Can I differentiate $f(z)= z^2 e^\overline{z}$ with z in complex numbers?

Question

$\overline{z}$ is the conjugate of $z$.

And $z=x+iy$.

I don't know if it can be differentiated in any point for every complex number $z$.

I've tried to proof if it is differentiable in an exact point. I have separated the real part and the imaginary part and I have tried if the partial derivates are equal or not. But they are equal in an infinite number of points.

So, I don't know what can I try now.

Answer 1

$$f(z) = z^2 e^{\overline{z}} = (x+iy)^2 e^{x-iy} = (x^2 -y^2 + 2ixy) e^{x-iy} = e^x(x^2 -y^2 + 2ixy)(\cos y - i \sin y)\\ =e^x[(x^2 - y^2)\cos y + 2xy \sin y] + i e^x[2xy\cos y - (x^2-y^2)\sin y].$$

Then apply the Cauchy Riemann equations, with $u(x,y) =e^x[(x^2 - y^2)\cos y + 2xy \sin y]$ and $v(x,y) = e^x[2xy\cos y - (x^2-y^2)\sin y]$.

This gives you a test of whether $f$ is differentiable.