$$\begin{align} e^{2 \pi i } &= 1 \\ e^{1 + 2 \pi i} &= e \\ e^{1 - 2 \pi i} &= e \\ {(e^{1 - 2 \pi i})}^{1 + 2 \pi i} &= e^{1 + 2 \pi i}=e\\ e^{1+4 \pi^2} &= e\\ e^{4 \pi^2} &= 1 \\ \pi &= 0 \end{align}$$
Where is an error?
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1How did you conclude $e^{4 \pi^2}=1 \implies \pi = 0$? – Sahiba Arora Jul 15 '17 at 19:28
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@SahibaArora take $\ln$ from both pieces of equality – Mykola Pochekai Jul 15 '17 at 19:33
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4This strikes me as a more involved version of$$e^{z} = e^{2\pi i(z/2\pi i)} = (e^{2\pi i})^{z/2\pi i} = 1^{z/2\pi i} = 1$$for all complex $z$. – Andrew D. Hwang Jul 15 '17 at 19:33
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1@kp9r4d Well that is the mistake. – Sahiba Arora Jul 15 '17 at 19:34
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4Exponentiation to complex powers does not follow the multiplication rule, i.e. $(a^b)^c$ need not be equal to $a^{bc}$. – Matt Samuel Jul 15 '17 at 19:37
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2No error. Pi does equal zero. It's the sad existential absurdity of the universe. ... Okay, logs of complex numbers are multivalued because exponents are periodic. This result if very similar to $-5 = \sqrt {(-5)^2} = 5$ so $5 = -5$ of that $\arccos (\cos 7\pi/4 ) = pi/4$ so $7 = 1$. $e^{2\pi i} = 3^0 = 1$. That doesn't meant $2\pi i = 0$. Because exponents are not 1 to 1 in reals and therefore logarithms are not unique. – fleablood Jul 15 '17 at 19:38
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@SahibaArora In my opinion mistake is think that operation $z \mapsto z^w$ is well defined for all complex $w$, i don't think that deduction $e^x = 1 \wedge x \in \mathbb{R} \to x = 0$ wrong. – Mykola Pochekai Jul 15 '17 at 19:41
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How did you get that $e^{4\pi^2} = 1$. It doesn't. $e^{4\pi^2} =139720110310117791.46581391647308$. But that's not the important mistake... or maybe it is. – fleablood Jul 15 '17 at 19:41
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$((a)^b)^c \ne a^{bc}$ in complex numbers. That's a result that only holds in reals. – fleablood Jul 15 '17 at 19:44
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Possibly of interest: For which complex $a$, $b$, $c$ does $(a^b)^c = a^{bc}$ hold? – Andrew D. Hwang Jul 15 '17 at 20:04
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Possible duplicate of 1 = e^2π. Where did I make a mistake? – Simply Beautiful Art Jul 15 '17 at 20:59
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Sahiba, if $\pi = 0$ then we get $e^0 = 1 \therefore \pi = 0$ – George N. Missailidis Aug 12 '17 at 00:35
1 Answers
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There are actually three mistakes in the proof:
No 1:
$$ e^{1 - 2 \pi i} = e \Rightarrow {(e^{1 - 2 \pi i})}^{1 + 2 \pi i} = e^{1 + 2 \pi i}$$
In the second equation, the LHS is a multi-valued function, while the RHS is a complex number.
No 2:
$$ e^{4 \pi^2} = 1 \Rightarrow \pi = 0 $$
As complex function, the exponential is NOT one to one.
No 3
Also note that the laws of exponentiation don't hold for complex exponentials, one should not use $${(e^{1 - 2 \pi i})}^{1 + 2 \pi i}= e^{(1 - 2 \pi i) (1 + 2 \pi i)}$$
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2But if we additionally knows that $4 \pi^2$ is real then No 2 is not really mistake. – Mykola Pochekai Jul 15 '17 at 19:48
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Am I the only one who hates thinking of complex exponentiation as a "multi-valued" function. It seems really stupid to me and makes it sound unnecessarily complicated and magical. – mathworker21 Jul 15 '17 at 20:29
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I think's so too, it's more reasonable to say that $z \mapsto z^w$ is not really well defined function $\mathbb{C} \to \mathbb{C}$. – Mykola Pochekai Jul 15 '17 at 20:34
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@mathworker21 When you get into stuff like complex analysis, you define complex exponentiation to be single valued (so an actual function), however, you lose common exponential rules that hold for real numbers sometimes. – Simply Beautiful Art Jul 15 '17 at 21:01
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1@kp9r4d One usually defines $z\mapsto z^w$ as $z\mapsto e^{w\ln(z)}$, where $\ln(z)$ is usually defined by $\ln|z|+i\operatorname{atan}(z)$, where $\operatorname{atan}(z)\in(-\pi,\pi]$ and $e^{x+iy}=e^x(\cos(y)+i\sin(y))$, a well-defined function. – Simply Beautiful Art Jul 15 '17 at 21:03
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@SimplyBeautifulArt I know complex analysis very very well. You define $\ln$ a certain way and then define exponentiation. But it seems very stupid to say it's a multi-valued function. Like there are many ways to choose a log and then you get exponentiation – mathworker21 Jul 15 '17 at 21:04
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@mathworker21: $e^z$ is well-defined and not multivalued; it is periodic. It is its inverse, the log, which is multivalued. – gary Jul 15 '17 at 21:05
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@gary No, $z\mapsto z^w$ is multivalued if left without restriction. Take $1^i$ for example. – Simply Beautiful Art Jul 15 '17 at 21:05
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@SimplyBeautifulArt: True, but I was referring to $e^z$. I just edited it.. – gary Jul 15 '17 at 21:06
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@mathworker21 My point was that its only multivalued if you treat it like a multivalued function. If you place restrictions and use good working definitions, then its not (but again, you'll likely fail to observe when common exponentiation rules fail) – Simply Beautiful Art Jul 15 '17 at 21:06
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@gary Then I see no reason to ping kp9r4d on $e^z$. Did you mean to ping mathworker21? – Simply Beautiful Art Jul 15 '17 at 21:07
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@SimplyBeautifulArt: A good reason is that I need a double-shot of espresso. Good point ;) , I just edited to ping Mathworker. – gary Jul 15 '17 at 21:08
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@SimplyBeautifulArt what I mean is that for each branch of the log, you get a function $f$ so that $z = e^{f(z)}$. Saying $z^i$, for example, is multi-valued just confused me and badly abuses notation in my opinion. – mathworker21 Jul 15 '17 at 21:15
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@mathworker21: Aaaarrrrgh, I keep writing stupid things. logz is multivalued. Let me delete that. I meant to write that $e^z$ is not injective, because it is periodic, and so it does not have a global inverse. – gary Jul 15 '17 at 21:15