Reducing the result of $z=e^{x^2-y^2}\cos(2xy)$ after a coordinate rotation by $45^\circ$

Question

Starting with $$z=e^{x^2-y^2}\cos(2xy)$$ I must make the substitution $$\begin{align} x &\to \phantom{-} x\cos 45^\circ + y\sin 45^\circ \\ y &\to -x\sin 45^\circ + y \cos 45^\circ \end{align}$$ where $$\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}$$

I tried reducing, but I am making a mistake. Please help me find it. I got

$$\begin{align} z&=e^{2}\cdot\cos(2\cdot\sqrt{2}\cdot 0) \\ z&=e^{2} \end{align}$$

Wolfram alpha will graph the function without reducing it:

Answer 1

Not sure what you are trying to do, but it looks like you are rotating the coordinate system.

let: $x = \frac{\sqrt {2}}{2} u + \frac{\sqrt {2}}{2} v\\ y = \frac{\sqrt {2}}{2} u - \frac{\sqrt {2}}{2} v$

$x^2 - y^2 = 2uv\\ 2xy = u^2 - v^2$

$z = e^{2uv}\cos (u^2 - v^2)$

Answer 2

$x^2-y^2\pm 2ixy=(x\pm iy)^2$ Therefore $e^{(x^2-y^2)}cos(2xy)=\frac{e^{(x+iy)^2}+e^{(x-iy)^2}}{2}$.

I presume this is what the question is seeking.