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I came across a perplexing thing while testing some assumptions in Wolfram|Alpha, and as I don't have a math education beyond college algebra, I thought this would be a good place to ask. I would just like to emphasize that this is not a homework question, and is purely curiosity.

Common faulty math proofs tend to use logic like this:

1^0 = 1
2^0 = 1
therefore 1 = 2

This is obviously false because $x^0$ is defined as 1 for all real values of $x$.

But in the case of $e^{2\pi i} = e^0$, the base is the same: $e$. So if $a^x = a^y$ is true, shouldn't it follow, then, that $x = y$ and therefore $2\pi i = 0$? If not, why?

Braden Best
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    Seems to be the same question as "why is $\sin(0)=\sin(2\pi)$ true but $0 = 2\pi$ false? – Michael Oct 19 '18 at 20:33
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    @Michael It is the exact same question, by equating imaginary parts of both sides. :P – Franklin Pezzuti Dyer Oct 19 '18 at 20:34
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    $(-1)^2=(1)^2$ but $-1\ne 1$. – hamam_Abdallah Oct 19 '18 at 20:36
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    @Frpzzd : How come your comment that affirms my comment gets 2 upvotes, but my comment gets none? =) – Michael Oct 19 '18 at 20:36
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    @Michael I have no clue. But if this comment starts getting upvotes, then we'll know something funny is going on. XD – Franklin Pezzuti Dyer Oct 19 '18 at 20:36
  • @Michael See I can at least visualize why $sin(0) = sin(2\pi)$ -- the sine and cosine functions are like waves with a period of $\pi$, and at $2\pi$, you arrive at the same "middle point" as with 0, because $sin(0) = sin(\pi)$. It follows, then, that $sin({\pi \over 2}) = 1$ and $sin({3\pi \over 2}) = -1$. $e^{2\pi i} = e^0$ is not as intuitive. So maybe I know a little more trig that I let on. It's not much, though. – Braden Best Nov 01 '18 at 05:11

4 Answers4

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That's because the exponential isn't an injective function on $\mathbb C$. If is an injective function on $\mathbb R$, and so the single-valued logarithm is well-defined on $\mathbb R^+$ but not on $\mathbb C$.

This isn't true for the same reason that $(-1)^2=1^2$ doesn't imply that $-1=1$.

Franklin Pezzuti Dyer
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  • Would you care to elaborate/add examples? For example, I don't see how the (-1)^2 = 1^2 example is equivalent, because -1 = 1 is false, unlike e = e. – Braden Best Oct 19 '18 at 20:34
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    @BradenBest It is equivalent because $f(x)=x^2$ is not injective on $\mathbb R$ just as $g(x)=e^x$ is not injective on $\mathbb C$. – Franklin Pezzuti Dyer Oct 19 '18 at 20:35
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Recall that

$$f(x)=f(y) \implies x=y$$

$\forall x,y$ in the domain requires that $f$ is an injective function.

user
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  • I don't know how policies go here on Math.SE, but on every other SE site I've been on, it's considered best practice to quote your links. I think a simplified definition below the link would help a lot, and in the event that Wikipedia goes down, would protect your answer from losing 80% of its substance. There's also the fact that Wikipedia has a tendency to make liberal use of jargon that only someone already in the field is likely to grasp. – Braden Best Oct 20 '18 at 05:34
  • @BradenBest For the policies, I never had any advice here about infragements for wiki links maybe because editing the answer we can read the full details and therefore it is self quoted in some sense. By the way I need to edit it for some typos I don't like. For your other considerations, in my opinion the definition of "injective function" is the key point to completely solve your doubt, we don't need to add much more to it at least as a first answer. For that reason I didn't add any other details to it, the link suffices and in case it goes down one can find everywhere details about it. – user Oct 20 '18 at 06:03
  • Of course if you had asked for any clarification I would have tried to give that but I was assuming that after that and other answers all was clear to you or to any future reader. Anyway you have posted a good question and the most important think is that now your doubt is solved, that's nice. Bye – user Oct 20 '18 at 06:03
  • My opinion is that, with or without the link, the answer hinges on a term that it does not directly define or elaborate on, and that hurts the answer. Whenever I include a link in an answer, I try to always quote the relevant passage, regardless of whether the source is Wikipedia, a help page on Google, or a dinky little blog from 2004. I.e. I treat links as a complement to my answer, such that it's not required to click on them. The same goes for jargon. If I mention something like "functional programming", and this term is important to my answer, I will give examples and whatnot. Cheers :) – Braden Best Oct 20 '18 at 06:35
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Simply because the exponential function is periodic with period $2\pi i$.

That means that for any $x$, the values of $e^x$ and $e^{x+2\pi i k}$ are equal for all integers $k$.

In particular, you are looking at $x=0$ and $k=0,1$; so $e^0=e^{2\pi i}$, but of course $0$ and $2\pi i$ are different numbers.

MPW
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I think the others have given fine answers, but since they haven't been accepted I'll offer one more. Perhaps the piece you're missing (I'm guessing) is this:

So if $a^x=a^y$ is true, shouldn't it follow, then, that $x=y$?

No, it doesn't follow! Suppose e.g. that $a=1$: clearly, $1^5 = 1^2$ does not imply $5 = 2$ (the former is true and the latter is false). So one cannot simply equate the exponents (unless one knows that some additional constraints hold).

What then is the mechanical ("algebraic") procedure for resolving expressions like $a^x=a^y$? Take $\log_a$ of both sides, i.e. the logarithm with base $a$. So for $e^x = e^y$, you would apply $\log_e = \ln$, the natural logarithm. And now here you apply what the others have said: just like taking a square root (to solve $x^2 = y^2$, yielding $y = \pm x$), or applying an inverse sine (to solve $\sin(x) = \sin(y)$, yielding $x = y \pm 2n\pi$), when applying the logarithm you have to account for the fact that the $e^{\text{a complex number}}$ is not one-to-one (multiple inputs can mapped to the same output), unlike $e^{\text{a real number}}$. In particular, you get $x = y \pm i2n\pi$. Is this consistent with $e^0 = e^{i2\pi}$? Phew, yes.

Quiggin
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  • Well, the reason none of them have been accepted yet is because I haven't been at my computer all day today. When I get back from work and read up on injective functions, then I'll decide which most thoroughly answers the question. Don't let whether a question is accepted influence whether you add an answer, though. If you have enough information to share that you feel the need to write a separate answer, then your answer is valuable regardless. Don't hold back. – Braden Best Oct 20 '18 at 00:37
  • After reading all the answers, it was a toss-up between yours and @gimusi's answers. I'm going to give it to you, because yours is the most "layman-friendly" and thorough with respect to addressing the flaw in my logic. Also, he links to another website without quoting it (best practice is to quote your links) or trying to give a basic definition, which means that if Wikipedia were to go down, his answer would lose 80% of its substance. – Braden Best Oct 20 '18 at 05:26