Question: Where is $f(z) = e^z(z-\bar{z})^2$ differentiable?
My attempts: By letting $z = x+iy$ and by using the Cauchy Riemann Equations, I was able to get to a point where $\sin y = 0$ and $\cos y = 0$. I've tried to question three times and I get the same result each time. For the real part of the function I got $u(x,y) = -4y^2\cos(y)e^x$ and for the imaginary part I got $v(x,y) = -4y^2\sin(y)e^x$. From there, using $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$. The equations I arrived at were:
$$ -4y^2\cos(y)e^x = -8y\sin(y)e^x - 4y^2\cos(y)e^x$$ and $$-8y\cos(y)e^x+4y^2\sin(y)e^x=4y^2\sin(y)e^x.$$
It was from these questions that I arrived that $\sin(y) = 0$ and $\cos(y) = 0$. Have I made a mistake? Any help or guidance would be greatly appreciated!
Hint: If $e^{z}(z-\overline z)^{2}$ is differentiable at some point then $(z-\overline z)^{2} =e^{-z} (e^{z}(z-\overline z)^{2})$ is also differentiable at the point. That should make the computations much simpler.