# How do you simplify i^-12?

Nov 10, 2015

$1$

#### Explanation:

You can use regular power rules:
${a}^{- n} = \frac{1}{a} ^ n$
${a}^{n + m} = {a}^{n} \cdot {a}^{m}$
${\left({a}^{n}\right)}^{m} = {a}^{n \cdot m}$

And in addition, whenever you have the term ${i}^{2}$, always use ${i}^{2} = - 1$.

So, let me break it down in very small steps:

${i}^{- 12} = \frac{1}{i} ^ 12 = \frac{1}{i} ^ \left(2 \cdot 6\right) = \frac{1}{{i}^{2}} ^ 6 = \frac{1}{- 1} ^ 6 = \frac{1}{- 1} ^ \left(2 \cdot 3\right) = \frac{1}{{\left(- 1\right)}^{2}} ^ 3 = \frac{1}{1} ^ 3 = 1$

Alternatively, you can use ${i}^{2} = - 1 \implies {i}^{4} = 1$ and get there a bit quicker:
${i}^{- 12} = \frac{1}{i} ^ 12 = \frac{1}{i} ^ \left(4 \cdot 3\right) = \frac{1}{{i}^{4}} ^ 3 = \frac{1}{1} ^ 3 = 1$