Trigonometric Form and the Surprising Result: 1=-1?

[springo]

Hi,
as I was studying complex numbers today I came across this, and I couldn't explain it:
$$1=e^{i0}$$
$$1=e^{i2\Pi}$$
$$1^{1/2}=e^{i2\Pi/2}$$
$$1=e^{i\Pi}$$
$$1=-1$$
Where is the mistake?
Thank you very much for your help.

 

[Hurkyl]

$$1=e^{i2\pi}$$
$$1^{1/2}=e^{i2\pi/2}$$

Why would you think that the latter equation follows from the former?

 

[springo]

$$1^{1/2}=(e^{i2\pi})^{1/2}$$
Is that wrong?

 

[HallsofIvy]

The question really was, "why do you think that 1^1/2= 1?". In the real number system we take functions to be "single valued". In the complex number system that is no longer possible.

In particular, in the real number system, we define x^1/2 to be "the positive number, a, such that a²= x". Since the complex numbers are not an ordered field that definition cannot be used.

 

[springo]

OK, so basically I would have to take -1, but why not take 1 too?
And if I take 1, I still get 1=-1, right?