= Solution
A <Seifert surface> for an <oriented knot> $K$ is a compact connected oriented surface $F\subset S^3$ whose oriented boundary is $K$. For homology classes represented by oriented curves $x,y\subset F$, the <Seifert form> is
$$
\theta_F([x],[y])=\operatorname{lk}(x^+,y),
$$
where $x^+$ is the positive normal push-off. Choosing a basis of $H_1(F;\mathbb Z)$ gives a <Seifert matrix> $A$.
For $\omega\in S^1\setminus\{1\}$, the <Levine-Tristram signature> is
$$
\sigma_\omega(K)=\operatorname{sign}\bigl((1-\omega)A+(1-\overline\omega)A^T\bigr).
$$
The determinant of this Hermitian matrix vanishes away from $\omega=1$ exactly at the unit roots of the <Alexander polynomial of a knot> $\Delta_K$. Consequently the signature is locally constant on their complement.
For $\omega=e^{i\theta}$ near $1$,
$$
(1-\omega)A+(1-\overline\omega)A^T
=i\theta(A^T-A)+O(\theta^2).
$$
The real skew-symmetric unimodular matrix $A^T-A$ has standard symplectic blocks, so the Hermitian matrix $i(A^T-A)$ has its positive and negative <eigenvalue>[eigenvalues] in opposite pairs and has signature zero. Thus $\sigma_\omega(K)=0$ near $1$. If $\Delta_K$ has no unit roots, then $S^1\setminus\{1\}$ contains no singular point of the signature form and is connected, so local constancy gives $\sigma_\omega(K)=0$ everywhere. With the usual convention $\sigma_1(K)=0$, the signature vanishes identically.
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