Question

Number of values of a complex number \( z \) satisfying the condition \( z = z^2 + i\text{Im}(\overline{z}) \) is:

• A. 0
• B. 2
• C. 1
• D. 3

Hint:

Always factorize the imaginary part first if it contains products. It usually splits the problem into simpler real-valued cases.

Answer 1

The Correct Option is B

Concept: For any complex number \( z = x + iy \):
• \( \overline{z} = x - iy \), so \( \text{Im}(\overline{z}) = -y \).
• Equating real and imaginary parts of a complex equation allows solving for \( x \) and \( y \).

Step 1: Substituting \( z = x+iy \) into the equation.
\[ x + iy = (x+iy)^2 + i(-y) \] \[ x + iy = x^2 - y^2 + 2ixy - iy \] Equating Real parts: \( x = x^2 - y^2 \quad \cdots (1) \) Equating Imaginary parts: \( y = 2xy - y \quad \Rightarrow \quad 2y - 2xy = 0 \quad \Rightarrow \quad 2y(1-x) = 0. \)

Step 2: Solving the cases.
From the imaginary part equation, either \( y = 0 \) or \( x = 1 \). Case 1: \( y = 0 \) Substitute into (1): \( x = x^2 \Rightarrow x(x-1) = 0 \Rightarrow x=0, x=1 \). Solutions: \( z_1 = 0+0i = 0 \), \( z_2 = 1+0i = 1 \). Case 2: \( x = 1 \) Substitute into (1): \( 1 = 1^2 - y^2 \Rightarrow y^2 = 0 \Rightarrow y=0 \). This yields \( z = 1 \), which is already found.

Step 3: Counting solutions.
The unique values are \( z = 0 \) and \( z = 1 \). There are 2 such values.