Consider a sequence of complex polynomials $f \in \mathbb{C}[z]$, $f(0) \neq 0$, that are composed of a negligible fraction $o(\deg{f})$ of monomials. Are the zeros of such polynomials necessarily equidistributed in angle, for the uniform measure $d\theta/2\pi$ on $S^1 = \mathbb{C}^{\times} / \mathbb{R}^{> 0}$?


Certainly the zeros are equidistributed in angle when the number of monomials is bounded while the degree goes to infinity. This follows from much more general results of A. G. Khovanskii exposed in his book *Fewnomials* (Transl. Math. Monographs, vol. 88). But growing like $o(\deg{f})$?


Not knowing the answer, it might make more sense to look at this question contrapositively. Clearly, for any $\varepsilon > 0$ we can construct an $\varepsilon$-sparse sequence with $\deg{f} \to \infty$ and not equidistributed in angle; just consider $h(z^n)$ as $n$ is fixed subject to $1/n < \varepsilon$ while $h$ ranges over the $\mathbb{R}$-split real polynomials. Can we do better than this?



Angular distribution of zero sets of sparse polynomials
A lot depends on how exactly to understand this question. The crudest form (for an $m$-nomial of degree $n$ with non-zero free term, the number of all zeroes in a sector of aperture $2\pi\theta$ is $\theta n$ with an error at most $m$) is an exercise in elementary complex analysis. Indeed, consider a sector of aperture $2\pi\theta$ with the vertex at $0$ and of huge radius $R$.
The increment of the argument of $p$ over the arc is then $2\pi\theta n$ for all practical purposes. On the other hand, each loop around the origin on the interval $\{zt:0\le t\le R]$ creates $2$ real zeroes of $\Re p(zt)$, so the "side" increment of the argument is at most $\pi$ times the total number of real roots of $2$ real $m$-nomials (one for each side). However, by the Descartes rule of signs, an $m$-nomial can have at most $m$ real roots, and the story is over. 


