Key Concepts and Formulas

• Complex Number Representation: $z = x + iy$, where $x$ is the real part (Re(z)) and $y$ is the imaginary part (Im(z)).
• Conjugate of a Complex Number: $\overline{z} = x - iy$
• Modulus Squared: $|z|^2 = x^2 + y^2$
• Equality of Complex Numbers: $a + bi = c + di$ if and only if $a = c$ and $b = d$.

Step-by-Step Solution

Step 1: Substitute $z = x + iy$ and Expand

The given equation is: $\overline{z} = i z^2 + z^2 - z$ We substitute $z = x + iy$ and $\overline{z} = x - iy$ into the equation. Also, we calculate $z^2 = (x + iy)^2 = x^2 + 2ixy - y^2 = (x^2 - y^2) + 2ixy$. Then, $iz^2 = i(x^2 - y^2 + 2ixy) = i(x^2 - y^2) - 2xy$. Substituting these into the original equation yields: $x - iy = i((x+iy)^2) + (x+iy)^2 - (x+iy)$ $x - iy = i(x^2 - y^2 + 2ixy) + (x^2 - y^2 + 2ixy) - (x + iy)$ $x - iy = i(x^2 - y^2) - 2xy + x^2 - y^2 + 2ixy - x - iy$ Now, we group the real and imaginary parts: $x - iy = (x^2 - y^2 - 2xy - x) + i(x^2 - y^2 + 2xy - y)$ This step is crucial to separate the complex equation into two real equations.

Step 2: Equate Real and Imaginary Parts

We equate the real parts on both sides of the equation: $x = x^2 - y^2 - 2xy - x$ Rearranging the terms, we get: $2x = x^2 - y^2 - 2xy \quad \dots(1)$ Next, we equate the imaginary parts: $-y = x^2 - y^2 + 2xy - y$ Rearranging the terms, we get: $0 = x^2 - y^2 + 2xy \quad \dots(2)$ Now we have a system of two real equations with two variables.

Step 3: Solve the System of Equations

We have the following system of equations: $2x = x^2 - y^2 - 2xy \quad \dots(1)$ $0 = x^2 - y^2 + 2xy \quad \dots(2)$ From equation (2), we can write: $x^2 - y^2 = -2xy \quad \dots(3)$ Substituting (3) into (1), we get: $2x = -2xy - 2xy$ $2x = -4xy$ $2x + 4xy = 0$ $2x(1 + 2y) = 0$ This gives us two cases:

• Case A: $x = 0$
• Case B: $1 + 2y = 0 \implies y = -\frac{1}{2}$

Case A: $x = 0$ Substituting $x = 0$ into equation (2): $0^2 - y^2 + 2(0)y = 0$ $-y^2 = 0$ $y = 0$ So, $z_1 = 0 + 0i = 0$.

Case B: $y = -\frac{1}{2}$ Substituting $y = -\frac{1}{2}$ into equation (2): $x^2 - \left(-\frac{1}{2}\right)^2 + 2x\left(-\frac{1}{2}\right) = 0$ $x^2 - \frac{1}{4} - x = 0$ $4x^2 - 4x - 1 = 0$ Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 4$, $b = -4$, and $c = -1$: $x = \frac{4 \pm \sqrt{16 - 4(4)(-1)}}{8}$ $x = \frac{4 \pm \sqrt{32}}{8}$ $x = \frac{4 \pm 4\sqrt{2}}{8}$ $x = \frac{1 \pm \sqrt{2}}{2}$ So we have two solutions for $x$: $x_2 = \frac{1 + \sqrt{2}}{2}$ $x_3 = \frac{1 - \sqrt{2}}{2}$ Thus, we have two more complex numbers: $z_2 = \frac{1 + \sqrt{2}}{2} - \frac{1}{2}i$ $z_3 = \frac{1 - \sqrt{2}}{2} - \frac{1}{2}i$ The solutions are $z_1 = 0$, $z_2 = \frac{1 + \sqrt{2}}{2} - \frac{1}{2}i$, and $z_3 = \frac{1 - \sqrt{2}}{2} - \frac{1}{2}i$.

Step 4: Calculate the Sum of Squares of Moduli

We need to find $|z_1|^2 + |z_2|^2 + |z_3|^2$.

• $|z_1|^2 = 0^2 + 0^2 = 0$
• $|z_2|^2 = \left(\frac{1 + \sqrt{2}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 = \frac{1 + 2 + 2\sqrt{2}}{4} + \frac{1}{4} = \frac{4 + 2\sqrt{2}}{4} = \frac{2 + \sqrt{2}}{2}$
• $|z_3|^2 = \left(\frac{1 - \sqrt{2}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 = \frac{1 + 2 - 2\sqrt{2}}{4} + \frac{1}{4} = \frac{4 - 2\sqrt{2}}{4} = \frac{2 - \sqrt{2}}{2}$

Therefore, the sum of the squares of the moduli is: $0 + \frac{2 + \sqrt{2}}{2} + \frac{2 - \sqrt{2}}{2} = \frac{4}{2} = 2$

Common Mistakes & Tips

• Carefully handle the signs and imaginary unit 'i' during expansion and simplification.
• Remember to consider both cases when factoring an equation like $2x(1+2y) = 0$.
• Double-check the quadratic formula calculations to avoid errors.

Summary

By substituting $z = x + iy$ into the given equation and equating the real and imaginary parts, we obtained a system of two real equations. Solving this system yielded three complex number solutions. Finally, calculating the sum of the squares of the moduli of these solutions gave us the answer.

The final answer is $\boxed{2}$.