# How do you simplify i^64?

Nov 15, 2015

I found: ${i}^{64} = 1$

#### Explanation:

I know that:
${i}^{1} = \sqrt{- 1} = i$
${i}^{2} = {\left(\sqrt{- 1}\right)}^{2} = - 1$
${i}^{3} = i \cdot {i}^{2} = i \cdot - 1 = - i$
${i}^{4} = {i}^{2} \cdot {i}^{2} = - 1 \cdot - 1 = 1$
${i}^{5} = i \cdot {i}^{2} \cdot {i}^{2} = i \cdot \left(- 1\right) \cdot \left(- 1\right) = i$ AGAIN
${i}^{6} = {i}^{2} \cdot {i}^{2} \cdot {i}^{2} = \left(- 1\right) \left(- 1\right) \left(- 1\right) = - 1$
now it repeats the same pattern: $i , - 1 , - i , 1$ every 4.

You have ${i}^{64} = {\left({i}^{8}\right)}^{2} = {\left({i}^{4} \cdot {i}^{4}\right)}^{2} = {\left(1 \cdot 1\right)}^{2} = 1$