$f(z) = \overline{z}^2\operatorname{Re}(z)$ is analytic in a subset of $\mathbb{C}$? Find whether the function $f(z) = \overline{z}^2Re(z)$ is analytic in a subset of $\mathbb{C}$. That's what I did:
By letting $z = x+iy$ we have:
$$f(z) = \overline{z}^2\operatorname{Re}(z) = (x^2-y^2-2xyi)x = $$
$$x^3-xy^2-2x^2yi = x^3-xy^2+i(-2x^2y)$$
Now, applying the Cauchy Riemann condition, we have:
$$\frac{\partial u}{\partial x} = 3x^2-y^2$$
$$\frac{\partial v}{\partial y} = -2x^2$$
$$\frac{\partial u}{\partial x}  = \frac{\partial v}{\partial y} \rightarrow 3x^2-y^2 = -2x^2 \rightarrow$$
$$5x^2 = y^2 \rightarrow \sqrt5x=y, \sqrt5x=-y$$
Therefore, the function can be continuous in the subset where $y=\sqrt5x$ or $y=-\sqrt5x$.
Am I right?
 A: As you have shown, $f(x+iy) = x^3-xy^2+i(-2x^2y)$. The Cauchy-Riemann equations are
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{ \partial u}{\partial y} = - \frac{\partial v}{\partial x}$$
and if they are both satisfied by $f$ on an open connected subset of $\mathbb{C}$ (a domain), $f$ is analytic there. The first equation gives
$$3x^2-y^2 = -2x^2 \implies 5x^2 = y^2,$$
which is the equation of a pair of lines in the plain, as you have concluded. The second equation gives
$$-2xy = -4xy \implies xy = 0.$$
Hence, the only points that satisfy both equations have one coordinate equal to $0$. The two lines from the first equation intersect at the origin and this is the only point satisfying both equations. $\{ 0 \}$ is not open. Hence, the function is nowhere analytic. In fact, we could see this without considering the second equation as the union of two straight lines is neither an open set.
In your attempt, you write continuous. I wish to emphazise that $f$ is continuous everywhere. It is however nowhere analytic.
A: We can also see some benefit of thinking in terms of $\partial/\partial z$ and $\partial/\partial \overline{z}$, and formulation of the Cauchy-Riemann equation(s) for $f$ as ${\partial\over \partial \overline{z}}f=0$. Since $\overline{z}^2\Re(z)={1\over 2}(\overline{z}^2z+\overline{z}^3)$, the $\overline{z}$-derivative is $\overline{z}z+{3\over 2}\overline{z}^2$.
This is $0$ at $z=\overline{z}=0$, indeed. Beyond that, it would be zero when $z+{3\over 2}\overline{z}=0$, but, on size considerations, this only happens at $z=\overline{z}=0$.
