# How do you simplify i^59?

Jun 30, 2018

${i}^{59} = - i$

#### Explanation:

${i}^{59} = {i}^{56} {i}^{3} = {\left({i}^{16}\right)}^{4} {i}^{3} = 1 {i}^{3} = {i}^{2} i = - i$

Jun 30, 2018

$- i$

#### Explanation:

Recall that

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

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

${i}^{4} = 1$ (Any multiple of $4$ exponent will also be $1$)

With this in mind, we can rewrite ${i}^{59}$, since the exponent is a prime number.

${i}^{59} = \textcolor{b l u e}{{i}^{56}} \cdot {i}^{3}$

Since $56$ is a multiple of $4$, ${i}^{56} = 1$. What we have simplifies to:

$1 \cdot {i}^{3} = - i$

Hope this helps!

Jun 30, 2018

${i}^{59} = {i}^{59} \times {i}^{2} / {i}^{2} = \frac{\left({i}^{60}\right) \left(i\right)}{- 1} = {\left({i}^{4}\right)}^{15} \cdot \frac{i}{-} 1 = {1}^{15} \cdot - i = - i$.