To find the number of distinct roots of the given complex conjugate equation:z ˉ − z 2 = i ( z ˉ + z 2 ) \bar{z} - z^2 = i(\bar{z} + z^2) z ˉ − z 2 = i ( z ˉ + z 2 )
Step 1: Simplify the equation Rearranging the terms to group z ˉ \bar{z} z ˉ z 2 z^2 z 2 z ˉ − i z ˉ = z 2 + i z 2 \bar{z} - i\bar{z} = z^2 + iz^2 z ˉ − i z ˉ = z 2 + i z 2 z ˉ ( 1 − i ) = z 2 ( 1 + i ) \bar{z}(1 - i) = z^2(1 + i) z ˉ ( 1 − i ) = z 2 ( 1 + i )
Divide both sides by ( 1 + i ) (1 + i) ( 1 + i ) z 2 = ( 1 − i 1 + i ) z ˉ z^2 = \left( \frac{1 - i}{1 + i} \right) \bar{z} z 2 = ( 1 + i 1 − i ) z ˉ
Simplifying the factor 1 − i 1 + i \frac{1 - i}{1 + i} 1 + i 1 − i 1 − i 1 + i = ( 1 − i ) 2 ( 1 + i ) ( 1 − i ) = 1 − 2 i − 1 1 − ( − 1 ) = − 2 i 2 = − i \frac{1 - i}{1 + i} = \frac{(1 - i)^2}{(1 + i)(1 - i)} = \frac{1 - 2i - 1}{1 - (-1)} = \frac{-2i}{2} = -i 1 + i 1 − i = ( 1 + i ) ( 1 − i ) ( 1 − i ) 2 = 1 − ( − 1 ) 1 − 2 i − 1 = 2 − 2 i = − i
Thus, the equation simplifies to:
z 2 = − i z ˉ — (1) z^2 = -i\bar{z} \quad \text{--- (1)} z 2 = − i z ˉ — (1)
Step 2: Solve using Cartesian coordinates Let z = x + i y z = x + iy z = x + i y x , y ∈ R x, y \in \mathbb{R} x , y ∈ R z ˉ = x − i y \bar{z} = x - iy z ˉ = x − i y z 2 = ( x 2 − y 2 ) + 2 i x y z^2 = (x^2 - y^2) + 2ixy z 2 = ( x 2 − y 2 ) + 2 i x y
Substitute these into Equation (1):( x 2 − y 2 ) + 2 i x y = − i ( x − i y ) (x^2 - y^2) + 2ixy = -i(x - iy) ( x 2 − y 2 ) + 2 i x y = − i ( x − i y ) ( x 2 − y 2 ) + 2 i x y = − y − i x (x^2 - y^2) + 2ixy = -y - ix ( x 2 − y 2 ) + 2 i x y = − y − i x
Equating the real and imaginary parts:
Real Part: x 2 − y 2 = − y ⟹ x 2 = y 2 − y — (2) x^2 - y^2 = -y \implies x^2 = y^2 - y \quad \text{--- (2)} x 2 − y 2 = − y ⟹ x 2 = y 2 − y — (2)
Imaginary Part: 2 x y = − x ⟹ x ( 2 y + 1 ) = 0 — (3) 2xy = -x \implies x(2y + 1) = 0 \quad \text{--- (3)} 2 x y = − x ⟹ x ( 2 y + 1 ) = 0 — (3)
Step 3: Analyze the cases from Equation (3) Case 1: x = 0 x = 0 x = 0 Substitute x = 0 x = 0 x = 0 0 = y 2 − y ⟹ y ( y − 1 ) = 0 0 = y^2 - y \implies y(y - 1) = 0 0 = y 2 − y ⟹ y ( y − 1 ) = 0 y y y
y = 0 ⟹ z 1 = 0 y = 0 \implies z_1 = 0 y = 0 ⟹ z 1 = 0 y = 1 ⟹ z 2 = i y = 1 \implies z_2 = i y = 1 ⟹ z 2 = i
Case 2: 2 y + 1 = 0 ⟹ y = − 1 2 2y + 1 = 0 \implies y = -\frac{1}{2} 2 y + 1 = 0 ⟹ y = − 2 1 Substitute y = − 1 2 y = -\frac{1}{2} y = − 2 1 x 2 = ( − 1 2 ) 2 − ( − 1 2 ) = 1 4 + 1 2 = 3 4 x^2 = \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} x 2 = ( − 2 1 ) 2 − ( − 2 1 ) = 4 1 + 2 1 = 4 3 x = ± 3 2 x = \pm \frac{\sqrt{3}}{2} x = ± 2 3
This gives two more roots:
z 3 = 3 2 − 1 2 i z_3 = \frac{\sqrt{3}}{2} - \frac{1}{2}i z 3 = 2 3 − 2 1 i z 4 = − 3 2 − 1 2 i z_4 = -\frac{\sqrt{3}}{2} - \frac{1}{2}i z 4 = − 2 3 − 2 1 i
Conclusion The distinct roots of the given equation are:z ∈ { 0 , i , 3 2 − 1 2 i , − 3 2 − 1 2 i } z \in \left\{ 0, \; i, \; \frac{\sqrt{3}}{2} - \frac{1}{2}i, \; -\frac{\sqrt{3}}{2} - \frac{1}{2}i \right\} z ∈ { 0 , i , 2 3 − 2 1 i , − 2 3 − 2 1 i }
Therefore, the number of distinct roots is 4 .