The original question is: Where is the function below differentiable:
$$\begin{cases} f(z) = e^{x^2-y^2}*[\cos(2xy) - i\sin(2xy)] \end{cases}$$
Using Cauchy-Riemann:
$$\begin{cases} \frac{\delta(u)}{\delta(x)} = 2xe^{x^2-y^2}cos(2xy) - 2ye^{x^2-y^2}sin(2xy),\\ \frac{\delta(u)}{\delta(x)} = 2xe^{x^2-y^2}cos(2xy) - 2ye^{x^2-y^2}sin(2xy) \end{cases}$$
$$\begin{cases} \frac{\delta(v)}{\delta(x)} = -2xe^{x^2-y^2}sin(2xy) - 2ye^{x^2-y^2}cos(2xy),\\ \frac{\delta(u)}{\delta(y)} = -2ye^{x^2-y^2}cos(2xy) - 2xe^{x^2-y^2}sin(2xy) \end{cases}$$
The above simplifies to this system of equations that I have no idea how to solve.
$$\begin{cases} x\cos(2xy)-y\sin(2xy) = 0,\\ x\sin(2xy) + y\cos(2xy) = 0 \end{cases}$$
All the online equation calculator don't want to solve this. I can see there is an obvious root x=0, y=0.
How to solve this properly?
Note that,
$$ y\sin(2xy)=x\cos(2xy)$$ $$x\sin(2xy)=- y\cos(2xy) $$
which leads to
$$x^2 = -y^2,\>\>\>\>\tan^2(2xy)=-1$$
Only $x^2 = -y^2$ gives the valid solutions $x=y=0$.