Limit of modulus of the difference quotient exists, implying complex differentiability. A question from a past qualifying exam at my university reads:
Suppose that $f(z)=u(x,y)+iv(x,y)$ is a function on a domain $D$ and $z_0\in D$. Show that if: a) $u$ and $v$ are differentiable at $z_0$ b) the limit
$$ \lim_{\Delta z\rightarrow 0}\Big| \frac{f(z_0-\Delta z)-f(z_0)}{\Delta z}\Big|$$
exists, then either $f(z)$ or $\overline{f}(z)$ are complex differentiable at $z_0$.
If one can show that $u$ and $v$ satisfy the Cauchy-Riemann equations, then we are done, and this seems to be the way to approach the problem. However, I do not see how to do this. Because we can approach $z_0$ from any direction, then one can conclude that $|u_x(z_0)+iv_x(z_0)|=|v_y(z_0)-iu_y(z_0)|$ or equivalently $u_x(z_0)^2+v_x(z_0)^2=v_y(z_0)^2+u_y(z_0)^2$. The solution set of this equation is some circle in the complex plane. How can I proceed?
 A: The taylor series for $f$ is given by
$$ f(z)= f(z_0)+\frac{\partial f}{\partial z}(z_0)(z-z_0)+\frac{\partial f}{\partial \overline{z}}(z_0)(\overline{z-z_0})+\mathcal{O}(|z-z_0|^2)$$
or equivalently
$$ \frac{f(z)-f(z_0)}{z-z_0}=\frac{\partial f}{\partial z}(z_0)+\frac{\partial f}{\partial \overline{z}}(z_0) \frac{\overline{z-z_0}}{z-z_0} +\frac{\mathcal{O}(|z-z_0|^2)}{z-z_0}$$
Let $z\rightarrow z_0$ ($\Delta z\rightarrow 0$). By hypothesis, the limit below exists. However, we may approach $z_0$ from any direction, so we get that
$$ \lim_{\Delta z\rightarrow 0}\Big|\frac{f(z)-f(z_0)}{z-z_0}\Big|=\lim_{z\rightarrow z_0}\Big|\frac{\partial f}{\partial z}(z_0)+\frac{\partial f}{\partial \overline{z}}(z_0) \frac{\overline{z-z_0}}{z-z_0}+\frac{\mathcal{O}(|z-z_0|^2)}{z-z_0} \Big |=\Big|\frac{\partial f}{\partial z}(z_0)+\frac{\partial f}{\partial \overline{z}}(z_0)e^{it}\Big|$$
for any $t\in \mathbb{R}$. Let $a=\frac{\partial f}{\partial z}$ and $b=\frac{\partial f}{\partial \overline{z}}$. We have that $|a+be^{it}|$ is independent of $t$. Suppose that $a$ is nonzero. For the sake of contradiction, suppose that $b$ is nonzero.  Then we can pick $t\in[0,2\pi)$ such that $\textrm{Arg}(a)=\textrm{Arg}(be^{it})$. Then, $be^{it}$ is a scalar multiple of $a$, so there is an $c>0$ such that $|a+e^{it}b|=|a+ca|=|a||1+c|$. But we have that $|a+e^{it}b|=|a+-1e^{it}b|=|a-ca|=|a||1-c|$. Thus $|a||1+c|=|a||1-c|$ or $|1+c|=|1-c|$. Thus $c=0$, a contradiction. Thus $a$ or $b$ must be zero. If $a$ is zero, then by the Cauchy-Riemann equations, we can conclude that $\overline{f}$ is analytic at $z_0$. If $b=0$, then $f$ is analytic.
