Complex exponential to a power

[Mr Davis 97]

Say I have $e^{2\pi i n}$, where $n$ is an integer. Then it's clear that $(e^{2\pi i})^n = 1^n = 1$.
However, what if replace $n$ with a rational number $r$? It seems that by the same reasoning we should have that $e^{2\pi i r} = (e^{2\pi i})^r = 1^r = 1$. But what if $r=1/2$ for example? Then $e^{2 \pi i (\frac{1}{2})} = e^{\pi i} = -1$. What am I doing wrong here?

 

[jedishrfu]

Isn't that Euler's identity you now have?

Add 1 to both sides to get $e^{i\pi} + 1 = -1 + 1$ hence $e^{i\pi} + 1 = 0$

So sqrt of 1 has two roots 1 and -1 right? squaring the answer you got brings you back to the original $e^{2\pi i} = 1$

@fresh_42 or @Mark44 can provide a better answer I think.

 

[fresh_42]

You applied rules for real numbers on complex numbers:
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

If in doubt, use $\mathbb{C} = \mathbb{R}[x]/\langle x^2+1 \rangle$ to see which formulas can be shifted from right to left, and which can not.

 

[jedishrfu]

Thanks @fresh_42 I had forgotten about that insight by micro.

 

[FactChecker]

Say I have $e^{2\pi i n}$, where $n$ is an integer. Then it's clear that $(e^{2\pi i})^n = 1^n = 1$.
However, what if replace $n$ with a rational number $r$? It seems that by the same reasoning we should have that $e^{2\pi i r} = (e^{2\pi i})^r = 1^r = 1$.

No more than the fact that $sin(2\pi n) == 0$ for all $n\in N$ implies $sin(2\pi r) == 0$ for all rational numbers $r$

But what if $r=1/2$ for example? Then $e^{2 \pi i (\frac{1}{2})} = e^{\pi i} = -1$. What am I doing wrong here?

The complex exponential function $e^{i\theta}, \theta \in R$ is very fundamental. I suggest that you study it in the context of complex analysis.