Finding all differentiable $f(z) = u(x) + iv(y)$ in $\mathbb{C}$ where $u(x),v(y)$ are real valued functions. 
Finding all differentiable $f(z) = u(x) + iv(y)$ in $\mathbb{C}$ where $u(x),v(y)$ are real valued functions.

I’m not sure what to do. Would $f$ be differentiable simply if and only if both $u$ and $v$ were differentiable? 
My thinking is that if the limit
$$
\lim_{h \to 0}\frac{f(z+h)-f(z)}{h}
$$ 
exists then it would be equal to 
$$
\lim_{h \to 0}\frac{u(x+h)-u(x)}{h} + i\lim_{h \to 0}\frac{v(y+h)-v(y)}{h}
$$ 
and therefore would exist if they existed. But that doesn’t seem quite right.
I also thought about using Cauchy-Riemann equations, but seeing that I’m trying to find ones that are differentiable rather than those that aren’t, I thought they wouldn’t be of much help.
 A: I am adapting this from the derivation of the CR equations, assuming that $f(z) = f(x + iy) = u(x) + iv(y)$. The limits taken for the complex derivative need to exist when considered going to $0$ along both the real and imaginary axes, i.e.
$$\lim_{t \rightarrow 0} \frac{f(z + t) - f(z)}{t} = \lim_{t \rightarrow 0} \frac{f(z + it) - f(z)}{it}
$$
exists. Plugging in $f$ we get
$$
\lim_{t \rightarrow 0} \frac{f(z + t) - f(z)}{t} = \lim_{t \rightarrow 0} \frac{u(x + t) - u(x)}{t} + i \lim_{t \rightarrow 0} \frac{v(y) - v(y)}{t} = \frac{\partial u}{\partial x}\\ 
=\lim_{t \rightarrow 0} \frac{f(z + it) - f(z)}{it} = \lim_{t \rightarrow 0} \frac{u(x) - u(x)}{it} + i \lim_{t \rightarrow 0} \frac{v(y + t) - v(y)}{it} = \frac{\partial v}{\partial y},
$$
so it seems like all of the complex differentiable functions in that form satisfy
$$\frac{\partial u}{\partial x}(x) = \frac{\partial v}{\partial y}(y).
$$
The left-hand side depends solely on $x$ whereas the right-hand side depends solely on $y$, so they are actually constants, say they are both equal to $A \in \mathbb{R}$, say (a real constant since both $u(x)$ and $v(y)$ are real-valued functions). Then
$$u(x) = Ax + B,\qquad v(y) = Ay + C,$$
from which $f(z)$ in full-generality is
$$f(z = x+iy) = Ax + B + i(Ay + C) = Az + (B + iC).
$$
A: Note that $f$ is differentiable on $\Bbb{C}$, so the partial derivatives exists everywhere.  By the Cauchy–Riemann equations, $f(z)$ is differentiable at $z \in \Bbb{C}$ if and only if


*
*\begin{aligned}(1a)\qquad &{\frac {\partial u}{\partial x}}={\frac {\partial v}{\partial y}}\\[6pt](1b)\qquad &{\frac {\partial u}{\partial y}}=-{\frac {\partial v}{\partial x}}\end{aligned} are satisfied at $z \in \Bbb{C}$, and
*$u$ and $v$ are real differentiable at $z \in \Bbb{C}$
Since $u$ and $v$ in this question are functions with single variables $x$ and $y$ respectively, $(1b)$ is always satisfied, and $(1a)$ becomes
$$u'(x) = v'(y).\tag{*}\label1$$
Integrate both sides of \eqref{1} with respect to $x$.
$$u(x) = xv'(y) + C$$
Integrate both sides of \eqref{1} with respect to $y$.
$$u(x)y = xv(y) + Cy \tag{#}\label2$$
When $x\ne0$ and $y \ne 0$, this gives $$\frac{u(x)-C}{x} = \frac{v(y)}{y} = k$$ for some $k \in \Bbb{C}$, so
\begin{cases}
u(x) &= kx + C \\
v(y) &= ky.
\end{cases}
Hence, we conclude that $f(z) = u(x) + iv(y) = kz + C$.
A: 
Would $f$ be differentiable simply if and only if both $u$ and $v$ were differentiable? 

No. The functions $u$ and $v$ must also satisfy the CR equations. 
The Looman–Menchoff theorem states that a continuous complex-valued function defined in an open set of the complex plane is holomorphic if and only if it satisfies the Cauchy–Riemann equations.
Now by setting
$$
u'(x)=v'(y)
$$
and since this is true for all $(x,y)\in{\mathbb R}\times{\mathbb R}$,
one has 
$$
u(x)=ax+b,\quad v(y)=ay+c,\quad a,b,c\in{\mathbb R}.
$$
Theses are the only possible candidates for $f$ being (complex) differentiable.
Such $f$ would be continuous for such $u$ and $v$, by the Looman-Menchoff theorem, $f$ is (complex) differentiable. So one has found all the $u$ and $v$ such that $f$ is (complex) differentiable.
