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Question #8e2b3

1 Answer

Oct 16, 2017

$- \frac{1}{i}$ $-1/i$

Explanation:

By the rule $a^{- b} = \frac{1}{a^{b}}$ $color(red)(a^(-b)=1/(a^b)$

$= i^{- 39} = \frac{1}{1^{39}}$ $=i^(-39)=1/(1^39)$

By the rule $a^{m \cdot n} = {(a^{m})}^{n}$ $color(red)(a^(m*n)=(a^m)^n$

$= \frac{1}{1^{39}} = \frac{1}{i {(i^{2})}^{19}}$ $=1/(1^39)=1/(i(i^2)^19$

By the rule $i^{2} = - 1$ $color(red)(i^2=-1$

$\frac{1}{i {(i^{2})}^{19}} = \frac{1}{{(- 1)}^{19} i} = \frac{1}{- 1^{19} i}$ $1/(i(i^2)^19)=1/((-1)^19i)=1/(-1^19i)$ [As 19 is odd]

As $1^{19} = 1$ $color(red)(1^19=1$

$\frac{1}{- 1^{19} i} = \frac{1}{- i} = - \frac{1}{i}$ $1/(-1^19i)=1/-i=-1/i$

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