Is $f(z) = z \bar{z}$ differentiable at $0$? I am trying to work out when $f(z) = z \bar{z}$ is complex differentiable. Looking at the Cauchy-Riemann equations, they are not satisfied away from $0$, but I think you just get $0 = 0$ and $0=0$ at $0$. Since $f$ depends on the conjugate of $z$, I am inclined to think it shouldn't be differentiable, but I can't seem to formalise this. Please help!

$f$ is real-differentiable at $(0,0)$ because $f(x,y)=x^{2}+y^{2}$ is a polynomial. Now $\dfrac{\partial f}{\partial\overline{z}}(z)=z$ and hence $\dfrac{\partial f}{\partial\overline{z}}(0,0)=0$, so Cauchy-Riemann is satisfied, so $f$ is complex-differentiable at $(0,0)$.
Perhaps one goes directly with definition that $\left|\dfrac{z\overline{z}-0}{z-0}\right|=|\overline{z}|=|z|\rightarrow 0$ as $z\rightarrow 0$.
