Let $f(z)=\frac{\bar z^2}{z}$ when $z\neq 0$ and $f(0)=0$. Show that $f'(0)$ does not exist.

Question

Let $f(z)=\frac{\bar z^2}{z}$ when $z\neq 0$ and $f(0)=0$. Show that $f'(0)$ does not exist.

$$f'(0)=\lim_{z \to 0} \frac{f(z)-f(0)}{z-0}=\lim_{z \to 0} \left( \frac{x-iy}{x+iy}\right)^2$$ Reaching the origin along the line $y=mx$, we get $$f'(0)=\lim_{x \to 0} \left( \frac{x-imx}{x+imx}\right)^2=\left( \frac{1-im}{1+im}\right)^2$$ Since this value depends on $m$, the derivative takes different values along different paths and hence does not exist. Is this right and what are other ways to do this?

Answer 1

Yes, that is correct. But in fact, taking only two directions is enough:

• if $z\in\mathbb R$, then $\frac{f(z)-f(0)}z=1$ and therefore $\lim_{z\to0}\frac{f(z)-f(0)}z=1$;
• if $z\in(1+i)\mathbb R$, then $\frac{f(z)-f(0)}z=-1$ and therefore $\lim_{z\to0}\frac{f(z)-f(0)}z=-1$.

Answer 2

What you did is good. Another way is to notice that $$\frac{\partial f}{\partial \bar z} = 2\frac{\bar z}{z}$$ doesn’t vanish at the origin. Hence the complex derivative can’t exist at the origin.

This avoids to look at the real parts of $z$.