The real and complex Jacobian

 Lecture topic

Let $f:\C^n\to \C^n$ be a holomorphic function. We will show that the the real Jacobian is the square of the complex Jacobian. Write $f = (f_1,\dots,f_n)$ with $f_i = u_i+\img v_i$, where the $u_i$ are functions of the $z_j = x_j+\img y_j$. By the Cauchy-Riemann equations
$$\frac{\dy u_i}{\dy x_j} = \frac{\dy v_i}{\dy y_j} \hspace{1cm} \text{and} \hspace{1cm} \frac{\dy u_i}{\dy y_j} = -\frac{\dy v_i}{\dy x_j}$$
and expanding, we have that
$$\begin{align*} \frac{\dy f_i}{\dy z_j} & = \frac 12\left(\frac{\dy f_i}{\dy x_j} - \img \frac{\dy f_i}{\dy y_j}\right) \\ & = \frac12\left(\frac{\dy u_i}{\dy x_j} + \img \frac{\dy v_i}{\dy x_j} - \img\left(\frac{\dy u_i}{\dy y_j} + \img \frac{\dy v_i}{\dy y_j}\right)\right) \\ & = \frac12 \left(\frac{\dy u_i}{\dy x_j}+ \frac{\dy v_i}{\dy y_j} + \img \left(\frac{\dy v_i}{\dy x_j} - \frac{\dy u_i}{\dy y_j}\right)\right) \\ & = \frac{\dy u_i}{\dy x_j} + \img \frac{\dy v_i}{\dy x_j}. \end{align*}$$
The complex Jacobian of $f$ is $J_\C f$ (or its determinant), with entries
$$(J_\C f)_{i,j} = \frac{\dy f_i}{\dy z_j},$$
and the real Jacobian of $f$ is $J_\R f$ (or its determinant), with entries
$$\begin{align*} \begin{bmatrix} (J_\R f)_{2i-1,2j-1} & (J_\R f)_{2i-1,2j} \\ (J_\R f)_{2i,2j-1} & (J_\R f)_{2i,2j} \end{bmatrix} & = \begin{bmatrix} \displaystyle \frac{\dy u_i}{\dy x_j} & \displaystyle \frac{\dy u_i}{\dy y_j} \\ \displaystyle \frac{\dy v_i}{\dy x_j} & \displaystyle \frac{\dy v_i}{\dy y_j} \end{bmatrix} \\ & \tov{R_{2i-1} + \img R_{2i} \to R_{2i-1}} \begin{bmatrix} \displaystyle \frac{\dy f_i}{\dy z_j} & \displaystyle \img\frac{\dy f}{\dy z} \\ \displaystyle \frac{\dy v_i}{\dy x_j} & \displaystyle \frac{\dy v_i}{\dy y_j} \end{bmatrix} \\ & \tov{C_{2j} - \img C_{2j-i} \to C_{2j}} \begin{bmatrix} \displaystyle \frac{\dy f_i}{\dy z_j} & 0 \\ \displaystyle \frac{\dy v_i}{\dy x_j} & \displaystyle \overline{\frac{\dy f_i}{\dy z_j}} \end{bmatrix}, \end{align*}$$
where the row and column operations have been performed for all rows $2i$ and all columns $2j$. Moving all the odd-indexed columns to the left and all odd-indexed rows to the top, we get that
$$J_\R f \simeq \begin{bmatrix} A & 0 \\ * & B \end{bmatrix} \hspace{1cm} \text{with} \hspace{1cm} A_{i,j} = \frac{\dy f_i}{\dy z_j},\ \ \ B_{i,j} = \overline{\frac{\dy f_i}{\dy z_j}}.$$
Since the number of operations to switch the columns is the same as the number of operations to switch the rows, the sign of the determinant of $J_\R f$ will not change. That is,
$$\det(J_\R f) = \det(A)\det(B) = \det(J_\C f) \overline{\det( J_\C f)} = |\det(J_\C f)|^2.$$