Why is continuous differentiability required? I have two questions.
My book proves that if $f:\mathbb{C}\rightarrow \mathbb{C}$ is a holomorphic function, then it satisfies the Cauchy-Riemann equations, and if we look at the function as $F: \mathbb{R}^2\rightarrow \mathbb{R}^2$, then this function is differentiable. The point is that we are looking at conditions when we can go from complex differentiability(holomorphic), to differentiability when we look at the function in terms of real variables only ,that is differentiability in multivariate calculus.
question 1:
The book also has the converse of the above[picture], and in the converse result they write
$ f = u(x,y)+i*v(x,y)$, where $z=x+i*y$, now in the converse result they have that the u and v must be continuously differentiable, why not only differentiable? Do you guys see in the proof where they need to use that it is continuously differentiable and it would fail if they were only differentible?
question 2:
A related question is: If we have a function $h: \mathbb{R}^2\rightarrow \mathbb{R}^2 $. And we know that the partial derivatives exist, but they are not continous, will it may be that h is not differentiable?

 A: The continuous differentiability of $u$ and $v$ is not required for the proof, for the representation
$$u(x+h_1,y+h_2) - u(x,y) = \frac{\partial u}{\partial x}(x,y) \cdot h_1 + \frac{\partial u}{\partial y}(x,y)\cdot h_2 + \lvert h\rvert\cdot \psi_1(h)\tag{1}$$
and the analogous for $v$, the differentiability of $u$ (and $v$) in $(x,y)$ [resp. $x+iy$] is enough. From these representations and the Cauchy-Riemann equations (in the single point $(x,y)$) follows the complex differentiability of $f$ in $x+iy$.
That shall be the case for all $x+iy \in \Omega$, hence $f$ is holomorphic on $\Omega$.
Since every holomorphic function is analytic, nothing is lost by imposing stricter conditions than the proof requires on $u$ and $v$, but the proof only uses the differentiability in each point.
Concerning question 2, yes, there are functions where the partial derivatives exist everywhere that are not differentiable at least in some points.
The mere existence of the partial derivatives does not imply a representation of the type $(1)$, hence the existence of the partial derivatives plus the Cauchy-Riemann equations does not suffice to deduce the complex differentiability of $f$ in $x+iy$.
A: \begin{align*}
&u(x_0+h_1,y_0+h_2)-u(x_0,y_0)\\
&= h_1 \frac{u(x_0+h_1,y_0+h_2)-u(x_0,y_0+h_2)}{h_1}+h_2\frac{u(x_0,y_0+h_2)-u(x_0,y_0)}{h_2}\\
&= h_1 \left(\frac{\partial u}{\partial x}(x_0,y_0+h_2) + \varphi_1(h_1)\right) + h_2 \left(\frac{\partial u}{\partial y}(x_0,y_0) + \varphi_2(h_2)\right) & \text{where $\lim\limits_{h_i \to 0}\varphi_i(h_i)=0$}\\
&= h_1 \left(\frac{\partial u}{\partial x}(x_0,y_0) + \varphi_1(h_1)+\widetilde{\varphi}_2(h_2)\right) + h_2 \left(\frac{\partial u}{\partial y}(x_0,y_0) + \varphi_2(h_2)\right)  & \text{$\lim\limits_{h_2 \to 0}\widetilde{\varphi}(h_2)\to 0$ by continuity of $\partial u / \partial x$}\\
\end{align*}
