Justification for a real number to a complex power

Question

I searched extensively online and found that everyone would agree with the following:

$2^i = e^{\log{2^i}} = e^{i*\log{2}}$. About the second equality (taking the $i$ out of the $\log$), it seems natural. But I cannot justify this myself.

I understand that, $\log(z1*z2) = \log(z1)+\log(z2)$, and therefore, if I have something like $\log(i^2)$, then it's just $\log(i*i) = \log(i) + \log(i) = 2\log(i)$, but this is because we have a rational exponent. How can I do the similar thing for $2^i$? Or is there some other way to justify the step pulling the $i$ out of the $log$?

Thanks very much!

Answer 1

Why is $\log(2^i)=i\log(2)? $

Let $x= \log(2^i)$

Then by definition $e^x=2^i$ but indeed this is an infinite set of numbers in the complex plane with this property and $y=i\log(2)$ is one such element that has this property because $e^y=e^{i \log(2)}=(e^{\log(2)})^{i}=2^i$

As is commented below and in the answer by Santos: we will need to decide exactly how to extend $\log$ to the complex plane to make any progress here.

Answer 2

By definition, if $a>0$, then $a^z=e^{z\times\log a}$. In particular, $2^i=e^{i\times\log2}$.

Of course, here $\log a$ is the only real logarithm of $a$. In general, a non-zero complex number has infinitely many logarithms, and therefore you shouldn't mention $\log z$ unless you state clearly which logarithm you have in mind. In particular, the equality $\log(z_1z_2)=\log(z_1)+\log(z_2)$ does't make sense in general.