Complex function and Jacobian matrix 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?
 A: $$\begin{pmatrix}
\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\
-\frac{\partial u}{\partial y}&\frac{\partial u}{\partial x}\\
\end{pmatrix}\begin{pmatrix}
x\\y\\
\end{pmatrix}
=\begin{pmatrix}
\frac{\partial u}{\partial x}x +\frac{\partial u}{\partial y}y\\
-\frac{\partial u}{\partial y}x+\frac{\partial u}{\partial x}y\\
\end{pmatrix}$$
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.
A: This is an old post but maybe someone will still benefit from this: a complex number is isomorphic to a $2\times2$ matrix (preserving all structure of the field because the matrix product). One can obtain the matrix form this way 
$$ Z = x I + y i $$ where $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $ i = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $
So you get
$$ \begin{pmatrix} x & y \\ -y & x \end{pmatrix} $$
Basic complex algebra can be done this way and there is a polar form too
$$
Z = rU = r e^{i \theta}
$$
where
$$
U = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}
$$
A: *

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

*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.
