Question

The number of solution of the equation ${z^2} + \bar z = 0$is
$
  (a){\text{ 2}} \\
  (b){\text{ 4}} \\
  (c){\text{ 6}} \\
  (d){\text{ 8}} \\
$

Answer 1

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.