# How do you simplify  i^-3?

Apr 4, 2016

${i}^{-} 3 = i = \sqrt{- 1}$

#### Explanation:

this is the correct explanation.

${i}^{2} = \left(- 1\right)$

$i = \sqrt{- 1}$

${a}^{-} n = \frac{1}{a} ^ n$

${i}^{-} 3 = {\left(\frac{1}{i}\right)}^{3}$

$= \frac{1}{\sqrt{- 1}} ^ 3$

$= \frac{1}{\sqrt{- 1} \cdot \sqrt{- 1} \cdot \sqrt{- 1}}$

$= \frac{1}{- 1 \cdot \sqrt{- 1}}$

$= - \frac{1}{\sqrt{- 1}}$

Rationatise the denominator by multiplying the numerator and denominator by $\sqrt{- 1}$

$= \frac{- 1 \cdot \sqrt{- 1}}{\sqrt{- 1} \cdot \sqrt{- 1}}$

$= - \frac{\sqrt{- 1}}{-} 1$

$= \sqrt{- 1}$

$= i$

Sep 3, 2016

$i$

#### Explanation:

First, note that:

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

Also, using the rule ${a}^{-} b = \frac{1}{a} ^ b$, we see that:

${i}^{-} 3 = \frac{1}{i} ^ 3$

Multiply it by $\frac{i}{i}$:

$\frac{1}{i} ^ 3 \cdot \frac{i}{i} = \frac{i}{i} ^ 4 = \frac{i}{1} = i$