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I've been stuck with this for a while now. I have this chain of reasoning that would imply $e^{-\pi}=e^\pi$, obviously false, since $e^\pi$ and $e^{-\pi}$ are two real distinct numbers and so I must have made an assumption somewhere that I cannot actually do. I know I have to be very careful when working with complex numbers, especially when they're in the exponents, and so I tried to make the steps as small as I could so that it would be easier to point out where it went wrong.

\begin{align} e^{-\pi}&= e^{\pi\cdot -1}\tag{1}\\ &=e^{\pi\cdot i^2}\tag{2}\\ &= e^{\pi\cdot i\cdot i}\tag{3}\\ &= \left(e^{\pi\cdot i}\right)^i\tag{4}\\ &= (-1)^i\tag{5}\\ &=\left(\tfrac{1}{-1}\right)^i\tag{6}\\ &=\left((-1)^{-1}\right)^i\tag{7}\\ &=(-1)^{-1\cdot i}\tag{8}\\ &=(-1)^{-i}\tag{9}\\ &=\left(e^{i\pi}\right)^{-i}\tag{10}\\ &=e^{i\pi\cdot -i}\tag{11}\\ &=e^{-i^2\pi}\tag{12}\\ &=e^\pi\tag{13} \end{align} I suspect it has something to do with changing the base from $e$ to $-1$, but what does that mean? Are complex powers only defined for positive bases? Any help is appreciated.

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    $\begingroup$ Steps 4, 8, 11 are flawed, because $(a^b)^c=a^{bc}$ isn't necessarily true for complex $a,b,c$. $\endgroup$ – Wojowu Feb 2 '16 at 17:07
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    $\begingroup$ It has to do with raising things that are not $e$ to a complex power, and expecting that all the usual exponent rules (like $(a^b)^c=a^{bc}$) still works. $\endgroup$ – Arthur Feb 2 '16 at 17:07
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    $\begingroup$ In order to define $(-1)^i$, you need to pick a branch of the complex logarithm. So $(-1)^i$ is not well-defined. (It is multi-valued.) $\endgroup$ – Cheerful Parsnip Feb 2 '16 at 17:08
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    $\begingroup$ $e^π \neq 1/e^π$ $\endgroup$ – Archis Welankar Feb 2 '16 at 17:10
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    $\begingroup$ i bet 100 dollars that this is duplicate question $\endgroup$ – user153330 Feb 2 '16 at 17:35
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Here is your "proof" presented differently:

We have $e^{i\pi}=-1=\frac{1}{-1}=\frac{1}{e^{i\pi}}=e^{-i\pi}$. So far everything is right. Now our idea is to take both sides to the power of $i$: $(e^{i\pi})^i=(e^{-i\pi})^i$. The erroneous conclusion would appear if you used the identity $(a^b)^c=a^{bc}$. And here lies the problem: this identity doesn't hold for all complex numbers. (EDIT: in fact, this identity isn't always true if we have real numbers $a,b,c$ as leonbloy mentions in the comment. Keep that in mind!)

One might also touch on the topic: What is $(e^{i\pi})^i$? Here we need to go back to definition of exponentiation of complex numbers: $a^b=e^{b\ln a}$. However, there is a serious problem here: complex logarithm is multivalued. Taking one branch of the logarithm, we have $\ln e^{i\pi}=i\pi$, so $(e^{i\pi})^i=e^{i\cdot i\pi}$ (NOTE: here we use the definition of complex exponentiation, not exactly the property $(a^b)^c=a^{bc}$), which is $e^{-\pi}$.

However, at the same time we have $e^{i\pi}=e^{-i\pi}$, so we could say $\ln e^{i\pi}=-i\pi$. That way we get $(e^{i\pi})^i=e^{i\cdot (-i\pi)}=e^\pi$.

So if you think about this for a while, the core of the problem is that complex logarithm, and hence also exponentiation, are multivalued.

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    $\begingroup$ "$(a^b)^c=a^{bc}$ And here lies the problem: this identity doesn't hold for all complex numbers." Even with real numbers we can get trouble: $-1=(-1) ^1=(-1)^{2 \frac{1}{2}}=1^{\frac{1}{2}}=1$. Related math.stackexchange.com/questions/1628759/… $\endgroup$ – leonbloy Feb 2 '16 at 17:20

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