Animated Solution for Mathematics - Complex Numbers: Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z}{)}^2+|z|=0,z\in \mathbb{C}$. Then $4({\alpha }^2+{\beta }^2)$ is equal to :

Welcome, fellow explorer of the complex plane! Today, we are going to dismantle a beautiful problem that might look intimidating at first glance, but reveals a stunning symmetry once we peel back the layers.

We are dealing with the equation $(\bar{z}{)}^2+|z|=0$. At first, it looks like a jumble of conjugates and moduli, but we are going to use the most reliable tool in our JEE arsenal: the Cartesian representation.

Whenever you see an equation involving $z$, $\bar{z}$, and $|z|$, do not panic. The safest, most systematic approach is to define $z=x+iy$, where $x$ and $y$ are real numbers.

This immediately gives us the conjugate $\bar{z}=x-iy$ and the modulus $|z|=\sqrt{x^2+y^2}$. By substituting these into our original equation, we transform a complex-variable problem into a system of real-variable equations.

Let us substitute our definitions into the equation:

Now, we expand the squared term. Since $(x-iy{)}^2=x^2-y^2-2ixy$, our equation becomes:

We group the real and imaginary parts:

For this complex number to be zero, both the real part and the imaginary part must vanish independently. This gives us two crucial equations:

Look at the imaginary part: $-2xy=0$. This is a gift! It tells us that either $x=0$ or $y=0$.

Case 1: $x=0$ Substituting this into our real part equation, we get $-y^2+\sqrt{y^2}=0$. Since $\sqrt{y^2}=|y|$, we have $|y|-|y{|}^2=0$.

Factoring this, we get $|y|(1-|y|)=0$. This gives us $|y|=0$ (which leads to $z=0$) or $|y|=1$. Thus, $y=\pm 1$, yielding non-zero solutions $z=i$ and $z=-i$.

Case 2: $y=0$ Substituting this into the real part equation, we get $x^2+\sqrt{x^2}=0$, which simplifies to $|x{|}^2+|x|=0$. Since $|x|\ge 0$, the only solution is $|x|=0$, which leads to $z=0$. We reject this as it is not a non-zero solution.

We have found our non-zero solutions: $z_1=i$ and $z_2=-i$. The problem asks for the sum $\alpha$ and the product $\beta$ of these solutions.

The sum $\alpha =i+(-i)=0$. The product $\beta =i\cdot (-i)=-i^2=1$.

Finally, we evaluate the expression $4({\alpha }^2+{\beta }^2)$:

We have arrived at the answer! The final result is 4.