= Solution
Each component is an <unknot>, so its closed exterior $S^3\setminus\operatorname{int}U_i$ is a <solid torus>. Choose the usual boundary <basis> $(\mu_i,\lambda_i)$, where $\mu_i$ is a meridian of the component and $\lambda_i$ its <Seifert longitude>. In the exterior's <fundamental group>, $\mu_i$ generates $\mathbb Z$ and $\lambda_i$ is trivial, since it bounds a meridian disk of that exterior.
Every essential embedded circle on a torus represents a primitive homology class $a\mu_i+b\lambda_i$ with $\gcd(a,b)=1$. If it bounds a disk in the exterior, it is null-homotopic there, so $a=0$. Primitivity then forces $b=\pm1$. Essential simple curves of the same primitive unoriented slope on a torus are isotopic: lift them to the plane and straighten their parallel primitive lines, or cut the torus along one curve to obtain an annulus. Consequently
$$
\boxed{\alpha_i\text{ is isotopic to the preferred longitude }\lambda_i.}
$$
This is the <preferred longitude of an unknot>. The disk is only required to lie in the complement of its own <solid torus>; it may meet the other link component, so no additional disjoint-disk assertion is being used.
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