# 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.

## 3 Answers

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 $$f(z = x+iy) = Ax + B + i(Ay + C) = Az + (B + iC).$$

• How do you prove that $u_x$ and $v_y$ are constants? – user330477 Sep 16 '20 at 11:01
• @user330477 it's because $u_{x}$ is a function purely of $x$ (or a constant), and likewise $v_{y}$ is a function purely of $y$ (or a constant); but as $u_{x} = v_{y}$, the only way for this to hold is if they are constant – BenCWBrown Oct 4 '20 at 22:54

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 $$f(z) = u(x) + iv(y) = kz + C$$.

• Satisfying the CR equations at a point does not imply complex differentiable at a point. You need to know that $u,v$ are continuously differentiable (or, to be more precise, continuous differentiability suffices when paired with the CR-equations to imply complex differentiability). The phrase "if and only if" ought not be in the answer as it currently stands. – James S. Cook Nov 9 '18 at 0:53
• @JamesS.Cook I've edited my answer to include the right equivalent conditions, which is weaker than continuously differentiability. – GNUSupporter 8964民主女神 地下教會 Nov 9 '18 at 1:38
• Downvote removed. Indeed, real differentiability will suffice. – James S. Cook Nov 9 '18 at 2:06
• Why did you assume that $\frac{u(x)-C}{x}=\frac{v(y)}{y}=k$? – user330477 Sep 16 '20 at 11:02
• @user330477 From the definition of a function, the 1st part only depends on $x$, the 2nd part only depends on $y$. – GNUSupporter 8964民主女神 地下教會 Sep 16 '20 at 11:25

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.