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

Answer: (B)

4

Answer 2

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 $4(\alpha^2 + \beta^2)$:
$4(\alpha^2 + \beta^2) = 4(0^2 + (-1)^2) = 4.$