# How do you simplify i14?

Dec 9, 2015

$- 1$

#### Explanation:

Rewrite ${i}^{14}$ as ${\left({i}^{4}\right)}^{3} \times {i}^{2}$.

If $i = \sqrt{- 1} ,$ then ${i}^{2} = - 1$. From here ${\left({i}^{2}\right)}^{2} = {\left(- 1\right)}^{2}$, so ${i}^{4} = 1$.

If ${i}^{4} = 1$, then we can say that:

${i}^{14} = {\left(1\right)}^{3} \times {i}^{2}$

$= {i}^{2}$

$= - 1$