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.

  • 11
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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.) $\endgroup$ Feb 2 '16 at 17:08
  • 1
    $\begingroup$ $e^π \neq 1/e^π$ $\endgroup$ Feb 2 '16 at 17:10
  • 3
    $\begingroup$ i bet 100 dollars that this is duplicate question $\endgroup$
    – user153330
    Feb 2 '16 at 17:35

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.

  • 5
    $\begingroup$ "$(a^b)^c=a^{bc}$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.