Questions regarding complex differentiablity of complex functions with differentiable real/imaginary parts

Question

I'm studying about Complex functions and I came across these two following questions which I haven't really been able to solve.

Let $f\left(z\right)=u\left(x,y\right)+iv\left(x,y\right)$ be defined in the open ball $B\left(z_{0},r\right)$ and real differntiable at $z_{0}$ (meaning $u,v$ are differentiable as real functions at $\left(x_{0},y_{0}\right)$ ). Prove the following:

1. The group of limit points of the ratio $\frac{f\left(z\right)-f\left(z_{0}\right)}{z-z_{0}}$ as $z\to z_{0}$ is either one point or a circle.

2. If the limit $\lim\limits _{z\to z_{0}}\left|\frac{f\left(z\right)-f\left(z_{0}\right)}{z-z_{0}}\right|$ exists then either $f\left(z\right)$ or $\overline{f\left(z\right)}$ are complex differentiable at $z_{0}$ .

Help would be appreciated!

Answer 1

Since it's too long for a comment:

The differentiability of $u$ and $v$ at $z_0$ implies that we have a relation

$$f(z) = f(z_0) + a(z-z_0) + b(\overline{z-z_0}) + R(z)\tag{1}$$

with $\lim\limits_{z\to z_0} \dfrac{R(z)}{\lvert z-z_0\rvert} = 0$, where $a = \frac{\partial f}{\partial z}(z_0)$ and $b = \frac{\partial f}{\partial\overline{z}}(z_0)$.

Then, writing $z = z_0 + \rho e^{i\varphi}$, $(1)$ becomes

$$\frac{f(z)-f(z_0)}{z-z_0} = a + b e^{-2i\varphi} + \frac{R(z)}{z-z_0}.\tag{2}$$

Letting $\rho\to 0$, for arbitrary fixed $\varphi\in\mathbb{R}$, we see that $a+be^{-2i\varphi}$ is a limit point of the difference quotient, and since $\lim\limits_{z\to z_0} \left\lvert \dfrac{R(z)}{z-z_0}\right\rvert = 0$, the set of limit points of the difference quotient $\frac{f(z)-f(z_0)}{z-z_0}$ is $L = \{ a + b e^{-2i\varphi} : \varphi \in \mathbb{R}\}$. If $b = 0$, that is the singleton $\{a\}$, and if $b\neq 0$, it is the circle with centre $a$ and radius $\lvert b\rvert$.

That shows the first proposition.

For the second, note that $\lim\limits_{z\to z_0} \left\lvert \frac{f(z)-f(z_0)}{z-z_0}\right\rvert$ exists if and only if the set $L$ is contained in a circle with centre $0$ in the plane, and that is the case if and only if $b = 0$ (then $f$ is complex differentiable in $z_0$) or $a = 0$ (in which case $\overline{f}$ is complex differentiable in $z_0$).