$e^{2\pi i x} = (e^{2\pi i})^x$: What happens if x is rational? I'm a bit embarrassed that I've had difficulty on getting around this one:
$$e^{2\pi i x}$$
Solving it by itself, we can reduce it down to $(e^{2\pi i})^x = 1^x$ such that $e^{2\pi i x} = 1$ for all $x$. 
However, directly plugging in non-integer rational numbers, chiefly $x = 1/2$, we get results that does not stay true to the above equality. 
I'm a bit perplexed of trying to explain the different results for this one. I admit that I haven't had too much experience with complex numbers compared to other fields.
 A: Actually, when we are talking about complex numbers, the law of exponentiation is a little different from what we used to see in non-complex case. You can see this if you try to express your equality in trigonometric form, using Euler's formula.
The paradoxes arising from exponentiation of complex numbers ultimately led to introducing the concept of complex logarithm and can be resolved via defining principal branch.
The explicit explanation of the failure of exponential and logarithmic identities for the case of complex numbers can be found here:

The identity $(e^x)^y = e^{xy}$ holds for real numbers x and y, but assuming its truth for complex numbers leads to the following paradox, discovered in 1827 by Clausen.
For any integer $n$, we have:

*

*$e^{1+2\pi i n} = e^1 e^{2 \pi i n} = e \cdot 1 = e$

*$\left(e^{1+2\pi i n}\right)^{1+2\pi i n} = e$

*$e^{1+4\pi i n - 4 \pi^2 n^2} = e$

*$e^{1} e^{4\pi i n} e^{- 4 \pi^2 n^2} = e$

*$e^{- 4 \pi^2 n^2} = 1$
but this is false when the integer $n$ is nonzero.

