# Why is $e^{a\pi i}\neq (-1)^a$?

Why are the following statements incorrect? I have trouble understanding my mistake. $$e^{a\cdot \pi i} = e^{\pi i^a} = (-1)^a$$ $$e^{a\cdot 2\pi i} = e^{2\pi i^a} = (1)^a =1$$

Any clues would be appreciated!

• When you write $e^{\pi i^a}$, do you really mean $(e^{i\pi})^a$? Commented Nov 9, 2017 at 16:14
• Related Commented Nov 9, 2017 at 16:36

Your mistake is believing that $(a^b)^c = a^{bc} = (a^c)^b$ is true when the exponents aren't real. It's not. Or, if we force it to be, then your argument shows that every non-zero number is in fact $1$. I know which set of consequences I'd rather live with.
• Never. When the exponents aren't real (or when the base is not a positive real), exponentiation is multivalued, and there is no good, consistent way to choose among the multiple values that plays nicely with the exponentiation rules like the one above, or $a^{b+c}=a^ba^c$. Sure, you can find specific values that work, but there is no general condition that makes it always work, except for "base is positive and exponents are real". Commented Nov 9, 2017 at 17:21
I'm going to presume that by $e^{2\pi i^a}$ you meant $(e^{2\pi i})^a.$ You seem to suggest that since $e^{2\pi i} = 1$ you should have $e^{2\pi ia} = (e^{2\pi i})^a = 1^a = 1$ for every value of $a.$
The problem is that although exponential functions are one-to-one functions when their arguments are real, they are not one-to-one with complex arguments, and that upsets these usual identities. It is not only when $b=1$ that one can have $b^x=1.$
In conventional usage, $a^{b^{\,c}}$ means $a^{\left( b^{\,c}\right)},$ not $(a^b)^c.$