A cop-out argument by Feynman in his fabled Lectures on Physics In the Algebra chapter of the Feynman Lectures on Physics,  Feynman introduces complex powers:

Thus $$10^{(r+is)}=10^r10^{is}\tag{22.5}$$
But $10^r$ we already know how to compute, and we can always multiply anything by anything else; therefore the problem is to compute only $10^{is}$. Let us call it some complex number, $x+iy$. Problem: given $s$, find $x$, find $y$. Now if
$$10^{is}=x+iy$$ then the complex conjugate of this equation must also be true, so that
$$10^{−is}=x−iy$$

I don't think it's all that easy to guess/infer intuitively, this fact (about complex conjugates). Especially when you are a beginner (that's who the author addresses this chapter to,  building steadily from arithmetic through algebra, logarithms, etc... guided by the intellectual beacons of Abstraction and Generalisation),  it isn't at all convincing why if $10^{is}$ equals some $x + iy$, then $10^{-is}$ must be $x-iy$.
The only definition of $i$ is that its square is $-1$ (Feynman's reason for this to be true). What am I missing here?
 A: Feynman in that whole section is using results that he knows to be true and trying to provide some intuition, but you've picked one of several things in that chapter that are not mathematically rigorous.  Even just with the equations you've shown, a mathematician would want to prove that $10^{r+is} = 10^r 10^{is}$.  It's true, but you cannot assume it's true just from the definition $i^2 = -1$. (For other structures, like matrix exponentiation, it's not generally true, just for example.)
Similarly, he's assuming that $(10^{is})^* = 10^{-is}$.  Again, that's definitely true but it also does not immediately follow from the definitions.  It should be proved if you want to be mathematically rigorous.
I don't know that this is a "cop out" because I also don't think he's claiming rigor here at all.  He's trying to give some quick intuition and, like other teachers, he's approaching it with foresight to the answers in a way that he thinks is useful to the student.  I think there's no more or less to it than that.
A: As you say, the only defining property of $i$ is $i^2=-1$. If you replace $i$ with $-i$, everything still works. Therefore, if $10^{is}=x+iy$ with $s,\,x,\,y\in\Bbb R$, $10^{-is}$ has to be $x-iy$, otherwise you could "tell apart" $i,\,-i$.
One can construct functions $f$ that don't satisfy $f(z^\ast)=f(z)^\ast$ but - without going into analytic functions, Cauchy-Riemann equations etc. - their definition must contain $i$, in order to break the symmetry. This is why, for example, that equation can be violated by $e^{iz}$ but not $e^z$.
