Question

The number of complex numbers \(z\) satisfying \[ \overline{z}=iz^2 \] is

• A. \(3\)
• B. \(4\)
• C. \(2\)
• D. \(5\)

Hint:

For equations involving \(z\) and \(\overline{z}\), always put \[ z=x+iy \] and compare real and imaginary parts separately.

Answer 1

The Correct Option is B

Step 1: Assume \[ z=x+iy \] where \[ x,y\in \mathbb{R} \] Then, \[ \overline{z}=x-iy \] Also, \[ z^2=(x+iy)^2 \] \[ =x^2-y^2+2ixy \] Hence, \[ iz^2=i(x^2-y^2+2ixy) \] \[ =i(x^2-y^2)-2xy \] Therefore, \[ iz^2=-2xy+i(x^2-y^2) \]

Step 2: Compare real and imaginary parts.
Given, \[ \overline{z}=iz^2 \] So, \[ x-iy=-2xy+i(x^2-y^2) \] Comparing real parts, \[ x=-2xy \] Comparing imaginary parts, \[ -y=x^2-y^2 \]

Step 3: Solve the equations.
From \[ x=-2xy, \] we get \[ x(1+2y)=0 \] Thus, \[ x=0 \] or \[ y=-\frac12 \]

Case 1: \(x=0\)
Substitute into \[ -y=x^2-y^2 \] Then, \[ -y=-y^2 \] \[ y^2-y=0 \] \[ y(y-1)=0 \] Hence, \[ y=0 \] or \[ y=1 \] So two solutions are \[ z=0 \] and \[ z=i \]

Case 2: \[ y=-\frac12 \] Substitute into \[ -y=x^2-y^2 \] Then, \[ \frac12=x^2-\frac14 \] \[ x^2=\frac34 \] \[ x=\pm \frac{\sqrt3}{2} \] Thus two more solutions are \[ z=\frac{\sqrt3}{2}-\frac{i}{2} \] and \[ z=-\frac{\sqrt3}{2}-\frac{i}{2} \]

Step 4: Count the solutions.
Total number of complex numbers satisfying the equation is \[ 4 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{4} \]