Let ω be the complex number cos(2π/3) + i sin(2π/3). Then the number of distinct complex numbers z s

Q: Why did we add all columns to the first column? Is there a rule for this?

It is a classic JEE strategy! When you see a determinant where the sum of elements in each row or column is the same, adding them together creates a common factor. It is not a rule, but a powerful observation technique to simplify complex matrices.

Q: Wait, why do we reject the negative root or other possibilities? Is $z^3=0$ really just $z=0$?

In the complex plane, $z^3=0$ implies $z \cdot z \cdot z = 0$. By the Zero Product Property, $z$ must be $0$. There are no other distinct roots here. The question asks for the number of distinct complex numbers, and since $z=0$ is the only solution, the answer is $1$.

Q: Can I just expand the determinant directly without the transformations?

You could, but it is a trap! Expanding a $3 \times 3$ determinant with complex variables directly leads to a massive algebraic mess where you are almost guaranteed to make a sign error. The transformations are designed to save you time and ensure accuracy.

Question

Let $\omega$ be the complex number $\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}$. Then the number of distinct complex numbers $z$ satisfying $\begin{vmatrix} z+1 & \omega & \omega^2 \ \omega & z+\omega^2 & 1 \ \omega^2 & 1 & z+\omega \end{vmatrix} = 0$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today we are going to unravel a problem that, at first glance, might seem like a daunting wall of complex numbers and determinants. By the end of this journey, you will see it for what it truly is: a beautiful dance of symmetry.

Let us begin with the star of our show, $\omega$. You know it as $\omega = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}$. In the complex plane, this is a rotation of $120^\circ$ and represents the complex cube root of unity.

Remember that the cube roots of unity are the vertices of an equilateral triangle inscribed in the unit circle. We rely on two fundamental properties: 
$$1 + \omega + \omega^2 = 0$$
$$\omega^3 = 1$$

### The Intimidating Matrix

Now, consider the determinant:
$$\Delta = \begin{vmatrix} z+1 & \omega & \omega^2 \ \omega & z+\omega^2 & 1 \ \omega^2 & 1 & z+\omega \end{vmatrix} = 0$$

Look at the rows and columns. In JEE problems, whenever you see a structure like this, there is almost always a hidden symmetry waiting to be exploited.

If we apply the column operation $C_1 	o C_1 + C_2 + C_3$, something magical happens. Each element in the first column becomes $(z+1) + \omega + \omega^2$. Since $1 + \omega + \omega^2 = 0$, this simplifies to just $z$.

The determinant becomes:
$$z \begin{vmatrix} 1 & \omega & \omega^2 \ 1 & z+\omega^2 & 1 \ 1 & 1 & z+\omega \end{vmatrix} = 0$$

### The Path to Simplification

To make the calculation even easier, we use row operations: $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$. This creates zeros in the first column, making the expansion trivial:
$$z \begin{vmatrix} 1 & \omega & \omega^2 \ 0 & z+\omega^2-\omega & 1-\omega^2 \ 0 & 1-\omega & z+\omega-\omega^2 \end{vmatrix} = 0$$

Expanding this, we obtain:
$$z \left[ (z+\omega^2-\omega)(z+\omega-\omega^2) - (1-\omega)(1-\omega^2) ight] = 0$$

This expression simplifies using the difference of squares. The first part is $(z + (\omega^2-\omega))(z - (\omega^2-\omega)) = z^2 - (\omega^2-\omega)^2$. The second part is $(1-\omega)(1-\omega^2) = 1 - \omega^2 - \omega + \omega^3 = 1 - (\omega^2+\omega) + 1 = 1 - (-1) + 1 = 3$.

### The Final Revelation

Let us calculate $(\omega^2-\omega)^2$:
$$(\omega^2-\omega)^2 = \omega^4 - 2\omega^3 + \omega^2 = \omega - 2 + \omega^2 = -3$$

Substituting these values back into our equation, we get:
$$z \left[ (z^2 - (-3)) - 3 ight] = 0$$
$$z(z^2 + 3 - 3) = 0$$
$$z^3 = 0$$

Thus, **$z=0$** is the only solution. The number of distinct complex numbers satisfying the equation is **one**. It was not a monster; it was a puzzle waiting to be solved.

Animated Solution for Mathematics - Matrices and Determinants: Let $ω$ be the complex number $cos \frac{2 π}{3} + i sin \frac{2 π}{3}$ . Then the number of distinct complex numbers $z$ satisfying $z + 1 ω ω^{2} ω z + ω^{2} 1 ω^{2} 1 z + ω = 0$ is equal to

Enter Numerical Value:

Visualized Solution

The Cube Roots of Unity

Properties of $ω$

The Determinant Equation

Column Transformation $C_{1} \to C_{1} + C_{2} + C_{3}$

Applying the Transformation

Simplifying $C_{1}$

Factoring out $z$

Row Transformations for Elegance

Expanding the Determinant

Algebraic Expansion

Simplifying the Terms

The Final Equation

Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Intimidating Matrix

The Path to Simplification

The Final Revelation

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