Jacobian
linear algebra
learn more here, and here
\[ \begin{bmatrix} a \\
b\end{bmatrix} = a + bi \]
\[ \hat{I} = \begin{bmatrix} 1 \\
0\end{bmatrix} = 1 \]
\[ \hat{j} = \begin{bmatrix} 0 \\
1\end{bmatrix} = i \]
\[ \begin{bmatrix}
a \quad b \\
c \quad d \\ \end{bmatrix}
\begin{bmatrix} e \\
f \end{bmatrix} = \begin{bmatrix} ae + bf \\
ce + df\end{bmatrix}\]
\[ A = \begin{bmatrix} ? \quad ? \\
? \quad ?\end{bmatrix} \]
\[ A \hat{I} = \begin{bmatrix} a \\
c\end{bmatrix} \]
\[ A \hat{j} = \begin{bmatrix} b \\
d\end{bmatrix} \]
\[ A = \begin{bmatrix} a \quad b \\
c \quad d\end{bmatrix} \]
\[ a + bi = \begin{bmatrix} ? \quad ? \\
? \quad ?\end{bmatrix} \]
\[ (a + bi) \hat{I} = (a + bi) 1 = a + bi = \begin{bmatrix} a \\
b\end{bmatrix} \]
\[ (a + bi) \hat{j} = (a + bi) i = -b + ai = \begin{bmatrix} -b \\
a\end{bmatrix} \]
\[ a + bi = \begin{bmatrix} a & -b \\
b & a\end{bmatrix} \]
complex functions
\[ z = x + yi =\begin{bmatrix} x \\
y\end{bmatrix} \]
\[ f(z) = u + vi = \begin{bmatrix} u \\
v\end{bmatrix} \]
\[ dz = dx + dyi = \begin{bmatrix} dx \\
dy\end{bmatrix} \]
\[ df = du + dvi = \begin{bmatrix} du \\
dv\end{bmatrix} \]
\[ f\prime (z) = \frac{df}{dz} \]
\[ df = f\prime (z) dz \]
\[ \begin{bmatrix} du \\
dv\end{bmatrix} = f\prime (z) \begin{bmatrix} dx \\
dy\end{bmatrix} \]
\[ df = f(z + dz) - f(z) \]
\[ \frac{\partial}{\partial x} f(x, t) = : \frac{f(x + dx, t) - f(x, t)}{dx} \]
\[ \partial f(x, t)_x = : f(x + dx, t) - f(x, t) \]
\[ f(z + dx \hat{I}) = f(z) + \begin{bmatrix} \partial u_x \\
\partial v_x\end{bmatrix} \]
\[ f(z + dy \hat{j}) = f(z) + \begin{bmatrix} \partial u_y \\
\partial v_y\end{bmatrix} \]
\[ f(z + dx + dyi) = f(z + dz) = f(z) + \begin{bmatrix} \partial u_x \\
\partial v_x\end{bmatrix} + \begin{bmatrix} \partial u_y \\
\partial v_y\end{bmatrix} \]
\[ f(z + dz) - f(z) = df = \begin{bmatrix} \partial u_x \\
\partial v_x\end{bmatrix} + \begin{bmatrix} \partial u_y \\
\partial v_y\end{bmatrix} = \partial u_x \hat{I} + \partial v_x \hat{j} + \partial u_y \hat{I} + \partial v_y \hat{j} = \frac{\partial u}{\partial x} dx \hat{I} + \frac{\partial v}{\partial x} dx \hat{j} + \frac{\partial u}{\partial y} dy \hat{I} + \frac{\partial v}{\partial y} dy \hat{j} = \begin{bmatrix} \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy \\
\frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy\end{bmatrix} = \begin{bmatrix} \frac{\partial u}{\partial x} \quad \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} \quad \frac{\partial v}{\partial y}\end{bmatrix} \begin{bmatrix} dx \\
dy\end{bmatrix} \]
\[ \begin{bmatrix} du \\
dv\end{bmatrix} = \begin{bmatrix} \frac{\partial u}{\partial x} \quad \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} \quad \frac{\partial v}{\partial y}\end{bmatrix} \begin{bmatrix} dx \\
dy\end{bmatrix} \]
\[ f\prime (z) = \begin{bmatrix} \frac{\partial u}{\partial x} \quad \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} \quad \frac{\partial v}{\partial y}\end{bmatrix} \]
clarification and cauchy-riemann equations
technically a term like \(\partial u_y\) seems like the change in \(u\) when \(z\) adds \(dy\), but actually it is the change in \(u\) when \(z\) moves up by \(dyi\), but this is just \(x + yi + dyi\), so \(x + (y + dy)i\) which means that it is a change in \(u\) when \(y\) increases
in short, \(\partial u_y\) is not the change in \(u\) when \(z\) adds \(dy\), but the change of \(u\) when you increase \(y\)
on another note, the equation or jacobian matrix is a matrix, but as you know, every complex number has a corresponding matrix but not every matrix has a corresponding complex number, so to find out if the jacobian matrix is a complex number or just a matrix, or said another way, if complex function \(f(z)\) has a derivative, we need the cauchy-riemann equations, lets go derive them!
so, if the jacobian matrix is a complex number \(a + bi\) which I have been saving for something like this, than the corresponding matrix is:
\[ \begin{bmatrix} a & -b \\
b & a\end{bmatrix} \]
so that means that
\[ \begin{bmatrix} \frac{\partial u}{\partial x} \quad \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} \quad \frac{\partial v}{\partial y}\end{bmatrix} = \begin{bmatrix} a & -b \\
b & a\end{bmatrix} \]
Well, what makes a matrix of that form?
for one, the top left equals the bottom right equals the real part
and for another, the top right equals the negative of the bottom left equals the negative of the imaginary part
in conclusion...
\[ f\prime (z) = \begin{bmatrix} \frac{\partial u}{\partial x} \quad \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} \quad \frac{\partial v}{\partial y}\end{bmatrix} \]
and to test if this is a matrix or a complex number
\[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \]
\[ -\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} \]
examples
\[ f(z) = z^2 \]
\[ f(x +yi) = (x^2 - y^2) + (2xy)i \]
\[ u = x^2 - y^2 \]
\[ v = 2xy \]
\[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (x^2 - y^2) = 2x \]
\[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (x^2 - y^2) = -2y \]
\[ \frac{\partial v}{\partial x} = \frac{\partial}{\partial x} (2xy) = 2y \]
\[ \frac{\partial v}{\partial y} = \frac{\partial}{\partial y} (2xy) = 2x \]
\[ 2x = 2x \]
\[ 2y = 2y \]
\[ f\prime (z) = 2x + 2yi = 2z \]
another one!
\[ f(z) = \text{ccong} (z) \]
\[ f(x + yi) = x - yi \]
\[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} x = 1 \]
\[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} x = 0 \]
\[ \frac{\partial v}{\partial x} = \frac{\partial}{\partial x} (-y) = 0 \]
\[ \frac{\partial v}{\partial y} = \frac{\partial}{\partial y} (-y) = -1 \]
\[ 1 \ne -1 \]
\[ 0 = 0 \]
\[ f\prime (z) = \begin{bmatrix} 1 & 0 \\
0 & -1 \end{bmatrix} \]
In conclusion, \(z^2\) has a derivative, and \(\text{ccong} (z)\) does not.