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Question: The number of solution of the equation \({z^2} + \bar z = 0\)is \( (a){\text{ 2}} \\\ (b){...

The number of solution of the equation z2+zˉ=0{z^2} + \bar z = 0is
(a) 2 (b) 4 (c) 6 (d) 8  (a){\text{ 2}} \\\ (b){\text{ 4}} \\\ (c){\text{ 6}} \\\ (d){\text{ 8}} \\\

Explanation

Solution

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
z2+zˉ=0{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+iyz = x + iy
And the conjugate of z is zˉ=x+iy‾=x−iy\bar z = \overline {x + iy} = x - iy
So, substitute this value in above equation we have,
⇒(x+iy)2+(x−iy)=0\Rightarrow {\left( {x + iy} \right)^2} + \left( {x - iy} \right) = 0
Now expand the square according to (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab.
⇒x2+i2y2+2ixy+(x−iy)=0\Rightarrow {x^2} + {i^2}{y^2} + 2ixy + \left( {x - iy} \right) = 0
Now as we know in complex the value of i2=−1{i^2} = - 1 so, substitute this value in above equation we have,
⇒x2−y2+2ixy+x−iy=0\Rightarrow {x^2} - {y^2} + 2ixy + x - iy = 0
Now separate real and imaginary terms we have,
⇒x2−y2+x+i(2xy−y)=0\Rightarrow {x^2} - {y^2} + x + i\left( {2xy - y} \right) = 0
Now compare real and imaginary terms we have,
⇒x2−y2+x=0 ..........(1) &  2xy−y=0.......................(2)  \Rightarrow {x^2} - {y^2} + x = 0{\text{ }}..........\left( 1 \right){\text{ \& }} \\\ 2xy - y = 0.......................\left( 2 \right) \\\
Now simplify equation (2) we have,
⇒y(2x−1)=0 ⇒y=0, 2x−1=0  \Rightarrow y\left( {2x - 1} \right) = 0 \\\ \Rightarrow y = 0,{\text{ }}2x - 1 = 0 \\\
⇒y=0 & x=12\Rightarrow y = 0{\text{ \& }}x = \dfrac{1}{2}
Now from equation (1)
If y = 0
⇒x2−0+x=0 ⇒x(x+1)=0 ⇒x=0, x+1=0 ⇒x=0,−1  \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=12x = \dfrac{1}{2}
⇒(12)2−y2+12=0 ⇒y2=14+12=34 ⇒y=±34=±32  \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), (12,32\dfrac{1}{2},\dfrac{{\sqrt 3 }}{2}), (12,−32\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.