Sum of squares of modulus of all the complex number z satisfying `bar(z)= iz^(2) + z^(2)- z` is equal to

Answer

To solve the problem, we need to find the sum of squares of the modulus of all complex numbers \( z \) satisfying the equation:

\[ \overline{z} = iz^2 + z^2 - z \]

### Step 1: Express \( z \) in terms of its real and imaginary parts
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. The conjugate of \( z \) is given by:

\[ \overline{z} = x - iy \]

### Step 2: Substitute \( z \) into the equation
Substituting \( z \) into the equation gives:

\[ x - iy = i(x + iy)^2 + (x + iy)^2 - (x + iy) \]

### Step 3: Expand the right-hand side
First, calculate \( (x + iy)^2 \):

\[ (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + 2xyi \]

Now, substituting this back into the equation:

\[ x - iy = i[(x^2 - y^2) + 2xyi] + [(x^2 - y^2) + 2xyi] - (x + iy) \]

This simplifies to:

\[ x - iy = i(x^2 - y^2) - 2xy + (x^2 - y^2 + 2xyi) - x - iy \]

### Step 4: Combine like terms
Combine the real and imaginary parts:

Real part:
\[ x = -2xy + (x^2 - y^2) - x \]

Imaginary part:
\[ -y = (x^2 - y^2) + 2xy - y \]

### Step 5: Rearranging the equations
From the real part, we have:

\[ 2x + 2xy = x^2 - y^2 \quad \text{(1)} \]

From the imaginary part, we have:

\[ 0 = x^2 - y^2 + 2xy \quad \text{(2)} \]

### Step 6: Solve the equations
From equation (2):

\[ x^2 - y^2 + 2xy = 0 \implies x^2 + 2xy = y^2 \implies x^2 + 2xy - y^2 = 0 \]

This is a quadratic in \( x \):

\[ x^2 + 2xy - y^2 = 0 \]

Using the quadratic formula:

\[ x = \frac{-2y \pm \sqrt{(2y)^2 + 4y^2}}{2} = \frac{-2y \pm 2\sqrt{2}y}{2} = -y \pm \sqrt{2}y \]

Thus, we have:

\[ x = (-1 + \sqrt{2})y \quad \text{or} \quad x = (-1 - \sqrt{2})y \]

### Step 7: Find the modulus of \( z \)
The modulus of \( z \) is given by:

\[ |z| = \sqrt{x^2 + y^2} \]

For each case of \( x \):

1. For \( x = (-1 + \sqrt{2})y \):

\[ |z|^2 = ((-1 + \sqrt{2})y)^2 + y^2 = (1 - 2\sqrt{2} + 2)y^2 + y^2 = (3 - 2\sqrt{2})y^2 \]

2. For \( x = (-1 - \sqrt{2})y \):

\[ |z|^2 = ((-1 - \sqrt{2})y)^2 + y^2 = (1 + 2\sqrt{2} + 2)y^2 + y^2 = (3 + 2\sqrt{2})y^2 \]

### Step 8: Sum of squares of moduli
The sum of squares of the moduli is:

\[ |z|^2 = (3 - 2\sqrt{2})y^2 + (3 + 2\sqrt{2})y^2 = 6y^2 \]

### Step 9: Conclusion
The sum of the squares of the modulus of all the complex numbers \( z \) satisfying the given equation is:

\[ \boxed{6} \]