do equal bases always means equal exponents?

Question

Recently I saw an explanation on why $ e ^ {\pi i} = -1 $, and having the reason in mind I came to the idea that $ e ^ {n \pi i} $ is also equal to -1 if n is odd. Then I remembered something I learned at school, that if the base of two equal numbers are the same, their exponents must also be the same, with the exception that the base is not 0 and not 1 or -1, I think that was it.

But we can see that

• $ e^{\pi i} = e^{3\pi i} $
• $ \pi i \neq 3\pi i $

in this example, even though the bases are equal, the exponents are not, so my question is:
Is the rule only true if the exponents are not complex numbers, and/or does it have more restrictions?

Answer 1

Assuming that $a>0$ and $a\neq1$, the implication $a^b=a^c\implies b=c$ holds for all real values of $b$ and $c$. In other words, the function $\mathbb R\to \mathbb R, x\mapsto a^x$ is one-to-one when $a>0$ and $a\neq1$ – no further hypotheses are needed. (I'm ignoring the case where $a$ is negative, since in general $a^x$ is not well-defined for real values of $x$.)

However, this fails badly over $\mathbb C$. Not only is $z\mapsto e^z$ many-to-one, it is a periodic function with period $2\pi i$. This result stems from the fact that $\sin$ and $\cos$ are periodic functions with period $2\pi$, and the complex exponential function is related to $\sin$ and $\cos$ by the formula $e^{iz}=\cos(z)+i\sin(z)$.

Answer 2

The rule is only true if there are no complex numbers involved, because complex number exponentiation involves some form of rotation, so rotating by another $360^{\circ}$ would give you the same base, different exponent, but same number. If the exponents are real and the bases don't have unit norm, same base implies same exponent.

In fact, complex exponentiation isn't even well-defined. $i^i$ can take on infinitely many different values. In fact, over the complex numbers, even $1^\pi$ can take on infinitely many different values depending on how many whole turns $1$ is considered to be.