System of two equations in three unknowns

Question

How can I solve this system of equations for x and y:

$$ e^{-2y} \cos(2x)= \frac{1-w^2}{1+w^2} \text{ and } e^{-2y} \sin(2x)= \frac{2w}{1+w^2},$$ where $w$ is any complex number?

Answer 1

HINT: Square and add both equations to get:$$e^{-4y}=\frac{(1-w^2)^2+4w^2}{(1+w^2)^2}$$ Therefore $e^{-4y}=1$. Rest you can simply get $\tan 2x $ by dividing both equations.

Answer 2

Square the two equations, add them together, and you get $e^{-4y}$ as a function of $w$, from which, when you take the logarithm, you get $y$. Divide the two equations and you get $\tan(2x)$, from where you can find $x$