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

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    Exponent Imaginary
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Discussion Overview

The discussion revolves around the expression (e^{i\alpha})^2 and its relationship to the value of 1. Participants explore whether this equality holds for all values of \alpha or if it is restricted to specific cases, particularly multiples of π. The scope includes conceptual clarification and mathematical reasoning.

Discussion Character

  • Exploratory, Conceptual clarification, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant questions the assertion that (e^{i\alpha})^2 equals 1 without assuming \alpha is a multiple of π.
  • Another participant argues that if (e^{i\alpha})^2 equals 1, then it follows that e^{i\alpha} must equal ±1, leading to the conclusion that \alpha = kπ for some integer k.
  • A different participant suggests that the original statement may have referred to the square of the modulus, |e^{i\alpha}|^2, which equals 1 for any real \alpha, thus allowing \alpha to be arbitrary.
  • A later reply expresses agreement with the latter interpretation, indicating a shift in understanding.

Areas of Agreement / Disagreement

Participants do not reach a consensus. There are competing views regarding the conditions under which (e^{i\alpha})^2 equals 1, with some asserting it is limited to multiples of π and others suggesting it can hold for any real \alpha when considering the modulus.

Contextual Notes

The discussion highlights potential ambiguity in notation, particularly between squaring the expression and squaring the modulus. There is also an implicit assumption about the nature of \alpha being real versus imaginary.

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

Since [tex](e^{i\alpha})^2=1[/tex]
[tex]e^{i\alpha} = \pm 1[/tex]
so [itex]\alpha = k \pi~ \text{for some integer k}[/itex]
 
Piano man said:
In solutions to a problem I was working on, I saw that when an expression such as [tex]e^{i\alpha}[/tex] , alpha being an angle (in polar coordinates), was squared, the expression goes to unity, ie [tex](e^{i\alpha})^2=1[/tex]
But I see no reason to think that [tex]\alpha[/tex] is a multiple of [tex]\pi[/tex].
Could there be any other reason why it would be 1?

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

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

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

[tex]|e^{i\alpha}|^2=(e^{i\alpha})(e^{i\alpha})^\star=e^{i\alpha}e^{-i\alpha}=1[/tex]

Thus [tex]\alpha[/tex] can still be an arbitrary real number (it cannot be an imaginary number).
 
Ah! That's probably right - thank you :D
 

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