why the relationship between A^i vs A^-i

Question

I am trying to get an intuitive idea of imaginary exponents. I've read this: Understanding imaginary exponents and one thing I am wondering is why if

$A^i=x+iy$

then

$A^{-i}=x-iy$

I read the comments in the link, but the answer someone gave may be incomplete, or at least it doesn't make sense to me.

Answer 1

For any real-valued angle $\theta$, we find that $e^{i\theta} = \cos(\theta)+i\sin(\theta)$, from the Taylor Series expansions of $e, \cos, \sin$.

Thus for any positive real number $A$, we have $A^i {= e^{i\ln A} \\= \cos(\ln A)+i\sin(\ln A)}$

So likewise $A^{-i} {= e^{-i\ln A} \\=\cos(-\ln A)+i\sin(-\ln A) \\= \cos(\ln A)-i\sin(\ln A)}$

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Then we are releived to find that $A^i\cdotp A^{-i} {= \cos^2\ln A+\sin^2\ln A \\= 1}$