Minkowski's Linear Forms Theorem is often stated about linear forms with real coefficients. However, in Narkiewicz's *Elementary and Analytic Theory of Algebraic Numbers*, the following generalization of Minkowski's Linear Forms Theorem is stated (the theorem is actually stated for a general lattice in real space, but I'm only interested in the case where the lattice is $\mathbb{Z}^N$):


Let $U = \{ L\_j(x\_1, \ldots, x\_N) = \sum\_{i=1}^N a\_{ij} x\_i \mid 1 \leq j \leq N \}$ be a system of $N$ linear forms with complex coefficients satisfying the following condition: if $L\_j \in U$, then the conjugate form, $\overline{L\_j} \in U$. Let $M = [a\_{jk}]\_{1\leq j,k \leq N}$. Let $\epsilon\_1, \ldots, \epsilon\_N \in \mathbb{R}^+$ such that $\prod\_{i=1}^N \epsilon\_i \geq | \det M |$ and $L\_i = \overline{L\_j}$ implies $\epsilon\_i = \epsilon\_j$. Then there exists a nonzero point $(x\_1, \ldots, x\_n) \in \mathbb{Z}^N$ such that $|L\_j(x\_1,\ldots,x\_N)| \leq \epsilon\_j$ for each $1 \leq j \leq N$ with strict inequality holding for all but one $j$.


However, instead of providing a proof, Narkiewicz refers to two books by Cassels, both of which only prove the case where the coefficients of the forms are real. The proof of the real version---which uses Minkowski's Convex Body Theorem on $M^{-1} ([-1,1]^N)$---does not easily generalize to the complex version. Does anyone have an idea of how the proof of the generalization might go? I've tried translating the problem to real-space (using the fact that $N$-dimensional complex space is isomorphic to $2N$-dimensional real space), but the set that I want to apply Minkowski's Convex Body Theorem to does not have large enough volume.


EDIT: A proof may appear in the 39th "chapter" in Minkowski's *Geometrie der Zahlen* (on page 113). However, I cannot read German. Does anyone know of an English translation of the work?



Minkowski's Linear Forms Theorem With Complex Coefficients
Credit where it's due to @so-calledfriendDon who found a source before I figured this out. However, I figure I'll post the proof because Google Books can be a finicky thing.


First, note that we can assume that all $\epsilon\_j = 1$. If not, replace $L\_j$ by $\frac{L\_j}{\epsilon\_j}$ and note that all of the hypotheses pertaining to forms and their conjugates still hold.


Suppose there are $r$ forms which have real coefficients and $s$ pairs of complex conjugate forms so that $r + 2s = N$. Without loss of generality, $L\_1, \ldots, L\_r$ have real coefficients and $\bar{L}\_{r+j} = L\_{r+s+j}$ for each $1 \leq j \leq s$. Then we can write $$L\_{r+j} = \frac{R\_j + i I\_j}{\sqrt{2}} \\ L\_{r+s+j} = \frac{R\_j - i I\_j}{\sqrt{2}}$$ for linear forms with real coefficients $R\_j$ and $I\_j$ and $1 \leq j \leq s$. Observe that if we look at the set of forms $L\_j$ for $1 \leq j \leq r$ and $R\_k, I\_k$ for $1 \leq k \leq s$, we have $r + 2s = N$ linear forms in $N$ variables, where all the linear forms have real coefficients. In particular, using the fact that $\bar{a}\_{r+j,k} = a\_{r+s+j,k}$ for all $1 \leq j \leq s$ and $1 \leq k \leq N$ (this is simply restating that $\bar{L}\_{r+j} = L\_{r+s+j}$), we can explicitly describe $R\_j$ and $I\_j$: $$ R\_j(x\_1, \ldots, x\_N) = \sqrt{2} \text{Re}[a\_{r+j,1}]x\_1 + \cdots + \sqrt{2} \text{Re}[a\_{r+j,N}] x\_N \\ I\_j(x\_1, \ldots, x\_N) = \sqrt{2} \text{Im}[a\_{r+j,1}]x\_1 + \cdots + \sqrt{2} \text{Im}[a\_{r+j,N}] x\_N$$


Let $A$ be the matrix whose entries are the coefficients of the forms $L\_j$ for $1 \leq j \leq r$ and $R\_k, I\_k$ for $1 \leq k \leq s$. I won't go into the gory details, but using elementary properties of determinants, you can show that $|\det(A)| = |\det(M)| \leq 1$. Now you can apply the version of Minkowski's Linear Forms Theorem where the coefficients of the linear forms are real (proofs of this theorem are plentiful) to find a nonzero point $\mathbf{x} = (x\_1, \ldots, x\_N) \in \mathbb{Z}^N$ where $|L\_1(\mathbf{x})| \leq 1$, $|L\_j(\mathbf{x})| < 1$ for $2 \leq j \leq r$ and $|R\_k(\mathbf{x})|, |I\_k(\mathbf{x})| < 1$ for each $1 \leq k \leq s$.


But then we have, for each $1 \leq j \leq s$ $$|L\_{r+j}(\mathbf{x})|^2 = \left|\frac{R\_j(\mathbf{x}) + i I\_j(\mathbf{x})}{\sqrt{2}} \right|^2 = \frac{R\_j(\mathbf{x})^2 + I\_j(\mathbf{x})^2}{2} < 1 \\ |L\_{r+s+j}(\mathbf{x})|^2 = \left|\frac{R\_j(\mathbf{x}) - i I\_j(\mathbf{x})}{\sqrt{2}} \right|^2 = \frac{R\_j(\mathbf{x})^2 + I\_j(\mathbf{x})^2}{2} < 1$$.


This shows that $|L\_1(\mathbf{x})| \leq 1$ and for each $2 \leq j \leq N$, we have $|L\_j(\mathbf{x})| < 1$, which is exactly what we wanted.


