Proving that $e^\pi=e^{-\pi}$

Solution 1:

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.