Questions & Answers

Question

Answers

$

(a){\text{ 2}} \\

(b){\text{ 4}} \\

(c){\text{ 6}} \\

(d){\text{ 8}} \\

$

Answer
Verified

Hint – In this question we need to find the number of solutions of the given equation. Use the basic concept that a complex number can be written in the form of x + iy. Substitute this value in the given equation and simplify according to the conditions given using algebraic identities to reach the right answer.

“Complete step-by-step answer:”

Given equation is

${z^2} + \bar z = 0$

We have to find out the number of solutions of this equation.

As we know z is the complex number.

So, let $z = x + iy$

And the conjugate of z is $\bar z = \overline {x + iy} = x - iy$

So, substitute this value in above equation we have,

$ \Rightarrow {\left( {x + iy} \right)^2} + \left( {x - iy} \right) = 0$

Now expand the square according to ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$.

$ \Rightarrow {x^2} + {i^2}{y^2} + 2ixy + \left( {x - iy} \right) = 0$

Now as we know in complex the value of ${i^2} = - 1$ so, substitute this value in above equation we have,

$ \Rightarrow {x^2} - {y^2} + 2ixy + x - iy = 0$

Now separate real and imaginary terms we have,

$ \Rightarrow {x^2} - {y^2} + x + i\left( {2xy - y} \right) = 0$

Now compare real and imaginary terms we have,

$

\Rightarrow {x^2} - {y^2} + x = 0{\text{ }}..........\left( 1 \right){\text{ \& }} \\

2xy - y = 0.......................\left( 2 \right) \\

$

Now simplify equation (2) we have,

$

\Rightarrow y\left( {2x - 1} \right) = 0 \\

\Rightarrow y = 0,{\text{ }}2x - 1 = 0 \\

$

$ \Rightarrow y = 0{\text{ \& }}x = \dfrac{1}{2}$

Now from equation (1)

If y = 0

$

\Rightarrow {x^2} - 0 + x = 0 \\

\Rightarrow x\left( {x + 1} \right) = 0 \\

\Rightarrow x = 0,{\text{ }}x + 1 = 0 \\

\Rightarrow x = 0, - 1 \\

$

When $x = \dfrac{1}{2}$

$

\Rightarrow {\left( {\dfrac{1}{2}} \right)^2} - {y^2} + \dfrac{1}{2} = 0 \\

\Rightarrow {y^2} = \dfrac{1}{4} + \dfrac{1}{2} = \dfrac{3}{4} \\

\Rightarrow y = \pm \sqrt {\dfrac{3}{4}} = \pm \dfrac{{\sqrt 3 }}{2} \\

$

So the required solutions of given equation is (x, y) = (0 ,0), (-1, 0), ($\dfrac{1}{2},\dfrac{{\sqrt 3 }}{2}$), ($\dfrac{1}{2}, - \dfrac{{\sqrt 3 }}{2}$)

So, the number of solutions is 4.

Hence option (b) is correct.

Note – Whenever we face such types of problems the key concept is simply to use the basic generalization formula for any complex number. It is always advisable to have a good gist of the algebraic identities some of them have been mentioned above while performing the solution. This will help in getting the right track to reach the answer.

“Complete step-by-step answer:”

Given equation is

${z^2} + \bar z = 0$

We have to find out the number of solutions of this equation.

As we know z is the complex number.

So, let $z = x + iy$

And the conjugate of z is $\bar z = \overline {x + iy} = x - iy$

So, substitute this value in above equation we have,

$ \Rightarrow {\left( {x + iy} \right)^2} + \left( {x - iy} \right) = 0$

Now expand the square according to ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$.

$ \Rightarrow {x^2} + {i^2}{y^2} + 2ixy + \left( {x - iy} \right) = 0$

Now as we know in complex the value of ${i^2} = - 1$ so, substitute this value in above equation we have,

$ \Rightarrow {x^2} - {y^2} + 2ixy + x - iy = 0$

Now separate real and imaginary terms we have,

$ \Rightarrow {x^2} - {y^2} + x + i\left( {2xy - y} \right) = 0$

Now compare real and imaginary terms we have,

$

\Rightarrow {x^2} - {y^2} + x = 0{\text{ }}..........\left( 1 \right){\text{ \& }} \\

2xy - y = 0.......................\left( 2 \right) \\

$

Now simplify equation (2) we have,

$

\Rightarrow y\left( {2x - 1} \right) = 0 \\

\Rightarrow y = 0,{\text{ }}2x - 1 = 0 \\

$

$ \Rightarrow y = 0{\text{ \& }}x = \dfrac{1}{2}$

Now from equation (1)

If y = 0

$

\Rightarrow {x^2} - 0 + x = 0 \\

\Rightarrow x\left( {x + 1} \right) = 0 \\

\Rightarrow x = 0,{\text{ }}x + 1 = 0 \\

\Rightarrow x = 0, - 1 \\

$

When $x = \dfrac{1}{2}$

$

\Rightarrow {\left( {\dfrac{1}{2}} \right)^2} - {y^2} + \dfrac{1}{2} = 0 \\

\Rightarrow {y^2} = \dfrac{1}{4} + \dfrac{1}{2} = \dfrac{3}{4} \\

\Rightarrow y = \pm \sqrt {\dfrac{3}{4}} = \pm \dfrac{{\sqrt 3 }}{2} \\

$

So the required solutions of given equation is (x, y) = (0 ,0), (-1, 0), ($\dfrac{1}{2},\dfrac{{\sqrt 3 }}{2}$), ($\dfrac{1}{2}, - \dfrac{{\sqrt 3 }}{2}$)

So, the number of solutions is 4.

Hence option (b) is correct.

Note – Whenever we face such types of problems the key concept is simply to use the basic generalization formula for any complex number. It is always advisable to have a good gist of the algebraic identities some of them have been mentioned above while performing the solution. This will help in getting the right track to reach the answer.

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