Question about Cauchy-Riemann equations

Question

The equations and related theorems confuse me. So I try to prove them myself. Can someone check if I am correct?

Suppose $f$ maps an open set $\Omega\subset\mathbb{C}$ into $\mathbb{C}$. Let $E$ be the set of all $(x,y)$ for which $x+iy\in\Omega$. Then $E$ is open in $\mathbb{R^2}$. Denote the real part and imaginary part of $f$ by $u$ and $v$, respectively; that is, $f=u+iv$. Define mappings $F:E\rightarrow\mathbb{R}^2$, $U:E\rightarrow\mathbb{R}$ and $V:E\rightarrow\mathbb{R}$ by $$U(x,y)=u(x+iy)\text{,}\quad V(x,y)=v(x+iy)\text{,}\quad\text F=(U,V)\text{.}$$

If $f$ is differentiable at $z\in\Omega$, then $$\tag{1}\frac{\lvert f(z+h)-f(z)-f'(z)h\rvert}{\lvert h\rvert}=\frac{\lvert[u(z+h)-u(z)-u'(z)h_1+v'(z)h_2]+i[v(z+h)-v(z)-v'(z)h_1-u'(z)h2]\rvert}{\lvert h_1+ih_2\rvert}$$ tends to $0$ as $h=h_1+ih_2\rightarrow0$.

If $F$ is differentiable at $X=(x,y)\in E$, then $$\tag{2}\frac{\lVert F(X+H)-F(X)-DF(X)(H)\rVert}{\lVert H\rVert}=\frac{\lVert(U(X+H)-U(X)-\partial_xU(X)h_1-\partial_yU(X)h_2,V(X+H)-V(X)-\partial_xV(X)h_1-\partial_yV(X)h_2)\rVert}{\lVert(h_1,h_2)\rVert}$$ tends to $0$ as $H=(h_1,h_2)\rightarrow 0$.

If we have $$z=x+iy\text{,}$$ $$\tag{3}\partial_xU(x,y)=\partial_yV(x,y)\text{,}$$ $$\tag{4}\partial_yU(x,y)=-\partial_xV(x,y)\text{,}$$ $$\tag{5}u'(z)=\partial_xU(x,y)\text{,}$$ $$\tag{6}v'(z)=\partial_xV(x,y)\text{,}$$ then (1) is equivalent to (2) by the definition of norms on $\mathbb{R}^2$ and $\mathbb{C}$, and hence $f$ is holomorphic at $z$ if and only if $F$ is differentiable at $(x,y)$.

Suppose $f'(z)$ exists, then $$\begin{aligned} f'(z)&=\lim_{h_1\rightarrow 0}\frac{f(z+h_1)-f(z)}{h_1}\\ &=\lim_{h_1\rightarrow0}\frac{u(z+h_1)-u(z)}{h_1}+&i\lim_{h_1\rightarrow0}\frac{v(z+h_1)-v(z)}{h_1}&=u'(z)&+iv'(z)\\ &=\lim_{h_1\rightarrow0}\frac{U(x+h_1,y)-U(x,y)}{h_1}+&i\lim_{h_1\rightarrow0}\frac{V(x+h_1,y)-V(x,y)}{h_1}&=\partial_xU(x,y)&+i\partial_xV(x,y)\\ &=\lim_{h_2\rightarrow 0}\frac{f(z+ih_1)-f(z)}{ih_2}\\ &=\lim_{h_2\rightarrow0}\frac{u(z+ih_2)-u(z)}{h_2}+&i\lim_{h_2\rightarrow0}\frac{v(z+ih_2)-v(z)}{h_2}\\ &=\lim_{h_2\rightarrow0}\frac{U(x,y+h_2)-U(x,y)}{ih_2}+&i\lim_{h_2\rightarrow0}\frac{V(x,y+h_2)-V(x,y)}{ih_2}&=\partial_yV(x,y)&-i\partial_yU(x,y)\text{.} \end{aligned}$$ This proved (3)-(6).

THEOREM If $f$ is holomorphic at $z=x+iy\in\Omega$, then $F$ is differentiable at (x,y)$.

THEOREM If $F$ is differentiable at $(x,y)$, $U$ and $V$ satisfy the Cauchy-Riemann equation, then $f$ is holomorphic at $x+iy$.

Answer 1

There's are conflicting definitions (as seen from the comments you received) regarding the phrase 'holomorphic at a point':

• Some people say $f$ is holomorphic at a point $z_0$ if there is an open neighborhood $\Omega_0$ of $z_0$ such that at each point $z\in \Omega_0$, $\lim\limits_{h\to 0}\frac{f(z+h)-f(z)}{h}$ exists (briefly, $f$ is complex-differentiable at each point of $\Omega_0$).
• Some people say $f$ is holomorphic at a point $z_0$ if the limit $\lim\limits_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h}$ exists (briefly, $f$ is complex-differentiable at $z_0$).

I learned and use the second, and find the first to be confusing language, but apparently it is standard. It is now a standard exercise (essentially as you've tried to show) to prove the following theorem:

With your notation, $f$ is complex differentiable at $z_0=x_0+iy_0$ if and only if $F$ is Frechet-differentiable at $(x_0,y_0)$ and satisfies the Cauchy-Riemann equations \begin{align} \frac{\partial U}{\partial x}(x_0,y_0)=\frac{\partial V}{\partial y}(y_0,y_0),\quad\text{and}\quad \frac{\partial U}{\partial y}(x_0,y_0)=-\frac{\partial V}{\partial x}(x_0,y_0). \end{align}

While your proof has the right idea, I dont like it in the sense that you use symbols before proving their existence. For instance, when proving "$f$ is differentiable at $z \implies$ $F$ is differentiable at $X=(x,y)$", you write down the symbol $DF(X)(H)$ even before proving that the linear transformation $DF(X)$ exists.