How do you simplify $i^{- 12}$ $i^-12$ ?

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How do you simplify $i^{- 12}$ $i^-12$ ?

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Answer 1 · 2015-11-10T14:21:51

$1$ $1$

Explanation:

You can use regular power rules:
$a^{- n} = \frac{1}{a^{n}}$ $a^(-n) = 1/a^n$
$a^{n + m} = a^{n} \cdot a^{m}$ $a^(n+m) = a^n * a^m$
${(a^{n})}^{m} = a^{n \cdot m}$ $(a^n)^m = a^(n*m)$

And in addition, whenever you have the term $i^{2}$ $i^2$ , always use $i^{2} = - 1$ $i^2 = -1$ .

So, let me break it down in very small steps:

$i^{- 12} = \frac{1}{i^{12}} = \frac{1}{i^{2 \cdot 6}} = \frac{1}{{(i^{2})}^{6}} = \frac{1}{{(- 1)}^{6}} = \frac{1}{{(- 1)}^{2 \cdot 3}} = \frac{1}{{({(- 1)}^{2})}^{3}} = \frac{1}{1^{3}} = 1$ $i^(-12) = 1/i^12 = 1/i^(2*6) = 1/(i^2)^6 = 1 / (-1)^6 = 1 / (-1)^(2*3) = 1 / ((-1)^2)^3 = 1 / 1^3 = 1$

Alternatively, you can use $i^{2} = - 1 \Rightarrow i^{4} = 1$ $i^2 = -1 => i^4 = 1$ and get there a bit quicker:
$i^{- 12} = \frac{1}{i^{12}} = \frac{1}{i^{4 \cdot 3}} = \frac{1}{{(i^{4})}^{3}} = \frac{1}{1^{3}} = 1$ $i^(-12) = 1/i^12 = 1/i^(4 *3) = 1/(i^4)^3 = 1/1^3 = 1$

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How do you simplify $i^{- 12}$ $i^-12$ ?

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