I am reading from Degree Theory by N. Lloyd, and in one section he writes about the degree of a holomorphic map of several complex variables. I am unsure about one of the steps in a proof he gives.
The setup:
$\mathbb{C}^n$ is the vector space of complex $n$-tuples $z=(z_1,\ldots,z_n)$. $D$ is a bounded, open subset of $\mathbb{C}^n$. $\mathscr{H}(\bar{D})$ is the vector space of holomorphic mappings of $\bar{D}$ into $\mathbb{C}$; that is, $\phi=(\phi_1,\ldots,\phi_n)$ is in $\mathscr{H}(\bar{D})$ if each $\phi_j = u_j+iv_j$ is a holomorphic function of $n$ complex variables, so they satisfy the Cauchy-Riemann equations on some neighborhood of $\bar{D}$:
$$
\frac{\partial u_k}{\partial x_j} = \frac{\partial v_k}{\partial y_j},~~~
\frac{\partial u_k}{\partial y_j} = -\frac{\partial v_k}{\partial x_j},~~~
(k,j=1,\ldots,n).
$$
Now, let $p\in D$, $p\notin \partial D$. Consider a point $\zeta$ such that $\phi(\zeta)=p$. The matrix of $\phi'(\zeta)$ as a mapping of $\mathbb{C}^n$ into itself with respect to the standard basis is given by $A=(\alpha_{kj})$, where
$$
\alpha_{kj} = \frac{\partial u_k}{\partial z_j} = \frac{\partial u_k}{\partial x_j}-i\frac{\partial u_k}{\partial y_j}.
$$
So far so good.
Now, Lloyd claims that the determinant $\det \phi'(\zeta) \geq 0$. There isn't much explanation as to why, other than to consider the Jordan normal form. In fact, this doesn't even seem true for $n=1$. I know of the result that the square of this determinant is the determinant of the corresponding real Jacobian matrix, but I'm not sure if that's what the author is referring to.
I don't have much of an idea of what the the eigenvalues of a Jacobian of a holomorphic map of several complex variables should be. Would someone please explain to me why the determinant should be nonnegative? (I have minimal knowledge of several complex variables, and it is not a focus in this book.)
(More context, in case it helps: we are proving the following theorem:
Suppose $D$ is an open bounded subset of $\mathbb{C}^n$, and $\phi\in\mathscr{H}(\bar{D})$. If $p\notin\phi(\partial D)$, then $\deg(\phi,D,p)\geq 0$.
Here $\deg(\phi, D, p)$ is the degree of $\phi$ with respect to $D$ and $p$.)
Best Answer
Earlier in that section, the author writes: "$d(\phi,D,p)$ is defined by viewing $\phi$ as a map from a subset of $\mathbb R^{2n}$ into $\mathbb R^{2n}$." Although $A=\phi'(\zeta)$ is introduced as a complex $n\times n$ matrix, Lloyd implicitly switches to the real point of view, because the purpose of computing the Jacobian is to find the degree of $\phi$.
One does not have to go as far as the Jordan form; bringing $A$ (over complex numbers) to triangular form is good enough. When considered as a real matrix, this matrix has $2\times 2$ blocks along the main diagonal: a complex diagonal entry $\lambda=\alpha+i\beta$ is now represented by the block $$\begin{pmatrix} \alpha & -\beta \\ \beta & \alpha \end{pmatrix}$$ It is easy to see that the real determinant of $A$ is the product of determinants of these blocks, which leads to $\det_{\mathbb R}A=|\det_{\mathbb C}A|^2$.
They can be any $n$-tuple of complex numbers: consider the map $(z_k)\mapsto (w_k)$ defined by $w_k=\lambda_k z_k$. More generally, every complex matrix is the Jacobian matrix of the complex linear transformation it defines.