If `omega` is a complex cube root of unity, then `((1+i)^(2n)-(1-i)^(2n))/((1+omega^(4)-omega^(2))(1-omega^(4)+omega^(4))` is equal to

Options

A. 0 if n is even
B. 0 for all ` n in Z`
C. `2^(n-1) cdot I ` for all `n in N`
D. None of these

Answer

To solve the given problem, we need to evaluate the expression:

\[ \frac{(1+i)^{2n} - (1-i)^{2n}}{(1 + \omega^4 - \omega^2)(1 - \omega^4 + \omega^4)} \]

where \(\omega\) is a complex cube root of unity.

### Step 1: Simplify the numerator

First, we simplify the numerator \((1+i)^{2n} - (1-i)^{2n}\).

Using the binomial theorem, we can express \((1+i)^{2n}\) and \((1-i)^{2n}\):

\[ (1+i)^{2n} = (1^2 + i^2)^{n} = (2i)^{n} = 2^n i^n \]

\[ (1-i)^{2n} = (1^2 + (-i)^2)^{n} = (2(-i))^{n} = 2^n (-i)^n \]

Thus, the numerator becomes:

\[ (1+i)^{2n} - (1-i)^{2n} = 2^n i^n - 2^n (-i)^n = 2^n (i^n - (-i)^n) \]

### Step 2: Analyze \(i^n - (-i)^n\)

We know that:

- If \(n\) is even, \(i^n = (-i)^n\) and thus \(i^n - (-i)^n = 0\).
- If \(n\) is odd, \(i^n = -(-i)^n\) and thus \(i^n - (-i)^n = 2i^n\).

So we can rewrite the numerator as:

\[ \text{Numerator} = \begin{cases} 0 & \text{if } n \text{ is even} \\ 2^{n+1} i^{n} & \text{if } n \text{ is odd} \end{cases} \]

### Step 3: Simplify the denominator

Now, we simplify the denominator \((1 + \omega^4 - \omega^2)(1 - \omega^4 + \omega^4)\).

Since \(\omega^3 = 1\), we have \(\omega^4 = \omega\). Thus, the denominator simplifies to:

\[ (1 + \omega - \omega^2)(1 - \omega + \omega) \]

The second part simplifies to \(1\):

\[ 1 - \omega + \omega = 1 \]

Now, we focus on the first part:

\[ 1 + \omega - \omega^2 \]

Using the property of cube roots of unity, we know that \(1 + \omega + \omega^2 = 0\), which implies:

\[ \omega + \omega^2 = -1 \implies 1 + \omega - \omega^2 = 1 + \omega + 1 = 2 + \omega \]

Thus, the denominator simplifies to:

\[ 2 + \omega \]

### Step 4: Final expression

Now we can write the entire expression:

\[ \frac{2^n (i^n - (-i)^n)}{(2 + \omega)} \]

### Step 5: Evaluate for specific values of \(n\)

1. **If \(n\) is even**: The numerator is \(0\), hence the entire expression is \(0\).
2. **If \(n\) is odd**: The numerator is \(2^{n+1} i^{n}\), and we can evaluate it further depending on the value of \(n\).

### Conclusion

The expression evaluates to \(0\) for all even \(n\) and a non-zero value for odd \(n\). Therefore, the answer is:

\[ \text{The expression is } 0 \text{ for even } n. \]