Where is $ze^{\overline{z}}$ differentiable?

Question

Define $$f(z)=ze^{\overline{z}}.$$ I have an argument that is differentiable only at $z=0$; however, I am unsure of my argument for non-zero $z$ is correct.

If $z\not=0$, we obtain $\log(f(z)/z)=\overline{z},$ and this is nowhere differentiable. Since $\overline{z}$ is not differentiable on any branch-cut we define for $\log$, this shows that $f$ is not differentiable on $\mathbb{C}-\lbrace0\rbrace.$

At $z=0$, writing $h=a+ib,$ we have $f'(0)=\lim_{h\to 0}\frac{he^{a-bi}}{h}=1.$ So $f$ is only differentiable at $0$.

If there are other proofs that don't require the Cauchy-Riemann equations, please share.

Answer 1

If $f$ is differentiable at some nonzero point, so is $g:z\mapsto e^z\frac{f(z)}{z}=e^{z+\bar{z}}=e^{2\Re(z)}$, but this is impossible because $g$ has non-equal directional derivatives. In fact, if we write $z=x+iy$, then along the $y$-direction, $g(z)=e^{2x}$ is a constant function and the directional derivative is $0$. Along the $x$-direction, however, the directional derivative is $\frac{de^{2x}}{dx}=2e^{2x}\ne0$.

Answer 2

This has a nice formulation in terms of differentials. If you have a smooth function of $z$ and $\bar{z}$, then it is complex differentiable with respect to $z$ if and only if, if you write its differential as a linear combination of $\mathrm{d}z$ and $\mathrm{d} \bar{z}$, the coefficient on $\mathrm{d} \bar{z}$ is zero.

Since $$ \mathrm{d}f(z) = e^{\bar{z}} \mathrm{d}z + z e^{\bar{z}} \mathrm{d}\bar{z} $$

we see that $f(z)$ is complex differentiable if and only if $z e^{\bar{z}} = 0$. Since $e^\bar{z}$ is everywhere nonzero, this happens if and only if $z=0$.