Is $f$ holomorphic?

Question

Consider the function $f : \mathbb C → \mathbb C$ defined as $f(z) = \bar{z} ^2 $

1. Show that $f$ is holomorphic (= complex differentiable) at the point $z = 0$.

2. Is $f$ holomorphic on $\mathbb C$?

My attempt:

1. I calculated the limit as $h\to0$ of $\frac{f(0+h) - f(0)}{h}$: $\lim_{h\to0} \frac{\bar{h}^2}{h} = \lim_{h→0}$ $\frac{h^2}{h}$ $=0$

   Thus $f$ is holomorphic

2. I calculated the same limit but for any $z_0$ in $\mathbb C$, i.e. $\lim_{h→0}$ $\frac{f(z_0 + h) - f(z_0)}{h}$. I got that the limit is infinity. So the limit exists and thus $f$ is holomorphic on $\mathbb C$

Are my attempts correct?

Answer 1

Your solution for a) is o.k.

b) is not o.k.

Show that

$$\lim_{h \to 0, h \in \mathbb R}\frac{f(z_0 + h) - f(z_0)}{h}=2 \overline{z_0}$$

and

$$\lim_{h \to 0, h \in i \mathbb R}\frac{f(z_0 + h) - f(z_0)}{h}=-2 \overline{z_0}.$$

If $z_0 \ne 0$ , these limits are not equal, thus $\lim_{h \to 0}\frac{f(z_0 + h) - f(z_0)}{h}$ does not exist.

Answer 2

Not sure how you got the limit is infinity. Anyway, just write the function in the form $u+iv$. We have $f(x+iy)=(x-iy)^2=(x^2-y^2)-2xyi$. Hence $u(x,y)=x^2-y^2, v(x,y)=-2xy$. Now it is easy to see that the Cauchy-Riemann equations hold only at the origin. So $f$ can't be differentiable at any other point.