# How do you simplify i^27?

Nov 9, 2015

${i}^{27} = - i$

#### Explanation:

${i}^{27} = {i}^{24 + 3} = {i}^{24} \cdot {i}^{3} = {i}^{4 \cdot 6} \cdot {i}^{2 + 1} = {\left({i}^{4}\right)}^{6} \cdot {i}^{2} \cdot i$

Now, we know that ${i}^{2} = - 1$ and so ${i}^{4} = {\left({i}^{2}\right)}^{2} = 1$

Substituting that in gives us

${i}^{27} = {1}^{6} \cdot \left(- 1\right) \cdot i = - i$

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The above is a formal way of going about it, but for quick mental calculations, remembering that ${i}^{4} = 1$ allows you to "pull out" the greatest multiple of 4 possible from the exponent.
This leaves you with an exponent of 0, 1, 2, or 3 (any greater and there is another multiple of 4 to remove).
Then it is easy to work out that ${i}^{0} = 1$, ${i}^{1} = i$, ${i}^{2} = - 1$, and ${i}^{3} = - i$.