If `omega (ne 1)` is a complex cube root of unity and `(1 + omega^(4))^(n)=(1+omega^(8))^(n)`, then the least positive integral value of n is

Options

A. 2
B. 3
C. 6
D. 12

Answer

To solve the equation \((1 + \omega^4)^n = (1 + \omega^8)^n\), where \(\omega\) is a complex cube root of unity (and \(\omega \neq 1\)), we can follow these steps:

### Step 1: Understand the properties of cube roots of unity
The complex cube roots of unity are given by:
- \(\omega = e^{2\pi i / 3}\)
- \(\omega^2 = e^{4\pi i / 3}\)
- \(\omega^3 = 1\)

From this, we can derive:
- \(\omega^4 = \omega\) (since \(\omega^3 = 1\))
- \(\omega^8 = \omega^2\)

### Step 2: Substitute \(\omega^4\) and \(\omega^8\) in the equation
Substituting these values into the original equation gives:
\[ (1 + \omega)^n = (1 + \omega^2)^n \]

### Step 3: Simplify the equation
Now we can rewrite the equation:
\[ (1 + \omega)^n = (1 + \omega^2)^n \]

### Step 4: Divide both sides by \((1 + \omega^2)^n\)
Assuming \(1 + \omega^2 \neq 0\), we can divide both sides:
\[ \left(\frac{1 + \omega}{1 + \omega^2}\right)^n = 1 \]

### Step 5: Analyze the fraction
The equation \(\left(\frac{1 + \omega}{1 + \omega^2}\right)^n = 1\) implies that:
\[ \frac{1 + \omega}{1 + \omega^2} = 1 \quad \text{or} \quad \frac{1 + \omega}{1 + \omega^2} = \omega^k \quad \text{for some integer } k \]

### Step 6: Find the values of \(1 + \omega\) and \(1 + \omega^2\)
Calculating these:
- \(1 + \omega = 1 + e^{2\pi i / 3} = 1 - \frac{1}{2} + i\frac{\sqrt{3}}{2} = \frac{1}{2} + i\frac{\sqrt{3}}{2}\)
- \(1 + \omega^2 = 1 + e^{4\pi i / 3} = 1 - \frac{1}{2} - i\frac{\sqrt{3}}{2} = \frac{1}{2} - i\frac{\sqrt{3}}{2}\)

### Step 7: Calculate the ratio
Now, we find:
\[ \frac{1 + \omega}{1 + \omega^2} = \frac{\frac{1}{2} + i\frac{\sqrt{3}}{2}}{\frac{1}{2} - i\frac{\sqrt{3}}{2}} = \frac{(1 + i\sqrt{3})}{(1 - i\sqrt{3})} \]

### Step 8: Simplify the ratio
Multiplying numerator and denominator by the conjugate of the denominator:
\[ = \frac{(1 + i\sqrt{3})(1 + i\sqrt{3})}{(1 - i\sqrt{3})(1 + i\sqrt{3})} = \frac{1 + 2i\sqrt{3} - 3}{1 + 3} = \frac{-2 + 2i\sqrt{3}}{4} = \frac{-1 + i\frac{\sqrt{3}}{2}}{2} \]

### Step 9: Determine the conditions for \(n\)
The expression \(\left(\frac{1 + \omega}{1 + \omega^2}\right)^n = 1\) holds true if \(n\) is a multiple of the order of the ratio. The order of \(\omega\) is 3, so \(n\) must be a multiple of 3.

### Step 10: Find the least positive integral value of \(n\)
The least positive integral value of \(n\) that satisfies this condition is:
\[ n = 3 \]

### Conclusion
Thus, the least positive integral value of \(n\) is \(3\).

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