proof verification - Is there a pair of complex numbers for which $|x^{y}| > |x|^{|y|}$?

This is part of larger program involving functions that preserve order over the positive reals.

But I'm trying to show that

$$\forall x,y \in \mathbb{C} \ |x^{y}| \le |x|^{|y|}$$

My work:

Let $x = a_x + b_xi$, Let $y = a_y + b_yi$, $\Re = \text{Real Part}, \Im = \text{Imaginary Part}$

$$|x^{y}| = |e^{\ln(x)y}| = |e^{\ln(a_x + b_xi)(a_y+b_yi)}|=e^{\Re(\ln(a_x + b_xi))a_y-\Im(\ln(a_x + b_xi))b_y }$$

This needs to be compared to $|x|^{|y|} = e^{\ln \left(\sqrt{a_x^2 + b_x^2} \right)\sqrt{a_y^2+b_y^2}}$

But i'm at a loss for how to compare these two objects. I do have some intuition though, for one as $b_y \rightarrow \infty$ the first expression tends towards 0, while the bottom expression grows without bound.

Answer

$i^{-i}=\exp(-i \ln i)$

Take the principal branch of the logarithm, then $\ln i=i\pi/2$, thus

$i^{-i}=\exp(\pi/2)$

and the requested inequality is satisfied as $|x|=|y|=1$ and $|rhs|>1$.