Tag: complex numbers and linear inequations

Questions Related to complex numbers and linear inequations

The number of solution of $z^2 + \bar{z} = 0$ is

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $5$

  2. $4$

  3. $2$

  4. $3$

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Correct Option: B
Explanation:
Let $z=x+iy$.
Now,
$z^2+\overline{z}=0$
or, $x^2-y^2+2ixy+(x-iy)=0$
or, $(x^2-y^2+x)+i(2xy-y)=0$
Now comparing the real and imaginary part both sides we get,
$x^2-y^2+x=0$.....(1) and $2xy-y=0$.....(2).
From (2) we get, $x=\dfrac{1}{2}$ or $y=0$.
Now $x=\dfrac{1}{2}$ gives from (1) we get, $y=\pm \dfrac{\sqrt{3}}{2}$.
And $y=0$ gives from (1) we get, $x=0, 1$.
So the solution s are $(0,0), (1,0), \left(\dfrac{1}{2},\pm \dfrac{\sqrt{3}}{2}\right)$.
So we have $4$ solutions.

If $z \neq 0$, then $ \overset{100}{\underset{0}{\int}}arg(-|z|)dx =$

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $0$

  2. Not defined

  3. $100$

  4. $100\pi$

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Correct Option: A

The complex no. $\dfrac{1+2i}{1-i}$ lies in which quadrant of the complex plane

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. first

  2. second

  3. third

  4. fourth

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Correct Option: B

If $|z^2-1|=|z^2|+1$, then z lies on?

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. The real axis

  2. The imaginary axis

  3. A circle

  4. An ellipse

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Correct Option: A

Which  of the following is correct ?

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

  1. $2 + 3i > 1 + 4i$

  2. $2 + 2i > 3 + 3i$

  3. $5 + 8i > 5 + 7i$

  4. None of these

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Correct Option: C
Explanation:

Option $A$

$2+3i>1+4i$
$2^2+3^2>1^2+4^2$
$4+9>1+16$
$13>17$

It is not correct.

Option $B$
$2+2i>3+3i$

$2^2+2^2>3^2+3^2$
$4+4>9+9$
$8>18$

It is not correct.


Option $C$
$5+8i>5+7i$

$5^2+^82>5^2+7^2$
$25+64>25+49$
$89>74$

It is correct.

Hence, this is the answer.

Find the value of $\begin{vmatrix} 2+i & 2-i \ 1+i & 1-i \end{vmatrix}$ if $i^2=-1$.

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

  1. A complex quantity

  2. real quantity

  3. $0$

  4. cannot be determined

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Correct Option: A
Explanation:

$\left| \begin{matrix} 2+i & 2-i \ 1+i & 1-i \end{matrix} \right| \ =(2+i)(1-i)-(2+i)(1+i)\ =2-2i+i-{ i }^{ 2 }-2-2i-i-{ i }^{ 2 }\ =-4i+2$

So it is complex quantity

Find the value of $\dfrac{i^{4n+1}-i^{4n-1}}{2}$.

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

  1. $-1$

  2. $1$

  3. $-i$

  4. $i$

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Correct Option: D
Explanation:
We have,
$\dfrac{i^{4n+1}-i^{4n-1}}{2}$
$=\dfrac{(i^{4})^{n}.i-(i^{4})^{n}.i^{-1}}{2}$
$=\dfrac{i-i^{-1}}{2}$ [ Since $i^2=1\Rightarrow i^4=1$]
$=\dfrac{i+i}{2}$ [ As $i^2=-1\Rightarrow i=-\dfrac{1}{i}$]
$=i$.

$i^{242}=$

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

  1. $i$

  2. $-i$

  3. $1$

  4. $-1$

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Correct Option: D
Explanation:

$(i)^{242}$

$=(i^{2})^{121}$

$=(-1)^{121}$                              since $i=\sqrt{-1}$.
Hence
$(-1)^{121}=-1$
Thus
$i^{242}=-1$

$\displaystyle i+\frac{1}{i}=$

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

  1. $1$

  2. $-1$

  3. $0$

  4. $2i$

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Correct Option: C
Explanation:

Let $Z= i+\dfrac{1}{i}$


Mutiplying numerator and denominator by i. We get,

          $=i+\dfrac{i}{i^{2}}$

          $=i+\dfrac{i}{-1} \quad \dots (i^2=-1)$

          $=i-i$

      $Z=0$

Hence, 

$i+\dfrac{1}{i}=0$.

Evaluate :

 $(-\sqrt{-1})^{4n+3}, n \in N$

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

  1. -$i$

  2. $i$

  3. $1$

  4. -$1$

Show answer
Correct Option: B
Explanation:

$(-\sqrt{-1})^{4n+3} = (-i)^{4n+3}
= (-i)^{4n}(-i)^3
= {(-i)^4}^n(-i)^3
= 1 \times (-i)^3 = i$