# How do you simplify i^1776?

Dec 18, 2016

1

#### Explanation:

there is a pattern to powers of $i$:

${i}^{1} = i$

${i}^{2} = - 1$

${i}^{3} = - i$

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

this cycle repeats for every power of $4$.

to simplify ${i}^{1776}$, we have to see if $1776$ is divisible by $4$, and if not, find the remainder:

$\frac{1776}{4} = 444$

there is no remainder, meaning that ${i}^{1776}$ is in the fourth part of the pattern.

this also means that ${i}^{1776}$ could be simplified as the answer to ${i}^{4}$, which is 1.