Simplify
\[\left( \frac{1 + i}{1 - i} \right)^{1000}.\]
First,
\[\frac{1 + i}{1 - i} = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i.\]So,
\[\left( \frac{1 + i}{1 - i} \right)^{1000} = i^{1000} = (i^2)^{500} = (-1)^{500} = \boxed{1}.\]
1
Level 1
Intermediate Algebra
Problem: Simplify
\[\left( \frac{1 + i}{1 - i} \right)^{1000}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To simplify the expression, first recognize that the complex number can be expressed in polar form. The numerator \(1 + i\) can be rewritten as \(\sqrt{2} \text{cis} \frac{\pi}{4}\) and the denominator \(1 - i\) as \(\sqrt{2} \text{cis} \left(-\frac{\pi}{4}\right)\). 

Next, divide the two complex numbers:
\[
\frac{1 + i}{1 - i} = \frac{\sqrt{2} \text{cis} \frac{\pi}{4}}{\sqrt{2} \text{cis} \left(-\frac{\pi}{4}\right)} = \text{cis} \left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right) = \text{cis} \left(\frac{\pi}{2}\right).
\]

Now, raise this result to the power of 1000:
\[
\left( \text{cis} \left(\frac{\pi}{2}\right) \right)^{1000} = \text{cis} \left(1000 \cdot \frac{\pi}{2}\right).
\]

Calculate \(1000 \cdot \frac{\pi}{2}\):
\[
1000 \cdot \frac{\pi}{2} = 500\pi.
\]

Since \(500\pi\) is an integer multiple of \(2\pi\), we can simplify:
\[
\text{cis}(500\pi) = \text{cis}(0) = 1.
\]

Thus, the final answer is \(1\).<|END_OF_LATENT|> The final answer is $1$.