Cauchy-Riemann Equations

The goal of this blog is for me to go through the derivation of Cauchy-Riemann Equations starting from derivability of a function on a complex variable. That said, this blog is aimed to improve my own understanding of the subject, and less aimed to convey the results to others, since I haven't mastered the concept.

Differentiability for $\mathbb{R}^2 \to \mathbb{R}^2$ and $\mathbb{C} \to \mathbb{C}$

Definition. Let $f \colon \Omega \subseteq \mathbb{R}^2 \to \mathbb{R}^2$ be a continuous function, where $\Omega$ is an open set. We say that $f$ is differentiable at $x_0 \in \mathbb{R}^2$ if there exists a linear map $\mathrm{d}f|_{x_0} \colon \mathbb{R}^2 \to \mathbb{R}^2$ such that $$f(x_0+h) = f(x_0) + \mathrm{d}f|_{x_0}(h) + \mathcal{o}(h)$$ as $\|h\| \to 0$ (for any norm $\|\cdot\|$).

To motivate the definition for complex differentials, we want to replace the linear map $\mathrm{d}f|_{x_0}$ with a complex number $\mathrm{d}f|_{x_0}$. This gives us the following definition.

Definition. Let $f \colon \Omega \subseteq \mathbb{C} \to \mathbb{C}$ be a continuous function, where $\Omega$ is an open set. We say that $f$ is differentiable at $z_0 \in \mathbb{C}$ (where we write $z_0 = x_0 + iy_0$ so that $x_0, y_0 \in \mathbb{R}$) if there exists a complex number $\mathrm{d}f|_{z_0} \in \mathbb{C}$ such that $$f(z_0+h) = f(z_0) + \mathrm{d}f|_{z_0}h + \mathcal{o}(h)$$ as $h \to 0$.

Well, then, suppose $f$ can be written as $u + iv$, so that $f(x+iy) = u(x, y) + iv(x, y)$ for all $x, y \in \mathbb{R}$ such that $x+iy \in \Omega$, and $u, v \colon \mathbb{R}^2 \to \mathbb{R}$. Then what we see is that from $$f(z_0 + h) = f(z_0) + \mathrm{d}f|_{z_0}h + \mathcal{o}(h),$$ this means $$\lim_{h \to 0, h \ne 0} \frac{f(z_0+h)-f(z_0)}{h} = \mathrm{d}f|_{z_0}$$ so when we restrict $h$ to be real, we have $$\lim_{h \to 0, h \ne 0, h \in \mathbb{R}} \frac{f(z_0+h)-f(z_0)}{h} = \mathrm{d}f|_{z_0}.$$ That is, $$\lim_{h \to 0, h \ne 0, h \in \mathbb{R}} \frac{u(x_0+h, y_0) + iv(x_0+h, y_0) - u(x_0, y_0) - iv(x_0, y_0)}{h} = \mathrm{d}f|_{z_0}.$$ If we take the real part of both sides, we have $$\lim_{h \to 0, h \ne 0, h \in \mathbb{R}} \frac{u(x_0+h, y_0) - u(x_0, y_0)}{h} = \Re(\mathrm{d}f|_{z_0}).$$ This is precisely $\partial_x u(x_0, y_0) = \Re(\mathrm{d}f|_{z_0})$. Once we do the same for the imaginary part, we have $$\partial_x v(x_0, y_0) = \Im(\mathrm{d}f|_{z_0}).$$ Now, instead of restricting $h$ to be real, let us restrict $h$ to be imaginary. We have $$\lim_{h \to 0, h \ne 0, h \in i\mathbb{R}} \frac{f(z_0+h)-f(z_0)}{h} = \mathrm{d}f|_{z_0}.$$ or equivalently, $$\lim_{h \to 0, h \ne 0, h \in \mathbb{R}} \frac{u(x_0, y_0+h) + iv(x_0, y_0+h) - u(x_0, y_0) - iv(x_0, y_0)}{ih} = \mathrm{d}f|_{z_0}.$$ Now, using the fact that $\frac{1}{i} = -i$, we have $$\lim_{h \to 0, h \ne 0, h \in \mathbb{R}} \frac{-iu(x_0, y_0+h) + v(x_0, y_0+h) + iu(x_0, y_0) - v(x_0, y_0)}{h} = \mathrm{d}f|_{z_0}.$$ Taking real part of this equation, we have $$\partial_y v(x_0, y_0) = \Re(\mathrm{d}f|_{z_0}),$$ and the imaginary part gives $$-\partial_y u(x_0, y_0) = \Im(\mathrm{d}f|_{z_0}).$$ Comparing these two the previous results, we have shown that $$\partial_x u(x_0, y_0) = \Re(\mathrm{d}f|_{z_0}) = \partial_y v(x_0, y_0),$$ and $$\partial_x v(x_0, y_0) = \Im(\mathrm{d}f|_{z_0}) = -\partial_y u(x_0, y_0).$$ Translating these back into the notion of $f$, we have that these two equations $$\begin{cases} \partial_x \Re(f(z_0)) = \partial_y \Im(f(z_0)) \\ \partial_x \Im(f(z_0)) = -\partial_y \Re(f(z_0)) \end{cases}$$ hold true whenever $f$ is complex-differentiable at $z_0 \in \Omega$.

These equations are, indeed, the Cauchy-Riemann equations.