Understanding the Expression \(\text{i}^{-2n}\)

The question asks for the number of distinct values the expression \(\text{i}^{-2n}\) can take, where \(\text{i} = \sqrt{-1}\) and \(n\) is any integer (\(n \in \mathbb{Z}\)).

First, let's recall the cycle of powers of \(\text{i}\):

• \(\text{i}^1 = \text{i}\)
• \(\text{i}^2 = -1\)
• \(\text{i}^3 = \text{i}^2 \cdot \text{i} = -1 \cdot \text{i} = -\text{i}\)
• \(\text{i}^4 = \text{i}^2 \cdot \text{i}^2 = (-1) \cdot (-1) = 1\)

The powers of \(\text{i}\) repeat every four integers. For any integer exponent \(k\), \(\text{i}^k\) can be determined by finding the remainder when \(k\) is divided by 4. \(\text{i}^k = \text{i}^{k \pmod 4}\).

Analyzing the Exponent

The exponent in our expression is \(-2n\). Since \(n\) is an integer, \(-2n\) must always be an even integer. Let \(k = -2n\). Then \(k\) can take values like ..., -6, -4, -2, 0, 2, 4, 6, ...

We can write any even integer \(k\) as \(2m\) for some integer \(m\). Since \(k = -2n\), we have \(2m = -2n\), which means \(m = -n\). As \(n\) can be any integer, \(m\) can also be any integer.

Evaluating \(\text{i}^{-2n}\)

Now, we can rewrite the expression \(\text{i}^{-2n}\) as \(\text{i}^{2m}\):

\(\text{i}^{-2n} = \text{i}^{2m}\)

Using the property of exponents \((a^b)^c = a^{bc}\), we can write:

\(\text{i}^{2m} = (\text{i}^2)^m\)

We know that \(\text{i}^2 = -1\). Substituting this value:

\((\text{i}^2)^m = (-1)^m\)

Determining the Distinct Values of \((-1)^m\)

The value of \((-1)^m\) depends on whether the integer \(m\) is even or odd:

• If \(m\) is an even integer (e.g., -2, 0, 2, 4, ...), \((-1)^m = 1\).
• If \(m\) is an odd integer (e.g., -3, -1, 1, 3, 5, ...), \((-1)^m = -1\).

Since \(m = -n\) and \(n\) can be any integer, \(m\) can also be any integer. Therefore, \(m\) can be either even or odd, depending on the value of \(n\).

For example:

• If \(n = 0\), \(m = 0\) (even), \(\text{i}^{-2(0)} = \text{i}^0 = (-1)^0 = 1\).
• If \(n = 1\), \(m = -1\) (odd), \(\text{i}^{-2(1)} = \text{i}^{-2} = (-1)^{-1} = \frac{1}{-1} = -1\).
• If \(n = 2\), \(m = -2\) (even), \(\text{i}^{-2(2)} = \text{i}^{-4} = (-1)^{-2} = \frac{1}{(-1)^2} = \frac{1}{1} = 1\).
• If \(n = -1\), \(m = 1\) (odd), \(\text{i}^{-2(-1)} = \text{i}^2 = (-1)^1 = -1\).

The expression \((-1)^m\) can only take two distinct values: 1 and -1.

Conclusion

Since \(\text{i}^{-2n}\) simplifies to \((-1)^m\) where \(m\) is an integer, and \((-1)^m\) can only be 1 or -1, there are exactly two distinct values that \(\text{i}^{-2n}\) can have for different integer values of \(n\).

The possible values are 1 and -1.

Therefore, the number of distinct values is two.