Continuously Differentiable vs Holomorphic I was wondering if being continuously differentiable is equivalent to being holomorphic on some open set? I believe holomorphic implies continuously differentiable but I don't know how to prove/disprove the other direction.
 A: Let $U\subset \Bbb{C}$ be a non-empty open set, $f:U\to\Bbb{C}$ a given function, and $f=u+iv$ denote the decomposition into real and imaginary parts. Here, we can view $\Bbb{C}\cong\Bbb{R}^2$ and correspondingly think of $U$ as being a non-empty open set in $\Bbb{R}^2$. Now there are several notions to be discussed:

*

*$f$ is analytic on $U$: i.e for each $z_0\in U$, there is an $r>0$ and a sequence of coefficients $\{a_n\}_{n=0}^{\infty}\subset \Bbb{C}$ such that the open disk $D_r(z_0)$ is contained in $U$, and for all $z\in D_r(z_0)$, we have $f(z)=\sum\limits_{n=0}^{\infty}a_n(z-z_0)^n$. In other words, about each point $f$ admits a local power series expansion.


*$f$ is holomorphic on $U$: meaning $f$ is (complex) differentiable at each point of $U$, i.e for each $z_0\in U$, $\lim\limits_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h}$ exists (which is what we denote as $f'(z_0)$).


*$f$ is continuously complex differentiable on $U$: i.e $f$ is holomorphic on $U$ and $f':U\to\Bbb{C}$ is continuous.


*$u,v:U\to\Bbb{R}$ are continuously differentiable in the sense that $u,v,\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}, \frac{\partial v}{\partial x},\frac{\partial v}{\partial y}$ all exist and are continuous on $U$, and furthermore the Cauchy-Riemann equations are satisfied at each point of $U$.


*$u,v:U\to\Bbb{R}$ are Frechet-differentiable on $U$ and satisfy the Cauchy-Riemann equations at each point of $U$.
These five conditions are all equivalent. Well, it is pretty easy to show $(1)$ implies $(3)$ which clearly implies $(2)$ (actually, $(1)$ and induction shows that $f$ has complex derivatives of all orders on $U$). Next, $(3)$ and $(4)$ are easily seen to be equivalent, and likewise $(2)$ and $(5)$ are easily seen to be equivalent.
So, at the very least so far we have the following relatively "easy" implications:
\begin{align}
(1)\implies (3)\implies (4)\implies (5)\implies (2)
\end{align}
Note that $(4)\implies (5)$ is a basic fact proved in multivariable calculus: continuity of partials implies Frechet differentiability; the proof uses the mean-value theorem from single-variable calculus (some of the reverse implications are easy but not all of them).
So, to complete everything, we just have to show $(2)$ implies $(1)$. This is the hardest part, and this is done by first establishing Cauchy's theorem (if $f$ is holomorphic on $U$, then for every $\gamma$ which is the boundary of a rectangle $R$ with $\overline{R}\subset U$ and with sides parallel to the coordinate axes, $\int_{\gamma}f(z)\,dz=0$), followed by Cauchy's integral formula (if $f$ is holomorphic on $U$, then for each $z\in U$ and sufficiently small $r>0$, $f(z)=\frac{1}{2\pi i}\int_{|\zeta-z|=r}\frac{f(\zeta)}{\zeta-z}\,d\zeta$), from which we can deduce (by expanding the integrand as a geometric series and uniform convergence) that $f$ admits locally a power series expansion.
Finally, the proof of this last implication $(2)\implies (1)$ also establishes the fact that the radius of convergence of a power series of a holomorphic function is the distance to the nearest singularity (i.e if $f$ is holomorphic on a disk $D_r(z_0)$, then $f$ admits a power series expansion centered at $z_0$ with radius of convergence $\geq r$), so that's an added bonus. As a particular case, if $f$ is entire, then the radius of convergence is infinity (this fails badly in $\Bbb{R}$).
A: The definitions:
If $O \subset \mathbb{R}^{2} \approx \mathbb{C}$ is open, a function $f : O \to \mathbb{R}^2 \approx \mathbb{C}$ is $C^1$ if all first order partials of $f$ exist and are continuous on $O$.
$f$ is holomorphic if $f$ is complex Frechet differentiable at each point $z \in O$, that is, for each $z \in O$, there is a linear map $f'(z) : \mathbb{C} \to \mathbb{C}$ such that for $w \in \mathbb{C}$ small,
$$f(z + w) = f(z) + f'(z)w + r(w), \hspace{20pt}\lim_{w \to 0}\frac{r(w)}{|w|} = 0.$$
Of course, linear maps from $\mathbb{C}$ to $\mathbb{C}$ are just elements of $\mathbb{C}$, so $f'(z) \in \mathbb{C}$.
Some comments:
Suppose $f$ is $C^1$. By a standard theorem of calculus, it is true that $f$ is real Frechet differentiable, that is, for each $z \in O$, there is a linear map $Df(z) : \mathbb{R}^2 \to \mathbb{R}^2$ such that for $w \in \mathbb{R}^2$ small,
$$f(z + w) = f(z) + Df(z)w + r(w), \hspace{20pt}\lim_{w \to 0}\frac{r(w)}{|w|} = 0.$$
Note that the only difference between the above statement and the definition of complex Frechet differentiability is that in the above $Df(z)$ is only real linear, while $f'(z)$ is complex linear. Hence a $C^1$ function $f$ is holomorphic if and only if for each $z \in O$, $Df(z)$ is a complex number, i.e. has the form
$$Df(z) = \begin{pmatrix}
a & -b \\
b & a \\
\end{pmatrix}.
$$
The above equation is known as the Cauchy-Riemann equation.
Now it is easy to find functions $f$ where $Df$ does not take the above form. For example, let $f(z) = \overline{z}$.
If $f$ is holomorphic, it is true that $f$ is $C^1$. This is known as Goursat's theorem. See page 99 of https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/complex.pdf for a proof. Once you know that $f$ is $C^1$, then you can deduce from the Cauchy integral theorem that $f$ is actually infinitely complex differentiable.
