finding complex function (contradicting answers)

Question

Please help. I understand the first part (i) in the image link , but i don't know how to find a function satisfying the (ii) part.

Is there any function $f$ such that $f$ is complex-differentiable at $z$ if and only if $|z|\le 1 $?

Any help regarding this part is appreciated. Can you give some examples and how do we find such complex functions satisfying this condition.

What I did: I thought of real function which is discontinuous at some points in $-1\leq z\le1$ but i don't think so it is correct to think this way.

Answer 1

I thought about something like for the first case: $$ f_1(z) = \left\{ \begin{array}{cc} z^2 & \textrm{if } z \in D_1 \\ 0 & \textrm{if } z \notin D_1 \end{array} \right. $$ where $D_1 = \{z \in \mathbb{C} : \operatorname{Re}(z) \in \mathbb{Q} \vee \operatorname{Im}(z) \in \mathbb{Q} \}$ .

For the second case $$ f_2(z) = \left\{ \begin{array}{cc} 0 & \textrm{if } z \in D_2 \\ (|z|-1)^2 & \textrm{if } z \notin D_2 \end{array} \right. $$ where $D_2 = \{z \in \mathbb{C} : |z| \leq 1 \} \cup D_1$ for the second case.

The idea is to have a discontinuous function everywhere except in the relevant points and that when we are near the boundary of the relevant points the function is (continuous and) differentiable.

Note that in the second case, the second branch of $f_2(z)$ can be represented by $f_2(z) = u(z) + i v(z)$, with $z = r e^{i \theta}$ and: $$ u(z) = \operatorname{Re}(f_2(z)) = (|z|-1)^2 = (r-1)^2 \\ v(z) = \operatorname{Im}(f_2(z)) = 0 $$

Therefore, $$ \frac{\partial u}{ \partial r} = 2(r-1); \quad \frac{\partial u}{ \partial \theta} = 0; \quad \frac{\partial v}{ \partial r} = 0; \quad \frac{\partial v}{ \partial \theta} = 0 $$ The Cauchy-Riemann equations in polar coordinates at the boundary $r=1$ give: $$ \begin{array}{rcccc} \dfrac{\partial u}{ \partial r} = 2(r-1) & = & 0 & = & \dfrac{1}{r} \dfrac{\partial v}{ \partial \theta} \\ \dfrac{\partial v}{ \partial r} & = & 0 & = & -\dfrac{1}{r} \dfrac{\partial u}{ \partial \theta} \end{array} $$

I'm uncertain regarding the second case, so I will appreciate any comments if I forgot something relevant.

Answer 2

Take $f(x,y)=f(r)=f(z,\bar{z})=(x^2+y^2-1)^N=(r^2-1)^N=(z\bar{z}-1)^N, |z| \ge 1$ and $f=0$ on the unit disc, where $N \ge 2$ is a positive integer; by taking partial derivatives, $f$ is clearly real-diferentiable $N-1$ times (with final $(N-1)$th continuos derivative) on the unit circle (and is real analytic both inside and outside the circle, while being complex analytic inside the circle).

$\partial_{\bar z}f=2Nz (z\bar{z}-1)^{N-1}$ (outside and on the unit circle) so it immediately follows that $f$ is complex differentiable on the unit circle too (though it is not analytic there!)

For point 1, $f(x,y)=f(r)=f(z,\bar{z})=x^2+y^2=r^2=z\bar{z}=|z|^2$ is real analytic and complex differentiable only at $0$ as $\partial_{\bar z}f=z$, but for point 2 we cannot have a real analytic function like that, only a real-differentiable one