Q:

Proof of index laws for complex numbers

Can someone give a proof that index laws (and hence log laws) apply for complex numbers in the same way they do to reals, specifically that:

$(a^{ix})^n = a^{ixn}$

Assuming $a, x, n$ are real and $i$ defined as $i^2 = -1$
I am trying to prove de Moivre's theorem from Euler's formula but have been told I can't just assume that index laws apply to complex numbers (but rather have to prove that they do).
 
Help would be greatly appreciated!

Thank you!!

A:

You have to be careful about multivalued functions.  By definition, $a^z = \exp(z \log(a))$, but $\log$ is multivalued: $\log(a) = \text{Log}(a) + 2 \pi i k$ where $\text{Log}$ is one particular branch of the logarithm, and $k$ can be any integer. 
Now 
$(a^{ix})^n = (\exp(i x \log(a))^n = \exp(n \log(\exp(i x \log(a)))$
and $\log(\exp(i x \log(a))) = i x \log(a) + 2 \pi i k$ for arbitrary integer $k$
so 
$$(a^{ix})^n = \exp(n (i x \log(a) + 2 \pi i k)) = a^{ixn} \exp(2 \pi i k n)$$
If $n$ is an integer, so is $k n$, so $\exp(2 \pi i k n) = 1$ and you do have
$(a^{ix})^n = a^{ixn}$ (i.e. every value of the left side is a value of the right side, and vice versa).
However, if $n$ is not an integer they can be different.  
For example, $(i^2)^{1/2} = (-1)^{1/2} = \pm i$, but $i^{2 \cdot 1/2} = i$.

