Animated Solution for Mathematics - Complex Numbers: Let $S=\{z\in \mathbb{C}:\bar{z}=i(z^2+\text{Re}(\bar{z}))\}$. Then ${\sum }_{z\in S}|z{|}^2$ is equal to

Welcome, fellow traveler, to the beautiful world of complex numbers. Today, we are going to dissect a problem that, at first glance, might seem like a tangled mess of variables.

But fear not! The beauty of JEE Advanced problems lies in their structure. They are not designed to break you; they are designed to reveal the hidden symmetry of mathematics. Let us embark on this journey together.

The Cartesian Foundation

We are given the set $S=\{z\in \mathbb{C}:\bar{z}=i(z^2+\text{Re}(\bar{z}))\}$. Our goal is to find the sum of the squared magnitudes of all elements in $S$.

The first step is always the most critical: how do we represent $z$? We choose the Cartesian form: $z=x+iy$, where $x,y\in \mathbb{R}$.

This immediately gives us $\bar{z}=x-iy$ and $\text{Re}(\bar{z})=x$. By defining $z$ this way, we are essentially mapping our complex problem onto the $xy$-plane, where we feel much more comfortable.

The Algebraic Expansion

Now, let us substitute these into our equation: $\bar{z}=i(z^2+x)$. Substituting our definitions, we get:

Take a deep breath. Expanding $(x+iy{)}^2$ is a classic moment where students often stumble. Remember, $i^2=-1$, so $(x+iy{)}^2=x^2-y^2+2ixy$.

Plugging this back in, the equation becomes:

Now, distribute that $i$. When $i$ hits $2ixy$, it becomes $2i^{2}xy$, which is $-2xy$. This is the magic moment where the real and imaginary parts begin to separate. Our equation now looks like this:

The Bifurcation

For two complex numbers to be equal, their real parts must match, and their imaginary parts must match. This gives us a system of two real equations:

1. Real part: $x=-2xy$ 2. Imaginary part: $-y=x^2-y^2+x$

Look at the first equation: $x+2xy=0$, which factors beautifully into $x(1+2y)=0$. This is our bifurcation point! We have two distinct paths to follow.

If $x=0$, the imaginary equation becomes $-y=0^2-y^2+0$, or $y^2-y=0$. This gives us $y=0$ or $y=1$.

Thus, we have two complex numbers: $z_1=0+0i=0$ and $z_2=0+i=i$.

If $y=-\frac{1}{2}$, the imaginary equation becomes $-(-1/2)=x^2-(-1/2{)}^2+x$. Simplifying this, we get $1/2=x^2-1/4+x$.

Multiplying by 4 to clear the fractions, we get $2=4x^2-1+4x$, or $4x^2+4x-3=0$. Factoring this quadratic, we find $(2x-1)(2x+3)=0$, giving us $x=1/2$ and $x=-3/2$.

This yields two more complex numbers: $z_3=\frac{1}{2}-\frac{1}{2}i$ and $z_4=-\frac{3}{2}-\frac{1}{2}i$.

The Final Summation

We have found our four points: $0,i,\frac{1}{2}-\frac{1}{2}i,-\frac{3}{2}-\frac{1}{2}i$. The problem asks for ${\sum }_{z\in S}|z{|}^2$. Let us calculate the squared magnitude for each:

- $|z_1{|}^2=0^2+0^2=0$ - $|z_2{|}^2=0^2+1^2=1$ - $|z_3{|}^2=(\frac{1}{2}{)}^2+(-\frac{1}{2}{)}^2=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$ - $|z_4{|}^2=(-\frac{3}{2}{)}^2+(-\frac{1}{2}{)}^2=\frac{9}{4}+\frac{1}{4}=\frac{10}{4}=\frac{5}{2}$

Adding these together: $0+1+\frac{1}{2}+\frac{5}{2}=1+\frac{6}{2}=1+3=4$.

And there you have it! The final answer is 4. Through careful decomposition and systematic solving, we turned a complex equation into a simple, elegant result.