Here, $a^b=e^{b\log a}$ for some suitable (but fixed in advance) branch of the $\log$ function. What is the most general formula for $|a^b|$ when both $a$ and $b$ are complex, and what are the conditions on it? The formula $|a^b|=a^{\Re b}$ only works when $a$ is positive real. A calculation from the definition yields:
\begin{align} |a^b|&=|e^{b\log a}|=\exp(\Re[b\log a])\\ &=\exp(\Re[b]\Re[\log a]-\Im[b]\Im[\log a])\\ &=\exp(\Re[b]\log|a|-\Im[b]\Im[\log a])\\ &=|a|^{\Re b}\exp(-\Im[b]\Im[\log a])\\ \end{align}
but now things get a little tricky because it's not clear what you will get when you take the imaginary part of the $\log$ function. What does this formula look like after it has been maximally simplified?
The reason I'm trying to characterize this is because I want to find for which values of $b$ we have $|a|\le|c|\implies |a^b|\le|c^b|$. It's easy to show that this is true for real $b\ge0$ and false for $b<0$, but off the reals I'm not so sure. My conjecture is that it is true when $\Re b\ge0$.
Expanding out your expression a bit it is equal to $$|a^b| = |a|^{b_x}e^{-b_y \arg a}$$
where $\Re(b) = b_x, \qquad \Im(b) = b_y$.
So in particular it is not true. A large argument will affect this power strongly.
Put $a = 1, b = 1+2i,c = 2 i $
you get that $|a|<|c|$ but $|a^b| =1 > 2e^{-\pi} = |c^b|$