Question

Let \( \alpha \) and \( \beta \) be the sum and the product of all the non-zero solutions of the equation \[(\bar{z})^2 + |z| = 0, \quad z \in \mathbb{C}.\] Then \( 4(\alpha^2 + \beta^2) \) is equal to:

• A. 6
• B. 4
• C. 8
• D. 2

Answer 1

The Correct Option is B

Express \(z\) as a Complex Number: 
Let \(z = x + iy\) where \(x, y \in \mathbb{R}\) and \(\overline{z} = x - iy\). 
\[ \overline{z}^2 = (x - iy)^2 = x^2 - y^2 - 2ixy. \] 
Given: \[ \overline{z}^2 + |z| = 0. \]
Here, \(|z| = \sqrt{x^2 + y^2}\), so the equation becomes:
\[ x^2 - y^2 - 2ixy + \sqrt{x^2 + y^2} = 0. \] 
Separating Real and Imaginary Parts: 
The real part: \[ x^2 - y^2 + \sqrt{x^2 + y^2} = 0. \] 

The imaginary part: \[ -2xy = 0. \] 
From the imaginary part, either \(x = 0\) or \(y = 0\). 

Case 1: \(x = 0\) 
Substituting \(x = 0\) into the real part: \[ -y^2 + |y| = 0 \implies |y| = y^2. \]
This gives \(y = 0\) or \(y = \pm 1\). Since we are looking for non-zero solutions, we have:
\[ y = \pm 1 \implies z = i \quad \text{or} \quad z = -i. \] 
Case 2: \(y = 0\) 
Substituting \(y = 0\) into the real part: \[ x^2 + \sqrt{x^2} = 0, \] which gives no non-zero solutions for \(x\). 

Solutions: 
Thus, the non-zero solutions are \(z = i\) and \(z = -i\). 

Calculating \(\alpha\) and \(\beta\): 
\[ \alpha = i + (-i) = 0, \quad \beta = i \cdot (-i) = -1. \] 
Computing \(4(\alpha^2 + \beta^2)\): 
\[ 4(\alpha^2 + \beta^2) = 4(0^2 + (-1)^2) = 4. \]