# $J=\lambda M$ where $M \in SO(2)$ and $\lambda \in \mathbb{R}$ instead of Cauchy-Riemann

In "visual complex analysis" (Tristan Needham), the author explains that an application $f : \mathbb{C} \mapsto \mathbb{C}$ is complex differentiable on some domain if and only if it is everywhere locally an "amplitwist" [rotation+scaling] on this domain. Then he proceeds to explain the Cauchy-Riemann equations. I found Cauchy-Riemann to be a very poor capture of the above intuition. Indeed, lets start from the beginning:

If $f : \mathbb{R}^2 \mapsto \mathbb{R}^2$ is everywhere differentiable on some open domain $U$ then it can be locally approximated by its jacobian matrix:

$$f(x) = f(x_0) + J_f(x_0) ( x - x_0 ) + o(\left\| x-x_0 \right\|)$$

Hence, when asking for complex differentiability:

$$f'(z_0) = \lim_{z \rightarrow z_0} \frac{f(z) - f(z_0) }{z-z_0}$$

$$f(z) = f(z_0) + f'(z_0) ( z - z_0 ) + o(\left\| z-z_0 \right\|)$$

So the question is in what case can we identify the jacobian with a $J_f(x_0)$ with a complex number $f'(z_0)$ ?

The action of the multiplication by a complex $re^{i\theta}$ number can be represented by an "amplitwist" hence, the jacobian matrix must be of the form: $$r \left( \begin{matrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \\ \end{matrix} \right)$$

Indeed, for being well defined $f'(z_0)$ must not depend on the way we approached $z_0$, for example:

$$f(z) = \overline{z}$$

has a well defined jacobian matrix

$$\left( \begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix} \right)$$

but depending on direction the local action of the matrix cannot be identified by a single complex number:

$$f'(z_0) = \lim_{h \rightarrow 0} \frac{f(z_0 + h) - f(z_0) }{h} = \frac{\overline{h}}{h} = \frac{re^{-i\theta}}{re^{i\theta} } = e^{-2i\theta}$$

Hence depending on the way you approach $z_0$ the local action of the jacobian is identified with some element of the unit circle.

Now, the Cauchy-Riemann equations are always established the same way, by considering that two directional derivatives ( approaching either of the Cartesian axes ) must lead to the same complex number:

$$f'(z_0) = \lim_{x \rightarrow 0} \frac{f(z_0 + ix) - f(z_0) }{ix} = \lim_{x \rightarrow 0} \frac{f(z_0 + x) - f(z_0) }{x}, x \in \mathbb{R} \tag{1}$$

this leads to the Cauchy-Riemann equations and force the jacobian matrix to be:

$$\left( \begin{matrix} a & -b \\ b & a \\ \end{matrix} \right)$$

but does not enforce any relationship between $a$ and $b$, this matrix only captures the fact that the images of $\vec{x}$ and $\vec{y}$ will be orthogonal with direction preserved. Which is what we enforced by (1) but it does not force every angle to be preserved (conformality). It is easy to build a matrix

$$r \left( \begin{matrix} cos(\theta) & -sin(\phi) \\ sin(\phi) & cos(\theta) \\ \end{matrix} \right)$$

so that Cauchy-Riemann is respected but the application is not conformal. Then using $Im(z)$ and $Re(z)$ one can make the complex function matching the above jacobian.

I beleive the three following are equivalent:

1. $f$ is conformal
2. $f$ has a jacobian matrix $J=\lambda M$ where $M \in SO(2)$ and $\lambda \in \mathbb{R}$
3. $f$ respects the Cauchy-Riemann equations and every of the four entries $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$, $\frac{\partial v}{\partial x}$, $\frac{\partial v}{\partial y}$ are continuous.

QUESTIONS

I'm not sure how to prove the equivalence between 2 and 3:

• why would continuity and and angle preservation of two vectors imply angle preservation in general ?
• the other way around is geometrically more intuitive, as the jacobian is itself the result of a limit, $\lambda$ and $\theta$ of "nearby" jacobians must tend to those of jacobian of $z_0$ but I'm not sure how to write it.

How to prove above equivalence, and if it is correct, then why using Cauchy-Riemann at all, as it requires the same computations and provides less information than the jacobian characterization ?

If $f'(z_o)$ exists then it follows that the Jacobian can be written in terms of complex multiplication. In particular, $J_fh= f'(z_o)h$ where on the lhs I make use of matrix column multiplication whereas on the rhs I make use of complex multiplication. In short, I have exploited the isomorphism $(a,b) = a+ib$. If $f'(z_o)=a+ib$ then $$J_fh= f'(z_o)h = (a+ib)(h_1+ih_2) = ah_1-bh_2+i(bh_1+ah_2) = \left[\begin{array}{cc} a & -b \\ b & a \end{array} \right]\left[ \begin{array}{c} h_1 \\ h_2 \end{array}\right]$$ And it is simple to show such a matrix is a dilation and rotation: $$\left[\begin{array}{cc} a & -b \\ b & a \end{array} \right] = (a^2+b^2)\left[\begin{array}{cc} \frac{a}{a^2+b^2} & \frac{-b}{a^2+b^2} \\ \frac{b}{a^2+b^2} & \frac{a}{a^2+b^2} \end{array} \right]$$ Of course, this algebraic slight of hand fails when $a^2+b^2=0$. But, that is also the case in which $f'(z_o)=0$ and $f$ is not generally conformal at such a point.
Conversely, if $f=(u,v)$ and $J_f = \left[\begin{array}{cc} a & -b \\ b & a \end{array} \right]$, which implicits $u_x=v_y$ and $v_x=-u_y$, and $f=(u,v)$ is continuously differentiable then $f$ is differentiable as a mapping on $\mathbb{R}^2$ and through the isomorphism to $\mathbb{C}$ given by $1 = (1,0)$ and $i=(0,1)$ we can show $f$ is complex differentiable.
• Indeed, I did not realize that $\lambda$ allow $a$ and $b$ to move freely... $a=\lambda cos(\theta)$ and $b=\lambda sin(\theta)$ always has a solution in simply adding the square on both sides!!!! My bad! Dec 22, 2015 at 16:10