A rigorous proof of Cauchy-Riemann equations

Question

I faced several proofs of Cauchy-Riemann theorem :

Issue

Let $f \in \mathbb{C}^\mathbb{C} $ holomorphic.

$$ u : (x,y)\to \Re(f)(x+iy)$$ $$ v : (x,y)\to \Im(f)(x+iy)$$

Then:

$$ \begin{cases} \partial_1u=\partial_2v\\ \partial_2u=-\partial_1 v \end{cases} $$

But I always found a lack of rigour related to the following expression :

" $ f(z)=f(x+iy)=u(x,y)+iv(x,y) $ "

The issues comes from treating $f$ as a function of ${\mathbb{R}^2}^{\mathbb{R}^2}$ in nebulously way instead of treating it as a function of $ \mathbb{C}^\mathbb{C}$ . We also find proof like this. But it doesn't deal in terms of function in the limit but in decomposition. By doing like this we are mixing up the 2 kinds of functions described above.

Question

I thought we could define $$ \phi : (x,y)\to x+iy$$ Then $$ F:(x,y)\to f[\phi(x,y)]$$ So we have :

$$ F :(x,y) \to u(x,y)+iv(x,y)= \Re(f)(x+iy)+\Im(f)(x+iy)$$

How could we clearly show the theorem by properly introduce coherent functions ? I mean we can't say $ f: z \to f(x+iy)$ it makes non sense in terms of function.

Which functions could we introduce to have a natural proof involving limits of functions ? And without using the expression mixing $x,y$ even if it is understandable (but unprecise and muffled) in that way.

Thanks , I hope I'm precise enough.

Answer 1

Cauchy Riemann equations inherently involve real and imaginary parts

• in the codomain (that's $u$ and $v$)
• in the domain as well (since you have partial derivatives).

So there is absolutely no way to avoid having them in the proof.

Answer 2

There is a canonical way of identifying functions from $\mathbb{R}^2$ to $\mathbb{R^2}$ with functions from $\mathbb{C}$ to $\mathbb{C}$. In particular, if $f: \mathbb{R^2} \rightarrow \mathbb{R}^2$ is given by $(x, y) \rightarrow (u(x,y), v(x,y))$, you can view $f$ as a function from $\mathbb{C}$ to $\mathbb{C}$ by the map $a + bi \rightarrow u(a, b) + iv(a, b).$ It can be shown that this gives a one-to-one correspondence between functions from $\mathbb{R}^2$ to $\mathbb{R}^2$ and from $\mathbb{C}$ to $\mathbb{C}$, and that this correspondence is preserved under scaling and addition. Hence, even if these different types of functions are in play, as long as the identification is properly made nothing is being "mixed up." This identification is probably being done under the covers without mention in the proofs that you have referenced.

Answer 3

I think I can answer my question like this :

As usual define:

$$u:(x,y)\to \Re(f)(x+iy)$$ $$v:(x,y) \to \Im(f)(x+iy) $$

Then we will write $f$ without any $(x,y)$ , because we will write it in terms of function(s)

$$ f : z\to u(\Re(z),\Im(z)) +iv(\Re(z),\Im(z)) $$

In other words :

$$f=u(\Re({I_d}_\mathbb{C}),\Im({I_d}_\mathbb{C}))+iv(\Re({I_d}_\mathbb{C}),\Im({I_d}_\mathbb{C})) $$

So it is rigorous to talk about our usual proof etc.

Let $z$ in $\mathbb{C}$ .

$$ \forall h \in \mathbb{C}^*, \frac{f(z+h)-f(z)}{h}= \frac{[ u(\Re(z+h),\Im(z+h))- u(\Re(z),\Im(z))] +i[v(\Re(z+h),\Im(z+h))-v(\Re(z),\Im(z))]}{h} $$

By composition of holomorphic functions :

$$u(\Re({I_d}_\mathbb{C}),\Im({I_d}_\mathbb{C})) $$ $$v(\Re({I_d}_\mathbb{C}),\Im({I_d}_\mathbb{C}))$$

are holomorphic.

And we have $u$ and $v$ in $\mathcal{C} ^1(\mathbb{R}^2)$ because of the holomorphy of $f$ and the linearity of $\Re$.

Now I'm questionning on the meaning of the differentiation of ( for instance) $$u(\Re({I_d}_\mathbb{C}),\Im({I_d}_\mathbb{C})) $$

Thanks