# What does i^4 equal?

${i}^{4} = 1$

#### Explanation:

By definition, $i = \sqrt{- 1}$.

${i}^{4} = {\sqrt{- 1}}^{4}$

When we have a number, say $\sqrt{2}$ and we multiply it by another $\sqrt{2}$, we get what's inside the square root sign:

$\sqrt{2} \times \sqrt{2} = 2$

So let's apply that to our problem:

${\sqrt{- 1}}^{2} = - 1$

But our problem has four $i$, not two. So let's break it down into two sets of two:

${\sqrt{- 1}}^{4} = {\sqrt{- 1}}^{2 + 2} = {\sqrt{- 1}}^{2} \times {\sqrt{- 1}}^{2} = - 1 \times - 1 = 1$

We could also have done it this way:

${\sqrt{- 1}}^{4} = {\sqrt{- 1}}^{2 \times 2} = {\left({\sqrt{- 1}}^{2}\right)}^{2} = {\left(- 1\right)}^{2} = 1$