Complex function and Jacobian matrix

Question

Given some complex-differentiable function $f:\mathbb{C}\rightarrow\mathbb{C}$ defined $f(x,y)=u(x,y)+iv(x,y)$, we know the Cauchy-Riemann equations hold, so: $$\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}\quad\textrm{and}\quad\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}$$

Then, we can write the Jacobian for the function: $$\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\-\dfrac{\partial u}{\partial y}&\dfrac{\partial u}{\partial x}\end{bmatrix}$$

At this point, my textbook claims that this matrix has the same effect on $\mathbb{C}$ as multiplication by the complex number $a=\dfrac{\partial u}{\partial x}-i\dfrac{\partial u}{\partial y}$ (therefore, $a$ is the derivative of $f$), but I'm having a hard time seeing why that's the case, and how this value of $a$ was reached in the first place. Any suggestions?

Answer 1

$$\begin{pmatrix} \frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\ -\frac{\partial u}{\partial y}&\frac{\partial u}{\partial x}\\ \end{pmatrix}\begin{pmatrix} x\\y\\ \end{pmatrix} =\begin{pmatrix} \frac{\partial u}{\partial x}x +\frac{\partial u}{\partial y}y\\ -\frac{\partial u}{\partial y}x+\frac{\partial u}{\partial x}y\\ \end{pmatrix}$$

On the other hand,

$$\left(\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}\right) (x+iy)= \frac{\partial u}{\partial x}x +\frac{\partial u}{\partial y}y+ \left(-\frac{\partial u}{\partial y}x+\frac{\partial u}{\partial x} y\right)i$$

Compare the terms and see they are the same.

Answer 2

1. The column vectors of the matrix are orthogonal to each other (check by taking the dot product).

2. The column vectors are the same size (check by comparing their norms).

Therefore, this is a rigid rotation + isotropic scaling matrix.

Multiplying by a complex number has the effect of rotating and scaling too.