Let $t:\mathbb C\to\mathbb R$ be the usual field trace map, then we have the following additive character for the locally compact group $(\mathbb C,+)$. $$\psi:\mathbb C\to S^1\cong \mathbb C^\ast$$ $$x\mapsto e^{2\pi i\,t(x)}$$
This is called the standard character of $\mathbb C$ because any other character can be obtained from it by the composition with the multiplication map in $\mathbb C$. In other words we have the correspondence
$$\mathbb C\to\widehat{\mathbb C}$$
$$a\mapsto\psi(a\cdot) $$
which is also a topological isomorphism if we endow $\widehat{\mathbb C}$ with the adequate topology.
Definition: Let $G$ be a locally compact group abelian and let $S$ be a subset. The orthogonal complement (or annihilator) of $S$ is defined as: $$S^\perp:=\{\chi\in\widehat G\colon \chi(S)=1\}$$
In our case, if $S\subset\mathbb C$, then $S^\perp$ can be viewed as a subset of $\mathbb C$ thanks to the above correspondence.
Question: If $S=\{z\in\mathbb C\colon |z|\le c\}$ is the ball of radius $c$ in the complex plane, what is $S^{\perp}\subseteq\mathbb C$?
So it looks like you could replace $\mathbb{C}$ by $\mathbb{R}^2$, which I'm more comfortable with. We have the correspondence $x\in\mathbb{R}^2 \leftrightarrow \psi: \xi\mapsto e^{2\pi i \xi\cdot x}$.
Suppose $\xi\mapsto e^{2\pi i \xi\cdot x}$ is in the annihilator. Then $e^{2\pi i\xi\cdot x}=1$ for all $|\xi| \leq c$. Take $\xi_0$ in this ball, then $(\xi_0\cdot x) /n$ is an integer for infinitely many $n$, so $x$ must be 0. That is, the annihilator must be trivial.