# How do you simplify i^38?

Feb 10, 2016

${i}^{38} = - 1$

#### Explanation:

Let's see what happens if we compute any power of $i$:

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

${i}^{5} = {i}^{4} \cdot i = 1 \cdot i = i$

... and so on, after that, the sequence $i$, $- 1$, $- i$ and $1$ repeats itself.

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How can you determine which one it is for ${i}^{38}$?
Let's start with factorizing $38 = 4 \cdot 9 + 2$:

${i}^{38} = {i}^{4 \cdot 9 + 2} = {i}^{4 \cdot 9} \cdot {i}^{2} = {\left({i}^{4}\right)}^{9} \cdot {i}^{2} = {1}^{9} \cdot \left(- 1\right) = - 1$

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Let me also additionally show you a general way to determine ${i}^{n}$ for any positive integer $n$.

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