Let z̄ denote the complex conjugate of a complex number z and let i = √-1. In the set of complex num

Q: Wait, why can't I just take the modulus of both sides immediately at the start?

That is a brilliant instinct! Taking the modulus is a powerful tool. If you take the modulus of both sides, you get $|z^2| = |i| |\bar{z} + z^2|$, which simplifies to $|z|^2 = |\bar{z} + z^2|$. While this gives you a geometric constraint, it doesn't immediately give you the roots. It tells you about the magnitude, but you still need the algebraic approach to find the specific values of $z$. Use the modulus as a check, not as the primary solver!

Q: Why do we get two cases from $2xy + x = 0$? Can't I just divide by $x$?

Never divide by a variable unless you are absolutely certain it is non-zero! If you divide by $x$, you assume $x \neq 0$, and you will lose the solutions where $x=0$. By factoring out $x$, you capture all possibilities: the case where $x=0$ and the case where $2y+1=0$. Always factor; never divide by variables.

Q: Is $z=0$ actually a root? I feel like I might have missed it.

You are absolutely right to check! Always test the trivial solution. If you plug $z=0$ into the original equation $\bar{z} - z^2 = i(\bar{z} + z^2)$, you get $0 - 0 = i(0 + 0)$, which is $0=0$. It works! It is a valid root. Never assume the origin isn't a solution.

Question

Let $\bar{z}$ denote the complex conjugate of a complex number $z$ and let $i = \sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation $\bar{z} - z^2 = i(\bar{z} + z^2)$ is _____________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The dance of the complex plane is a journey into symmetry. When you first look at the equation $\bar{z} - z^2 = i(\bar{z} + z^2)$, it is natural to feel a momentary spike of anxiety.

However, in the world of JEE Advanced, clutter is often just a mask for hidden symmetry. Let us peel back that mask together.

### The Art of Grouping

The first rule of complex algebra is to organize your battlefield. We have $\bar{z}$ on both sides and $z^2$ on both sides.

By expanding the right-hand side, we get:
$$\bar{z} - z^2 = i\bar{z} + iz^2$$

Now, move the terms to group the variables:
$$\bar{z} - i\bar{z} = z^2 + iz^2$$

Factoring gives us:
$$\bar{z}(1 - i) = z^2(1 + i)$$

Suddenly, the equation is breathing. It is no longer a chaotic jumble; it is a clean relationship between the conjugate and the square.

### The Magic of Rationalization

Now, we isolate $z^2$ by dividing by $(1 + i)$:
$$z^2 = \bar{z} \left( \frac{1 - i}{1 + i} \right)$$

That fraction is a classic trap for the unprepared. We multiply the numerator and denominator by the conjugate of the denominator, $(1 - i)$:
$$\frac{1 - i}{1 + i} \cdot \frac{1 - i}{1 - i} = \frac{1 - 2i + i^2}{1^2 - i^2} = \frac{-2i}{2} = -i$$

The entire fraction collapses into a single, beautiful imaginary unit: $-i$. Our equation is now simply:
$$z^2 = -i\bar{z}$$

### The Cartesian Dive

With the equation simplified, we bring in the heavy artillery: the Cartesian form. Let $z = x + iy$, where $x, y \in \mathbb{R}$. Then $\bar{z} = x - iy$.

Substituting these into our simplified equation gives:
$$(x + iy)^2 = -i(x - iy)$$

Expanding both sides:
$$x^2 - y^2 + 2ixy = -ix + i^2y$$
$$x^2 - y^2 + 2ixy = -y - ix$$

Now, we equate the real and imaginary parts:
1. Real parts: $x^2 - y^2 = -y$
2. Imaginary parts: $2xy = -x$

### The Branching Path

Look at the imaginary part equation: $2xy = -x$. This is equivalent to $x(2y + 1) = 0$.

This is a fork in the road. Either $x = 0$ or $y = -1/2$. We must explore both paths.

**Case 1: If $x = 0$**
The real part equation $x^2 - y^2 = -y$ becomes $-y^2 = -y$, or $y^2 - y = 0$. This gives $y(y - 1) = 0$, so $y = 0$ or $y = 1$. 
This yields two roots: $z_1 = 0$ and $z_2 = i$.

**Case 2: If $y = -1/2$**
The real part equation $x^2 - y^2 = -y$ becomes:
$$x^2 - \left(-\frac{1}{2}\right)^2 = -\left(-\frac{1}{2}\right)$$
$$x^2 - \frac{1}{4} = \frac{1}{2} \Rightarrow x^2 = \frac{3}{4}$$

Thus, $x = \pm \frac{\sqrt{3}}{2}$. This yields two more roots: $z_3 = \frac{\sqrt{3}}{2} - \frac{1}{2}i$ and $z_4 = -\frac{\sqrt{3}}{2} - \frac{1}{2}i$.

### Final Conclusion

We have found four distinct roots:
**$0, \quad i, \quad \frac{\sqrt{3}}{2} - \frac{1}{2}i, \quad -\frac{\sqrt{3}}{2} - \frac{1}{2}i$**

Notice the beauty of these results. If you take the modulus of our simplified equation $z^2 = -i\bar{z}$, you get $|z|^2 = |-i| |\bar{z}|$, which means $|z|^2 = |z|$.

This implies $|z| = 0$ or $|z| = 1$. Our roots are either at the origin or on the unit circle, demonstrating the elegance of complex geometry.

Animated Solution for Mathematics - Complex Numbers: Let $\overset{z}{ˉ}$ denote the complex conjugate of a complex number $z$ and let $i = - 1$ . In the set of complex numbers, the number of distinct roots of the equation $\overset{z}{ˉ} - z^{2} = i (\overset{z}{ˉ} + z^{2})$ is _____________.

Enter Numerical Value:

Visualized Solution

The Equation

Rearranging Terms

Isolating $z^{2}$

Simplifying the Fraction

Simplified Equation

Cartesian Substitution

Expanding Both Sides

Equating Parts

Solving Equation 2

Case 1: $x = 0$

Case 2: $y = - 1/2$

Geometric Interpretation

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Art of Grouping

The Magic of Rationalization

The Cartesian Dive

The Branching Path

Final Conclusion

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