# 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 $(e^{i\pi})^a$? – Arthur Nov 9 '17 at 16:14
• Related – Giuseppe Negro Nov 9 '17 at 16:36

## 2 Answers

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.

• Alright, thanks! So i which cases are the exponents interchangeable (except for when they are real)? – Skydiver Nov 9 '17 at 16:45
• 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". – Arthur Nov 9 '17 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 $(a^b)^c.$