The imaginary unit $i$ is defined as $i = \sqrt{-1}$. Powers of $i$ follow a cyclical pattern every four terms:

$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1$

To simplify $i^{14}$, we use the fact that $i^n$ repeats every 4 powers, so we find the remainder when 14 is divided by 4:

$14 \div 4 = 3 \; \text{remainder} \; 2$

Thus, $i^{14} \equiv i^2$.

From the cycle, $i^2 = -1$. Therefore:

$i^{14} = -1$

Final Answer:

$i^{14} = -1$

Would you like more details or have questions about powers of $i$?

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5 Related Questions:

1. How do you simplify $i^{20}$?
2. What is the general formula for simplifying $i^n$?
3. How is the cycle of powers of $i$ derived?
4. Can powers of $i$ have real values?
5. How is $i^0$ defined?

1 Tip:

Always reduce powers of $i$ modulo 4 for quick simplification.