G. Myerson's argument can be used recursively to establish bounds for the sum of $N>10$ $n$-th roots of unity. For instance, start from $N=10$. Let us denote $\omega=\exp\frac{2i\pi}{n}$. GM's construction uses only the roots $\omega^k$ for $1\le k\le18$ and $\frac{n}{2}\le k\le\frac{n}{2}+19$ (say that $n$ is even). The corresponding sum is $z_n\ne0$ such that $|z_n|\le Cn^{-5}$. Now, say that $n$ is a multiple of $38$ ($n=38m$) and let us cover the complex plane by $m$ disjoint sectors of angle $\frac{\pi}{m}$. Each sector can be used to construct an other point, and the $m$ points obtained that way form a regular $m$-agon. Here is the induction argument: we may sum $10$ such points in order to obtain a point $z'$ with $z'=z_nz_m$. Now, $z'$ is the sum of $N'=100$ distinct $n$-th roots of unity, and we have $$|z'|\le Cn^{-5}\left(\frac{n}{38}\right)^{-5}=C'n^{-10}.$$ More generally, if $N=10^r$, we obtain a sum of $N$ $n$-th roots of unity ($n$ a multiple of $38^{r-1}$) of the form $Cn^{-\alpha}$ with $\alpha=5r=5\log_{10}N$.

Edit. Alternate description (but this is the same construction). Let $J$ be the set of exponents used by GM when $N=10$, that is $J=\{1,5,9,17,18\}\cup\left(\frac{n}{2}+\{2,3,11,15,19\}$. For $N=100$ and $n$ a multiple of $38$, set $$z':=\sum_{i\in J}\sum_{j\in J}\omega^{i+38j}.$$ If $n$ is large enough, this is a sum of distinct $n$th roots of unity, such that $z'=z_nz_m$.

G. Myerson's argument can be used recursively to establish bounds for the sum of $N>10$ $n$-th roots of unity. For instance, start from $N=10$. Let us denote $\omega=\exp\frac{2i\pi}{n}$. GM's construction uses only the roots $\omega^k$ for $1\le k\le18$ and $\frac{n}{2}\le k\le\frac{n}{2}+19$ (say that $n$ is even). The corresponding sum is $z_n\ne0$ such that $|z_n|\le Cn^{-5}$. Now, say that $n$ is a multiple of $38$ ($n=38m$) and let us cover the complex plane by $m$ disjoint sectors of angle $\frac{\pi}{m}$. Each sector can be used to construct an other point, and the $m$ points obtained that way form a regular $m$-agon. Here is the induction argument: we may sum $10$ such points in order to obtain a point $z'$ with $z'=z_nz_m$. Now, $z'$ is the sum of $N'=100$ distinct $n$-th roots of unity, and we have $$|z'|\le Cn^{-5}\left(\frac{n}{38}\right)^{-5}=C'n^{-10}.$$ More generally, if $N=10^r$, we obtain a sum of $N$ $n$-th roots of unity ($n$ a multiple of $38^{r-1}$) of the form $Cn^{-\alpha}$ with $\alpha=5r=5\log_{10}N$.

Edit. Alternate description (but this is the same construction). Let $J$ be the set of exponents used by GM when $N=10$, that is $J=\{1,5,9,17,18\}\cup\left(\frac{n}{2}+\{2,3,11,15,19\}\right)$. For $N=100$ and $n$ a multiple of $38$, set $$z':=\sum_{i\in J}\sum_{j\in J}\omega^{i+38j}.$$ If $n$ is large enough, this is a sum of distinct $n$th roots of unity, such that $z'=z_nz_m$.

G. Myerson's argument can be used recursively to establish bounds for the sum of $N>10$ $n$-th roots of unity. For instance, start from $N=10$. Let us denote $\omega=\exp\frac{2i\pi}{n}$. GM's construction uses only the roots $\omega^k$ for $1\le k\le18$ and $\frac{n}{2}\le k\le\frac{n}{2}+19$ (say that $n$ is even). The corresponding sum is $z_n\ne0$ such that $|z_n|\le Cn^{-5}$. Now, say that $n$ is a multiple of $38$ ($n=38m$) and let us cover the complex plane by $m$ disjoint sectors of angle $\frac{\pi}{m}$. Each sector can be used to construct an other point, and the $m$ points obtained that way form a regular $m$-agon. Here is the induction argument: we may sum $10$ such points in order to obtain a point $z'$ with $z'=z_nz_m$. Now, $z'$ is the sum of $N'=100$ distinct $n$-th roots of unity, and we have $$|z'|\le Cn^{-5}\left(\frac{n}{38}\right)^{-5}=C'n^{-10}.$$ More generally, if $N=10^r$, we obtain a sum of $N$ $n$-th roots of unity ($n$ a multiple of $38^{r-1}$) of the form $Cn^{-\alpha}$ with $\alpha=5r=5\log_{10}N$.

G. Myerson's argument can be used recursively to establish bounds for the sum of $N>10$ $n$-th roots of unity. For instance, start from $N=10$. Let us denote $\omega=\exp\frac{2i\pi}{n}$. GM's construction uses only the roots $\omega^k$ for $1\le k\le18$ and $\frac{n}{2}\le k\le\frac{n}{2}+19$ (say that $n$ is even). The corresponding sum is $z_n\ne0$ such that $|z_n|\le Cn^{-5}$. Now, say that $n$ is a multiple of $38$ ($n=38m$) and let us cover the complex plane by $m$ disjoint sectors of angle $\frac{\pi}{m}$. Each sector can be used to construct an other point, and the $m$ points obtained that way form a regular $m$-agon. Here is the induction argument: we may sum $10$ such points in order to obtain a point $z'$ with $z'=z_nz_m$. Now, $z'$ is the sum of $N'=100$ distinct $n$-th roots of unity, and we have $$|z'|\le Cn^{-5}\left(\frac{n}{38}\right)^{-5}=C'n^{-10}.$$ More generally, if $N=10^r$, we obtain a sum of $N$ $n$-th roots of unity ($n$ a multiple of $38^{r-1}$) of the form $Cn^{-\alpha}$ with $\alpha=5r=5\log_{10}N$.

Edit. Alternate description (but this is the same construction). Let $J$ be the set of exponents used by GM when $N=10$, that is $J=\{1,5,9,17,18\}\cup\left(\frac{n}{2}+\{2,3,11,15,19\}$. For $N=100$ and $n$ a multiple of $38$, set $$z':=\sum_{i\in J}\sum_{j\in J}\omega^{i+38j}.$$ If $n$ is large enough, this is a sum of distinct $n$th roots of unity, such that $z'=z_nz_m$.