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What's the restriction (for $a, p, q$) when I use the following in proof?

$a^{p/q}=(a^{1/q})^{p}$

The calculus book give a list said:

$a$ $real$, $q$ $odd$ --------------- $a^{1/q}$, called the $q$th root of $a$, is the number $b$ such that $b^{q}=a$

$a$ $nonnegative$, $q$ $even$ ---- $a^{1/q}$ is the nonnegative number $b$ such that $b^{q}=a$

$rational$ $exponents$ -------- $a^{p/q}=(a^{1/q})^{p}$

I'm mess up by these restrictions.

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    can this help? http://math.stackexchange.com/questions/1495532/when-is-abc-abc-true?rq=1 – 2017-01-11
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    @juniven In the book it said that $a^{1/q}$ when $q$ is even and $a$ $nonnegative$, your link said the restriction is $a>0$, which is different – 2017-01-11
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    Why not put that comment in your question? – 2017-01-11
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    @juniven added. – 2017-01-11

1 Answers 1

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In the context of complex numbers, the left side is $$\exp\left(\frac{p}{q} \log(a)\right)$$ and the right is $$\eqalign{\exp\left(p \log(a^{1/q})\right) &= \exp\left(p \log\left(\exp\left(\frac{\log(a)}{q}\right)\right)\right)\cr &= \exp\left(p \left(2 \pi i n + \frac{\log(a)}{q}\right)\right) \cr &= \exp(2\pi i np) \exp\left( \frac{p}{q} \log(a)\right)}$$ where $n$ is any integer. There is always at least one branch ($n=0$) of the right side that makes this true, while if $p$ is an integer it is always true.

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    btw, from which topics could I learn that the $2\pi i$ transformation you do in your formula? – 2017-01-11
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    Complex analysis. See e.g. Chapter 3 of [these notes](http://people.math.gatech.edu/~cain/winter99/complex.html) – 2017-01-11