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