If a,b,c are distinct integers and `omega(ne 1)` is a cube root of unity, then the minimum value of `|a+bomega+comega^(2)|+|a+bomega^(2)+comega|` is

Options

A. (a)`sqrt(3)`
B. (b)3
C. (c)`6sqrt(2)`
D. (d)2

Answer

To solve the problem, we need to find the minimum value of the expression

\[ |a + b\omega + c\omega^2| + |a + b\omega^2 + c\omega| \]

where \( \omega \) is a cube root of unity, specifically \( \omega = e^{2\pi i / 3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \) and \( \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \).

### Step 1: Express the terms in terms of \( \omega \)

We know that \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \). Therefore, we can rewrite the expressions:

\[ z_1 = a + b\omega + c\omega^2 \]
\[ z_2 = a + b\omega^2 + c\omega \]

### Step 2: Calculate the magnitudes

Using the properties of \( \omega \), we can express the magnitudes:

\[ |z_1| = |a + b\omega + c\omega^2| \]
\[ |z_2| = |a + b\omega^2 + c\omega| \]

### Step 3: Combine the magnitudes

We want to minimize:

\[ |z_1| + |z_2| \]

### Step 4: Use the identity involving squares

We can use the identity for the magnitudes of complex numbers:

\[ |z_1|^2 + |z_2|^2 = (a^2 + b^2 + c^2) + (b^2 + c^2 + a^2) - (ab + ac + bc) \]

This simplifies to:

\[ |z_1|^2 + |z_2|^2 = 2(a^2 + b^2 + c^2 - ab - ac - bc) \]

### Step 5: Express in terms of differences

We can express this in terms of the differences between the integers \( a, b, c \):

\[ = 2\left(\frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{2}\right) \]

### Step 6: Find the minimum value

Since \( a, b, c \) are distinct integers, the minimum value of the squared differences occurs when \( a, b, c \) are consecutive integers. Let's take \( a = p, b = p+1, c = p+2 \):

\[ |z_1| + |z_2| = \sqrt{2} \cdot \sqrt{(1)^2 + (1)^2 + (1)^2} = \sqrt{6} \]

### Final Step: Calculate the minimum value

Thus, the minimum value of

\[ |a + b\omega + c\omega^2| + |a + b\omega^2 + c\omega| \]

is

\[ \sqrt{6} \]

### Conclusion

The minimum value of the expression is \( 2\sqrt{2} \).