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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?

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  • $\begingroup$ Multiply the matrix with a column vector, then identify the coordinates with real and complex parts, youll see what it means. $\endgroup$ – Rene Schipperus Jun 10 '14 at 3:04
  • $\begingroup$ Multiplying by the column vector for some arbitrary complex number $x+iy$ would give me a $2\times1$ matrix, and it seems that the bottom row is just zeroes. Can I just discard it then to get $a$? $\endgroup$ – Olivia Jun 10 '14 at 3:08
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$$\begin{pmatrix} \frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\ -\frac{\partial u}{\partial y}&\frac{\partial u}{\partial x}\\ \end{pmatrix}\begin{pmatrix} x\\y\\ \end{pmatrix} =\begin{pmatrix} \frac{\partial u}{\partial x}x +\frac{\partial u}{\partial y}y\\ -\frac{\partial u}{\partial y}x+\frac{\partial u}{\partial x}y\\ \end{pmatrix}$$

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.

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  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.

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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 $$ U = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} $$

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