# How do you simplify i^17?

Dec 22, 2015

You should write ${i}^{17}$ as ${i}^{16} \cdot i$ this is same as ${\left({i}^{2}\right)}^{8} \cdot i$ simplifying this you would get the answer as $i$

#### Explanation:

Any problem of type ${i}^{n}$

Case 1: when $n$ is even

Rewrite it as ${\left({i}^{2}\right)}^{\frac{n}{2}}$ use the rule ${i}^{2} = - 1$ and then you would get $- 1$ when $\frac{n}{2}$ is odd and $+ 1$ when $\frac{n}{2}$ is even
Example to understand this case

${i}^{6} = {\left({i}^{2}\right)}^{3} = {\left(- 1\right)}^{3} = - 1$ here n" is even and n/2# is odd

${i}^{8} = {\left({i}^{2}\right)}^{4} = {\left(- 1\right)}^{4} = 1$ here $n$ is even and $\frac{n}{2}$ is even.

Case 2: when $n$ is odd

Rewrite ${i}^{n}$ as ${i}^{n - 1} \cdot {i}^{1}$
Apply the case(1) rules this time using $n - 1$ instead of $n$ the final answer would be $- i$ or $i$ depending on ${i}^{n - 1}$

Dec 22, 2015

${i}^{17} = i$

#### Explanation:

First note that ${i}^{4} = {i}^{2} \cdot {i}^{2} = - 1 \cdot - 1 = 1$

In general note that if $k$ and $c$ are integers and $n = 4 k + c$ then:

${i}^{n} = {i}^{4 k + c} = {i}^{4 k} {i}^{c} = {\left({i}^{4}\right)}^{k} {i}^{c} = {1}^{k} {i}^{c} = {i}^{c}$

In our case we can take $k = 4$ and $c = 1$ to find:

${i}^{17} = {i}^{16 + 1} = {i}^{16} {i}^{1} = {\left({i}^{4}\right)}^{4} i = {1}^{4} i = i$