# Kinda confused can someone explain how to calculate I^9, i^15, and i^-4?

Oct 3, 2016

${i}^{9} = i$, ${i}^{15} = - i$ and ${i}^{- 4} = 1$

#### Explanation:

One of the most basic thing to keep in mind is that ${i}^{2} = - 1$. Now if we have any power of $i$, what we need to do is to write it as a product or division of ${i}^{2}$ and then replace it with $- 1$.

Now ${i}^{9} = {i}^{2 + 2 + 2 + 2 + 1} = {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times i$

= $- 1 \times - 1 \times - 1 \times - 1 \times i = i$

Similarly

${i}^{15} = {i}^{2 + + 2 + 2 + 2 + 2 + 2 + 2 + 1}$

= ${i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times i$

= $- 1 \times - 1 \times - 1 \times - 1 \times - 1 \times - 1 \times - 1 \times i = - i$ and

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

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

Also note that ${i}^{1} = i$, ${i}^{2} = - 1$, ${i}^{3} = {i}^{2} \times i = - 1 \times i = - i$, ${i}^{4} = {\left({i}^{2}\right)}^{2} = {\left(- 1\right)}^{2} = 1$, ${i}^{5} = {i}^{4} \times i = i$, ${i}^{6} = {i}^{4} \times {i}^{2} = 1 \times - 1 = - 1$, ${i}^{7} = {i}^{4} \times {i}^{3} = 1 \times - i = - i$ and ${i}^{8} = {\left({i}^{4}\right)}^{2} = {\left(1\right)}^{2} = 1$

Hence successive powers of $i$ appear in a cycle of $i$, $- 1$, $- i$ and $1$. Thus, ${i}^{9} = i$, ${i}^{15} = - i$ and ${i}^{- 4} = 1$.