Why does (e^{i\alpha})^2 always equal 1?

[Piano man]

In solutions to a problem I was working on, I saw that when an expression such as $$e^{i\alpha}$$ , alpha being an angle (in polar coordinates), was squared, the expression goes to unity, ie $$(e^{i\alpha})^2=1$$
But I see no reason to think that $$\alpha$$ is a multiple of $$\pi$$.
Could there be any other reason why it would be 1?

 

[Mark44]

In solutions to a problem I was working on, I saw that when an expression such as $$e^{i\alpha}$$ , alpha being an angle (in polar coordinates), was squared, the expression goes to unity, ie $$(e^{i\alpha})^2=1$$
But I see no reason to think that $$\alpha$$ is a multiple of $$\pi$$.
Could there be any other reason why it would be 1?

Since $$(e^{i\alpha})^2=1$$
$$e^{i\alpha} = \pm 1$$
so $\alpha = k \pi~ \text{for some integer k}$

 

[SpectraCat]

In solutions to a problem I was working on, I saw that when an expression such as $$e^{i\alpha}$$ , alpha being an angle (in polar coordinates), was squared, the expression goes to unity, ie $$(e^{i\alpha})^2=1$$
But I see no reason to think that $$\alpha$$ is a multiple of $$\pi$$.
Could there be any other reason why it would be 1?

I don't think it actually said: $$(e^{i\alpha})^2=1$$

I'll bet what it said was: $$|e^{i\alpha}|^2=1$$

That notation indicates the square-modulus of a complex quantity, rather than just the square. In other words:

$$|e^{i\alpha}|^2=(e^{i\alpha})(e^{i\alpha})^\star=e^{i\alpha}e^{-i\alpha}=1$$

Thus $$\alpha$$ can still be an arbitrary real number (it cannot be an imaginary number).

 

[Piano man]

Ah! That's probably right - thank you :D