How do you simplify i^15?

Feb 8, 2016

${i}^{15} = - i$

Explanation:

Remember that ${i}^{2} = - 1$.

Thus,

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

Also, remember the power rule

${a}^{m} \cdot {a}^{n} = {a}^{m + n}$

Thus, you have

${i}^{15} = {i}^{4 + 4 + 4 + 3} = {i}^{4} \cdot {i}^{4} \cdot {i}^{4} \cdot {i}^{3} = 1 \cdot 1 \cdot 1 \cdot {i}^{3} = {i}^{3} = {i}^{2} \cdot i = - 1 \cdot i = - i$

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Also, I'd like to offer you a more general solution for ${i}^{n}$, with $n$ being any positive integer.

Try to recognize the pattern:

$i = i$
${i}^{2} = - 1$
${i}^{3} = {i}^{2} \cdot i = - 1 \cdot i = - i$
${i}^{4} = {i}^{3} \cdot i = - i \cdot i = - {i}^{2} = 1$
${i}^{5} = {i}^{4} \cdot i = 1 \cdot i = i$
${i}^{6} = {i}^{4} \cdot {i}^{2} = - 1$
...

So, basically, the power of $i$ is always $i$,$- 1$, $- i$, $1$, and then repeat.

Thus, to compute ${i}^{n}$, there are four possibilites:

• if $n$ can be divided by $4$, then ${i}^{n} = 1$
• if $n$ can be divided by $2$ (but not by $4$), then ${i}^{n} = - 1$
• if $n$ is an odd number but $n - 1$ can be divided by $4$, then ${i}^{n} = i$
• if $n$ is an odd number but $n + 1$ can be divided by $4$, then ${i}^{n} = - i$

Described in a more formal way,

${i}^{n} = \left\{\begin{matrix}1 & \text{ " n= 4k \\ i & " " n = 4k + 1 \\ -1 & " " n = 4k + 2 \\ -i & " } n = 4 k + 3\end{matrix}\right.$

for $k \in {\mathbb{N}}_{0}$.