# How do you simplify i^18?

It is ${i}^{18} = {\left({i}^{2}\right)}^{9} = {\left(- 1\right)}^{9} = - 1$

Assuming that $i$ is the imaginary unit with ${i}^{2} = - 1$

Nov 14, 2015

${i}^{18} = - 1$

#### Explanation:

Consider the following:

${a}^{2} \times {a}^{3} = a \times a \times a \times a \times a = {a}^{5} = {a}^{2 + 3}$

i^18 = i ^(2 + 16) = i^(2) times i^(16

$i = \sqrt{- 1} \text{ so } {i}^{2} = - 1$

${i}^{16}$ is 8 lots of ${i}^{2}$ multiplied together.

An even number of negatives multiplied together yields a positive number so:

${i}^{16} = + 1$

But we have

${i}^{2} \times {i}^{16} = \left(- 1\right) \times \left(+ 1\right) = - 1$

So ${i}^{18} = - 1$