Animated Solution for Mathematics - Complex Numbers: Sum of squares of modulus of all the complex numbers $z$ satisfying $\bar{z}=i{z}^2+z^2-z$ is equal to \_\_\_\_\_.

Imagine you are standing on the complex plane, looking at the equation $\bar{z}=i{z}^2+z^2-z$. At first glance, it looks like a tangled mess of variables.

In the world of JEE Advanced, complexity is often just a mask for hidden symmetry. We are not just solving an equation; we are uncovering the geometric footprint of these complex numbers.

The first instinct for many students is to panic. How do we handle $z^2$ and $\bar{z}$ simultaneously?

The secret lies in the property of the conjugate. We know that for any complex number $z=x+iy$, the sum $\bar{z}+z$ is simply $2x$.

By rearranging our given equation to $\bar{z}+z=(1+i)z^2$, we have transformed a chaotic expression into something that speaks the language of real and imaginary parts. We have effectively isolated the real part of the equation on the left side.

Now, let us step into the Cartesian realm. We substitute $z=x+iy$ into our equation. The left side becomes $2x$.

The right side, $(1+i)z^2$, requires careful expansion. We know $z^2=(x+iy{)}^2=x^2-y^2+2ixy$. When we multiply this by $(1+i)$, we get:

This is the heart of the problem. We have successfully separated the real and imaginary components. The equation now stands as:

Since the left side is purely real, the imaginary part of the right side must be zero. This gives us our first constraint:

This is a beautiful, homogeneous equation. It tells us that the imaginary part vanishes under specific conditions. Simultaneously, the real part must satisfy:

By substituting the condition $x^2-y^2=-2xy$ into this real part equation, we get $2x=-2xy-2xy$, which simplifies to $2x=-4xy$. This leads us to the elegant factorization:

The equation $2x(1+2y)=0$ tells us that either $x=0$ or $y=-\frac{1}{2}$.

If $x=0$, substituting back into our imaginary constraint gives $y=0$. Thus, $z_1=0$ is our first solution. It sits right at the origin, the anchor of our complex plane.

If $y=-\frac{1}{2}$, we substitute this into our imaginary constraint $x^2-y^2+2xy=0$ to get:

This quadratic equation gives us two more values for $x$, corresponding to two more complex numbers, $z_2$ and $z_3$. These numbers lie on the horizontal line $y=-\frac{1}{2}$.

Finally, we calculate the sum of the squares of the moduli. For $z_1=0$, $|z_1{|}^2=0$. For $z_2$ and $z_3$, we have $|z{|}^2=x^2+y^2=x^2+\frac{1}{4}$.

The sum of the squares of the moduli is:

Using the sum and product of roots for $4x^2-4x-1=0$, we know $x_2+x_3=1$ and $x_2{x}_3=-\frac{1}{4}$. Thus:

Adding the $\frac{1}{2}$ from our previous step, we get $\frac{3}{2}+\frac{1}{2}=2$. The final result is 2.