4

I have learned that one can derive the Cauchy-Riemann equations by using the "typical" definition of differentiation (i.e. $f'(z)=\lim_{h \rightarrow 0} \frac{f(z+h)-f(z)}{h}$) by letting $h=h_1+i0$ and calculating the limit and then letting $h=0+ih_2$ and calculating the limit.

I know that besides the (what I called typical) definition of differentiability there also is a equivalent formulation for it:

A function $f:(a,b) \rightarrow \mathbb{R}$ is differentiable in the point $p \in (a,b)$ if and only if $f(p+h)-f(p)=f'(p)h+r(h)$ with $\lim_{h \rightarrow 0} \frac{r(h)}{h}=0$.

I would like to use this definition of differentiability to derive the Cauchy-Riemann equations.

Here is my attempt:

Let $f$ be a complex-valued function defined on some open set $U \subseteq \mathbb{C}$.

If $f$ complex differentiable at some point $z \in U$, then we can write

$f(z+h)-f(z)=f'(z)h+r(h)$ with $\lim_{h \rightarrow 0} \frac{r(h)}{h}=0$. (with $h=h_1+ih_2,r(h):\mathbb{C} \rightarrow \mathbb{C}$)

Let $f(x+iy)=u(x,y)+iv(x,y)$, $f'(z)=a+ib$ and $r(h)=r_1(h_1,h_2)+ir_2(h_1,h_2)$.

Now let $g(x,y):=(u(x,y),v(x,y))$, where we view g as a function $\mathbb{R}^2 \rightarrow \mathbb{R}^2$.

For a function $g:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ to be differentiable in the point $(x,y)$ is equivlanet that $g(x+h_1,y+h_2)-g(x,y)=A \cdot (h_1,h_2)+(r_1(h_1,h_2),r_2(h_1,h_2))$ with $\lim_{h \rightarrow 0} \frac{(r_1(h_1,h_2),r_2(h_1,h_2))}{\|(h_1,h_2)\|}=0$, where $A$ is the Jacobi Matrix of $g$.

Then I did the following calculations:

$f'(z)h=(a+ib)(h_1+ih_2)=h_1 a-h_2 b+i(h_1b+h_2a)$

$A \cdot (h_1,h_2)=(h_1 u_x+h_2 u_y,h_1 v_x+h_2v_y)$.

Doing now the comparison we get the CR equations $u_x=v_y$ and $u_y=-v_x$.

Now I am stuck on showing that if $\lim_{h \rightarrow 0} \frac{r(h)}{h}=0$ then so does $\lim_{h \rightarrow 0} \frac{(r_1(h_1,h_2),r_2(h_1,h_2))}{\|(h_1,h_2)\|}=0$ (in $\mathbb{R^2}$).

How can I argue here? (And also, is my approach this far even correct?)

RobPratt
  • 50,938
user1493768
  • 359
  • 4
  • 3
    You might also be interested in one more way to do it. Since $dz=dx+idy$ and $d\bar{z}=dx-idy$, $dx=(dz+d\bar{z})/2$ and $dy=(dz-d\bar{z})/(2i)$. Substitute these in $df=f_xdx+f_ydy$ to write $df$ as a linear combination of $dz$ and $\bar{d}z$, and impose the condition that the second coefficient is $0$. (It should be obvious this is equivalent to $f_y=if_x$.) – J.G. Dec 22 '24 at 16:02
  • Yet another way to do it: As $f$ is real-differentiable you can write $f(p+h)-f(p)=Df(p)h+o(\lVert h\rVert)$ with $Df(p):\mathbb R^2\to\mathbb R^2$ the real-linear map that is there derivative of $f$ as a function of two real variables into $\mathbb R^2$, with $p,h\in\mathbb R^2$. Then $f$ is complex-differentiable if and only if $Df(p)$ is also $\mathbb C$-linear, which is equivalent to $Df(p)$ commuting with the standard complex structure on $\mathbb R^2$, i.e. $\begin{pmatrix}0&-1\1&0\end{pmatrix}$, which one then easily checks is equivalent to the Cauchy-Riemann equations – Lorago Dec 22 '24 at 19:31

1 Answers1

5

Your approach is correct so far. Then, argue that $\|(h_1,h_2)\|=|h|$ and $\|(r_1(h_1,h_2),r_2(h_1,h_2))\|=|r(h)|$.

Anne Bauval
  • 49,005