a, b, c are integers, not all simultaneously equal and ω is cube root of unity (ω ≠ 1), then minimum

Q: Wait, why can't the answer be 0? If $a=b=c=1$, isn't the sum $1(1+\omega+\omega^2) = 0$?

Excellent catch! You are absolutely right that if $a=b=c$, the sum is zero. But read the problem statement again carefully: '$a, b, c$ are not all simultaneously equal.' This is the 'trap' door. We are forbidden from choosing the case where the sum is zero. We must find the *next* smallest possible value.

Q: Why do we bother squaring the modulus? Can't we just work with $|a + b\omega + c\omega^2|$?

You *could*, but it would be a nightmare of square roots and trigonometry. In JEE Advanced, we value elegance. Squaring the modulus turns a complex geometric problem into a clean algebraic one involving integers. Always look for the path that simplifies the algebra!

Q: Is there a geometric way to see why the minimum is 1?

Yes! Think of $a, b, c$ as coordinates of a point in 3D space. The expression $\frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]$ is related to the squared distance of the point $(a, b, c)$ from the line $a=b=c$. Since $a, b, c$ are integers, the point cannot lie on the line $a=b=c$, so it must be at least a certain distance away.

Question

$a, b, c$ are integers, not all simultaneously equal and $\omega$ is cube root of unity ($\omega 
eq 1$), then minimum value of $|a + b\omega + c\omega^2|$ is

Options:

(A) 0

(B) 1

(C) $\sqrt{3}/2$

(D) 1/2

Sigma Percentile · Accepted Answer

### The Geometry of the Complex Plane

Welcome, future engineer. Today, we are going to peel back the layers of a problem that seems to be about complex numbers, but is actually a beautiful dance of algebra and geometry.

Imagine you are standing in the complex plane. We are looking at the expression $z = a + b\omega + c\omega^2$. Here, $\omega$ and $\omega^2$ are the cube roots of unity.

If you plot them, $1, \omega,$ and $\omega^2$ form a perfect equilateral triangle inscribed in the unit circle. This is our playground.

We are given that $a, b, c$ are integers, and they are not all equal. This constraint is the heartbeat of the problem. It tells us that we are not at the origin, but we are as close as we can possibly get.

### The Algebraic Transformation

To find the minimum of $|z|$, we could try to use trigonometry, but that is a path fraught with peril. Instead, we use the most powerful tool in our complex number toolkit: $|z|^2 = z \cdot \bar{z}$.

By squaring the modulus, we transform a geometric distance problem into an algebraic one. When we compute the conjugate $\bar{z} = a + b\omega^2 + c\omega$, we are essentially reflecting our vector across the real axis.

When we multiply $z$ by $\bar{z}$, we are performing a beautiful expansion. We get:

$$|z|^2 = a^2 + b^2 + c^2 + (ab + bc + ca)(\omega + \omega^2)$$

### The 'Aha!' Moment

Now, we invoke the fundamental property of the cube roots of unity: $1 + \omega + \omega^2 = 0$. This implies that $\omega + \omega^2 = -1$.

Substituting this into our expression, the complex components vanish entirely, leaving us with the purely real expression:

$$|z|^2 = a^2 + b^2 + c^2 - ab - bc - ca$$

This is a classic symmetric form. To reveal its true nature, we multiply and divide by $2$, yielding:

$$|z|^2 = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]$$

This is the 'Aha!' moment. We have expressed the squared modulus as a sum of squares of differences.

### The Integer Constraint

Finally, we return to our constraints. $a, b, c$ are integers, and they are not all equal. If they were all equal, the sum of squares would be zero.

But we are forbidden from that. To minimize this sum, we must make the differences as small as possible. Since they are integers, the smallest non-zero difference is $1$.

By setting two variables equal and the third differing by $1$ (e.g., $a=1, b=0, c=0$), we get:

$$|z|^2 = \frac{1}{2}[1^2 + 0^2 + (-1)^2] = 1$$

Thus, the **minimum value is 1**. You have just navigated the intersection of geometry, algebra, and number theory. Keep this elegance in your toolkit; it will serve you well in the exam hall.

Animated Solution for Mathematics - Complex Numbers: $a, b, c$ are integers, not all simultaneously equal and $ω$ is cube root of unity ( $ω \neq = 1$ ), then minimum value of $∣ a + bω + c ω^{2} ∣$ is

Select Answer:

Visualized Solution

Visualizing the Roots of Unity

The Modulus Squared Property

Finding the Complex Conjugate

Setting Up the Product

Expanding the Expression

Grouping the Terms

Applying the Sum of Roots

The Sum of Squares Identity

Analyzing the Integer Constraint

Finding the Minimum Value

Final Calculation and Conclusion

The Sigma Insight: Cube Roots and nth Roots of Unity

The Geometry of the Complex Plane

The Algebraic Transformation

The 'Aha!' Moment

The Integer Constraint

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