Distinct exponential characters are linearly independent on a nonempty open set, and the real-part grouping rule

Reduction to the scalar case. Both assertions are componentwise in ${\mathbb{C}}^q$: $F(z)=0$ holds if and only if every component $F_k(z)={\sum }_j(c_j{)}_k{e}^{i{\lambda }_j\cdot z}$ vanishes, and $\mathrm{Re}F(z)=0$ if and only if $\mathrm{Re}{F}_k(z)=0$ for every $k$, while the vector $C_{\nu }$ of clause (b) vanishes if and only if each of its components ${\sum }_{{\lambda }_j=\nu }(c_j{)}_k+{\sum }_{{\lambda }_j=-\bar{\nu }}\overline{(c_j{)}_k}$ vanishes. Applying the scalar assertions $q=1$ to each component function therefore yields the general assertions, so we may and do assume $c_1,\dots ,c_m\in \mathbb{C}$ in the rest of the proof.

Conjugation of the exponential. For every $w\in \mathbb{C}$, $\overline{\exp w}=\exp \bar{w}$. By [F1] the complex numbers $\exp w$ and $\exp \bar{w}$ are the sums of the everywhere convergent series ${\sum }_n{w}^n\mathrm{/}n!$ and ${\sum }_n{\bar{w}}^n\mathrm{/}n!$, with partial sums $S_N(w)={\sum }_{n\le N}{w}^n\mathrm{/}n!$ and $S_N(\bar{w})={\sum }_{n\le N}{\bar{w}}^n\mathrm{/}n!$; by [F2] conjugation fixes each real number $n!$ and satisfies $\overline{zw}=\bar{z}\bar{w}$ and $\overline{z+w}=\bar{z}+\bar{w}$, so $\overline{S_N(w)}=S_N(\bar{w})$ for every $N$. Also by [F2], $\bar{u}-\bar{v}=\overline{u-v}$ and, by the coordinate formula, $\mathrm{\mid }\bar{u}\mathrm{\mid }=\sqrt{(\mathrm{Re}u{)}^2+(-\mathrm{Im}u{)}^2}=\mathrm{\mid }u\mathrm{\mid }$, whence $\mathrm{\mid }\bar{u}-\bar{v}\mathrm{\mid }=\mathrm{\mid }u-v\mathrm{\mid }$. Writing $d=\overline{\exp w}-\exp \bar{w}$, for every $N$ we have $d=(\overline{\exp w}-S_N(\bar{w}))+(S_N(\bar{w})-\exp \bar{w})=(\overline{\exp w}-\overline{S_N(w)})+(S_N(\bar{w})-\exp \bar{w})$, so $\mathrm{\mid }d\mathrm{\mid }\le \mathrm{\mid }\exp w-S_N(w)\mathrm{\mid }+\mathrm{\mid }{S}_N(\bar{w})-\exp \bar{w}\mathrm{\mid }$; given $\epsilon >0$, the definition of convergence of the two series supplies $N$ with both summands below $\epsilon \mathrm{/}2$, so $\mathrm{\mid }d\mathrm{\mid }<\epsilon$, and since $\epsilon >0$ was arbitrary $\mathrm{\mid }d\mathrm{\mid }=0$, which forces $d=0$ by the definiteness of the modulus in [F2].

Characters and the annihilating operator. For $\mu \in \mathbb{C}$ let ${\phi }_{\mu }(t):=\exp (i\mu t)$ for $t\in \mathbb{R}$, the one-variable restriction of the complex exponential along the real-linear map $t\mapsto i\mu t$. By [F5], the chain rule applied to that map and the entire exponential of [F1] gives ${\phi }_{\mu }^{\mathrm{\prime }}(t)=i\mu \exp (i\mu t)=i\mu {\phi }_{\mu }(t)$, and the sum, scalar-multiple and product rules of [F5] give, for any finite combination ${\sum }_{\mu }{a}_{\mu }{\phi }_{\mu }$ with complex coefficients, derivative ${\sum }_{\mu }{a}_{\mu }i\mu {\phi }_{\mu }$, hence $(d\mathrm{/}dt-i\mu ){\phi }_{\mu }=0$ and $(d\mathrm{/}dt-i\mu ){\phi }_{\nu }=i(\nu -\mu ){\phi }_{\nu }$ for all $\nu \in \mathbb{C}$. Consequently, for pairwise distinct ${\mu }_1,\dots ,{\mu }_m$ and a fixed $j_0$, the composed operator $L_{j_0}:={\prod }_{j\mathrm{\ne }{j}_0}(d\mathrm{/}dt-i{\mu }_j)$ is complex-linear on finite combinations and acts on ${\phi }_{{\mu }_k}$ as multiplication by $0$ if $k\mathrm{\ne }{j}_0$ and by the constant ${\prod }_{j\mathrm{\ne }{j}_0}i({\mu }_{j_0}-{\mu }_j)$ if $k=j_0$; that constant is nonzero because $i\mathrm{\ne }0$ and ${\mu }_{j_0}-{\mu }_j\mathrm{\ne }0$ for $j\mathrm{\ne }{j}_0$ by pairwise distinctness, and a product of nonzero field elements is nonzero by [F3].

Induction step: reduction to a product box. Let $n\ge 2$ and assume assertion (a) established in dimension $n-1$ for every number of frequencies and every nonempty open subset of ${\mathbb{R}}^{n-1}$. Let $U\subseteq {\mathbb{R}}^n$ be nonempty and open, fix $z^{\ast }\in U$ and, by openness, $\epsilon >0$ with the open product box $B:={\prod }_{k=1}^n(z_k^{\ast }-\epsilon ,z_k^{\ast }+\epsilon )\subseteq U$; put $B^{\mathrm{\prime }}:={\prod }_{k=2}^n(z_k^{\ast }-\epsilon ,z_k^{\ast }+\epsilon )\subseteq {\mathbb{R}}^{n-1}$. Write $z=(z_1,z^{\mathrm{\prime }})$ with $z^{\mathrm{\prime }}\in {\mathbb{R}}^{n-1}$ and ${\lambda }_j=({\lambda }_{j1},{\lambda }_j^{\mathrm{\prime }})$. For each value $\mu$ in the finite set $\{{\lambda }_{11},\dots ,{\lambda }_{m1}\}$ put $I_{\mu }:=\{j:{\lambda }_{j1}=\mu \}$ and $P_{\mu }(z^{\mathrm{\prime }}):={\sum }_{j\in I_{\mu }}{c}_j{e}^{i{\lambda }_j^{\mathrm{\prime }}\cdot z^{\mathrm{\prime }}}$; then, by the splitting identity of [F4], the exponential addition law of [F1] and the distributivity of the finite sum, $F(z_1,z^{\mathrm{\prime }})={\sum }_{\mu }{e}^{i\mu z_1}{P}_{\mu }(z^{\mathrm{\prime }})$ for every $(z_1,z^{\mathrm{\prime }})\in B$, the outer sum running over the pairwise distinct values $\mu$.

Base case, dimension one. Let $U\subseteq \mathbb{R}$ be nonempty and open, so that it contains a nonempty open interval $J$; suppose $F={\sum }_{j=1}^m{c}_j{\phi }_{{\lambda }_j}\equiv 0$ on $U$. Fix $j_0$; then $L_{j_0}F\equiv 0$ on $J$, and by step 1.3 and the linearity of $L_{j_0}$ we have $L_{j_0}F=c_{j_0}({\prod }_{j\mathrm{\ne }{j}_0}i({\lambda }_{j_0}-{\lambda }_j)){\phi }_{{\lambda }_{j_0}}$ on $J$. Choose $t_0\in J$. Evaluating there gives $c_{j_0}({\prod }_{j\mathrm{\ne }{j}_0}i({\lambda }_{j_0}-{\lambda }_j))\exp (i{\lambda }_{j_0}{t}_0)=0$; cancelling the nonzero exponential by [F1] gives $c_{j_0}{\prod }_{j\mathrm{\ne }{j}_0}i({\lambda }_{j_0}-{\lambda }_j)=0$; the second factor is a product of nonzero complex numbers, hence nonzero by [F3], so $c_{j_0}=0$ by the field property in [F3]. As $j_0$ was arbitrary, $c_1=\cdots =c_m=0$; this proves (a) for $n=1$.

The conjugate sum. Suppose now that $\mathrm{Re}F\equiv 0$ on $U$, and define $\tilde{F}(z):={\sum }_{j=1}^m\overline{c_j}{e}^{-i\overline{{\lambda }_j}\cdot z}$ on $U$. For real $z$ and every $j$, step 1.2 applied to $w=i{\lambda }_j\cdot z$ gives $\overline{e^{i{\lambda }_j\cdot z}}=\exp \left(\overline{i{\lambda }_j\cdot z}\right)$, and by [F2] and the reality of $z$ we have $\overline{i{\lambda }_j\cdot z}=\bar{i}\overline{{\lambda }_j\cdot z}=(-i)(\overline{{\lambda }_j}\cdot z)=-i\overline{{\lambda }_j}\cdot z$; therefore $\overline{F(z)}={\sum }_j\overline{c_j}{e}^{-i\overline{{\lambda }_j}\cdot z}=\tilde{F}(z)$ on $U$ by [F2]. Since $w+\bar{w}=2\mathrm{Re}w$ by [F2], the hypothesis $2\mathrm{Re}F\equiv 0$ reads $F+\tilde{F}\equiv 0$ on $U$.

One-dimensional slices. Suppose $F\equiv 0$ on $U$, fix $z^{\mathrm{\prime }}\in B^{\mathrm{\prime }}$ and define $G(t):={\sum }_{\mu }{e}^{i\mu t}{P}_{\mu }(z^{\mathrm{\prime }})$ for $t$ in the interval $J:=(z_1^{\ast }-\epsilon ,z_1^{\ast }+\epsilon )$. By step 1.4, $G(t)=F(t,z^{\mathrm{\prime }})=0$ for every $t\in J$, and $G$ is a finite combination of the pairwise distinct characters ${\phi }_{\mu }$ with complex coefficients $P_{\mu }(z^{\mathrm{\prime }})$; the base case of step 2.1 applied on the nonempty open interval $J$ therefore gives $P_{\mu }(z^{\mathrm{\prime }})=0$ for every $\mu$. As $z^{\mathrm{\prime }}\in B^{\mathrm{\prime }}$ was arbitrary, each $P_{\mu }$ vanishes identically on the nonempty open set $B^{\mathrm{\prime }}$.

Induction hypothesis on each group. Fix a value $\mu$. The vectors ${\lambda }_j^{\mathrm{\prime }}$, $j\in I_{\mu }$, are pairwise distinct: if ${\lambda }_j^{\mathrm{\prime }}={\lambda }_k^{\mathrm{\prime }}$ and ${\lambda }_{j1}={\lambda }_{k1}=\mu$, then ${\lambda }_j={\lambda }_k$, hence $j=k$ by pairwise distinctness of the ${\lambda }_j$. The function $P_{\mu }={\sum }_{j\in I_{\mu }}{c}_j{e}^{i{\lambda }_j^{\mathrm{\prime }}\cdot z^{\mathrm{\prime }}}$ vanishes identically on the nonempty open set $B^{\mathrm{\prime }}\subseteq {\mathbb{R}}^{n-1}$, so assertion (a) in dimension $n-1$ gives $c_j=0$ for every $j\in I_{\mu }$. Every index lies in $I_{{\lambda }_{j1}}$, so $c_1=\cdots =c_m=0$; this completes the induction and proves (a) in every dimension $n\ge 1$.

Frequency grouping and application of part (a). Let $\mathrm{\Phi }\subseteq {\mathbb{C}}^n$ be the finite set of distinct values among ${\lambda }_1,\dots ,{\lambda }_m,-\overline{{\lambda }_1},\dots ,-\overline{{\lambda }_m}$, and for $\nu \in \mathrm{\Phi }$ put $C_{\nu }:={\sum }_{j:{\lambda }_j=\nu }{c}_j+{\sum }_{j:-\overline{{\lambda }_j}=\nu }\overline{c_j}={\sum }_{{\lambda }_j=\nu }{c}_j+{\sum }_{{\lambda }_j=-\bar{\nu }}\overline{c_j}$. Regrouping the finite sum of step 2.2 over the distinct frequencies and using the bilinearity of the extended inner product in [F4] gives $F+\tilde{F}={\sum }_{\nu \in \mathrm{\Phi }}{C}_{\nu }{e}^{i\nu \cdot z}$ on $U$; the frequencies $\nu \in \mathrm{\Phi }$ are pairwise distinct by construction, so assertion (a) as proved in steps 2.1 and 4.1 applied to this finite combination gives $C_{\nu }=0$ for every $\nu \in \mathrm{\Phi }$. For $\nu \mathrm{\notin }\mathrm{\Phi }$ both defining sums are empty, so $C_{\nu }=0$ trivially; hence $C_{\nu }=0$ for every $\nu \in {\mathbb{C}}^n$.

Consequences and scope. (i) If $\nu =-\bar{\nu }$ is self-paired, the condition $-\overline{{\lambda }_j}=\nu$ is equivalent to ${\lambda }_j=\nu$, so both sums of $C_{\nu }$ run over the same index set and $C_{\nu }={\sum }_{{\lambda }_j=\nu }(c_j+\overline{c_j})=2\mathrm{Re}{\sum }_{{\lambda }_j=\nu }{c}_j$ by [F2]; at $\nu =0$ this constrains only the real part, and the constraint is genuinely not stronger: for $m=1$, ${\lambda }_1=0$, $q=1$ and $c_1=i$ one has $F\equiv i$, so $\mathrm{Re}F\equiv 0$ and $C_0=c_1+\overline{c_1}=2\mathrm{Re}i=0$, a nonzero purely imaginary coefficient at frequency $0$. (ii) If ${\lambda }_j\mathrm{\ne }-\overline{{\lambda }_j}$ and ${\lambda }_j\mathrm{\notin }\{{\lambda }_k,-\overline{{\lambda }_k}:k\mathrm{\ne }j\}$, then $\{k:{\lambda }_k={\lambda }_j\}=\{j\}$ and $\{k:{\lambda }_k=-\overline{{\lambda }_j}\}=\mathrm{\varnothing }$ (a member $k=j$ would give ${\lambda }_j=-\overline{{\lambda }_j}$, and $k\mathrm{\ne }j$ is excluded by hypothesis), so $C_{{\lambda }_j}=c_j+0=c_j=0$ by step 5.1, which is the stated consequence. (iii) The scope (c) holds: steps 2.1, 1.4 and 4.1 used only pairwise distinctness of the ${\lambda }_j$ and the nonemptiness and openness of $U$, through the interval $J$ and the product box $B$; the frequencies were never assumed nonzero, ordered or normalized; the proof is a finite construction using complex arithmetic, the series for the exponential and its derivative, and no choice principle. No assertion is made for sets of empty interior: the openness of $U$ is used when a nonempty open interval and a product box are extracted from it, and the statement (c) records that distinct characters can agree on such smaller sets.