Question

The equation solution $ {z^2} = \bar z $ where $ z $ is a complex number has
a) 4 solutions
b) 2 solutions
c) No solution
d) Infinitely many solutions

Answer 1

Hint: We have the equation $ {z^2} = \bar z $ we will use the elementary properties of a complex number and reduce this equation to a simpler form, we will also use the property of conjugate of a complex number and equality of two complex numbers. Then we will conclude how many solutions the equation has. Clearly, we can see that the trivial solution $ z = 0 $ is satisfied so the option ‘c’ can be ignored, Similarly will find other solutions to justify our answer as to how many solutions the equation has.

Complete step-by-step answer:
The given equation is
 $ {z^2} = \bar z $ --(1)
Let $ z = x + iy $
Now, the square of $ z $ is $ {z^2} = {x^2} - {y^2} + i(2xy) $
Also the conjugate of $ z $ is $ \bar z = x - iy $
Using these results in (1) we get,
 $ {x^2} - {y^2} + i(2xy) = x - iy $
Clearly, the solutions are $ x = y = 0 $ and $ x = 1,y = 0 $
Let’s check t further,
The equality of two complex numbers exists if the real and imaginary part are equal.
Using this fact we have,
 $ {x^2} - {y^2} = x $ and $ 2xy = - y $ --(2)
Clearly, we have $ 2xy = - y \Rightarrow x = \dfrac{{ - 1}}{2} $
Now putting this value in $ {x^2} - {y^2} = x $ we have,
 $ {\left( {\dfrac{{ - 1}}{2}} \right)^2} - {y^2} = \dfrac{{ - 1}}{2} $
 $ \Rightarrow \dfrac{1}{4} + \dfrac{1}{2} = {y^2} $
 $ \Rightarrow {y^2} = \dfrac{3}{4} $
 $ \Rightarrow y = \dfrac{{ \pm \sqrt 3 }}{2} $
The solutions we obtained in coordinate forms are $ (0,0),(1,0),\left( {\dfrac{{ - 1}}{2},\dfrac{{\sqrt 3 }}{2}} \right),\left( {\dfrac{{ - 1}}{2},\dfrac{{ - \sqrt 3 }}{2}} \right) $ .
Therefore, we conclude that the given complex equation has only four roots. This implies option ‘a’ is correct among all other choices.
So, the correct answer is “Option A”.

Note: While solving equations always check if it is satisfied for common points like $ (0,0),(1,0),(0,1), $ etc because these points can’t be found out by solving the equations while they too contribute to the solution of the equation. The points get missed while solving equations like $ 2xy = - y $ . So, make sure to do this thing first always.