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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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.