$


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

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

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

  1. i

  2. i-1

  3. -i

  4. 0
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $i^{13} + i^{14} = i - 1$

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

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

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $x=3$