Let `omega(omega ne 1)` is a cube root of unity, such that `(1+omega^(2))^(8)=a+bomega` where a, b in R, then `|a+b|` is equal to

Options

A. 1
B. 3
C. 0
D. 2

Answer

To solve the problem, we need to evaluate \( (1 + \omega^2)^8 \) where \( \omega \) is a cube root of unity (and \( \omega \neq 1 \)). The cube roots of unity are given by:

\[ 1, \omega, \omega^2 \]

where \( \omega = e^{2\pi i / 3} = -\frac{1}{2} + i \frac{\sqrt{3}}{2} \) and \( \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - i \frac{\sqrt{3}}{2} \).

### Step 1: Calculate \( 1 + \omega^2 \)

Using the value of \( \omega^2 \):

\[ 1 + \omega^2 = 1 + \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = \frac{1}{2} - i \frac{\sqrt{3}}{2} \]

### Step 2: Express \( 1 + \omega^2 \) in polar form

The modulus \( r \) of \( 1 + \omega^2 \) is:

\[ r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

The argument \( \theta \) is given by:

\[ \theta = \tan^{-1}\left(\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \]

Thus, we can express \( 1 + \omega^2 \) in polar form as:

\[ 1 + \omega^2 = e^{-i \frac{\pi}{3}} \]

### Step 3: Raise \( 1 + \omega^2 \) to the power of 8

Now we compute \( (1 + \omega^2)^8 \):

\[ (1 + \omega^2)^8 = \left(e^{-i \frac{\pi}{3}}\right)^8 = e^{-i \frac{8\pi}{3}} = e^{-i \left(2\pi + \frac{2\pi}{3}\right)} = e^{-i \frac{2\pi}{3}} = \omega \]

### Step 4: Express \( \omega \) in terms of \( a + b\omega \)

We know \( \omega = 0 + 1\omega \). Thus, we can compare this with the expression \( a + b\omega \):

\[ a = 0, \quad b = 1 \]

### Step 5: Calculate \( |a + b| \)

Now we need to compute \( |a + b| \):

\[ |a + b| = |0 + 1| = |1| = 1 \]

### Final Answer

Thus, the value of \( |a + b| \) is:

\[ \boxed{1} \]