On Magnitudes of Sums of Roots of Unity and a Simple Trigonometric Inequality The Problem
Let $r,q,m$ be positive integers such that $4 \leq r$ and $1<m,q\leq r/2$. Is it the case that 
$$\left |  \sum_{k=0}^{q-1} \zeta^{km}\right | < \left |\sum_{k=0}^{q-1} \zeta^{k}\right |,$$
where $\zeta$ is a primitive $r^{th}$ root of unity.
Continuous Variant
Through use of a non-standard trigonometric identity which may be found here we can change the previous inequality to
$$|\sin(mq\pi/r)\sin(\pi/r)| < |\sin(m\pi/r)\sin(q\pi/r)|.$$
Then we may substitute real variables. Let $\theta \in (0,\pi/4]$ and $a,b \in \mathbb R$ so that $0< a\theta, b \theta \leq \pi$. The we have the inequality 
$$| \sin(ab \theta) \sin(\theta) |< | \sin(a \theta)\sin(b \theta) |. $$
Fix $\theta$ and consider the function 
$$f(x,y)=\frac{\sin(xy\theta)\sin(\theta)}{\sin(x\theta)\sin(y\theta)}$$
where $(x,y) \in [1,\pi/(2\theta)]^2=X$. So if we can show that $f$ has a unique maximum on $X$ at $(1,1)$ we would also have our result. One can show with a somewhat painful computation that the critical points of $f$ are on the line $x=y$, so we can consider instead the function
$$g(x)=\frac{\sin(x^2\theta)\sin(\theta)}{\sin^2(x\theta)}$$
and the corresponding inequality
$$|\sin(x^2\theta)\sin(\theta)|<|\sin^2(x\theta)|.$$
I have some partial results in this case where essentially I can control the inequality depending on where $x^2\theta$ is modulo $2\pi$. Which comes down to everything except when the residue of $x^2\theta$ modulo $2\pi$ is in the interval $(2x\theta-\theta,2x\theta)$. 
Motivation
Let the generalized binomial coefficient $C_q(n,k)$ be the coefficient of $x^k$ in the polynomial $(1+x+\cdots+x^{q-1})^k$, sometimes this appears in the literature as $\binom{n}{k}_q$. Define
$$PC_q(n,r,k)=\sum_{j \in \mathbb Z} C_q(n,k+rj)$$
then the sums in the statement of the problem arise naturally as the coefficients of a discrete Fourier expansion of $PC_q$. In particular proving the desired inequality shows that $PC_q$ behaves as a sine function as $n \rightarrow \infty$. There's some more background on $C_q$ and $PC_q$ is this answer. 
 A: Actually, I believe something a bit stronger is true: For any $S \subseteq \{1,\dots,r\}$ with $|S|=q$, we have 
$$ (*) |\sum_{j \in S} \zeta^j| \leq |\sum_{k=0}^{q-1} \zeta^k|.$$ 
Consider an arbitrary $S$, and let $v$ be the unit vector in the direction of $\sum_{j \in S} \zeta^j$ (here and later on we're thinking of the complex numbers as a $2$-dimensional real vectors).  Let $S'$ be the set consisting of the $q$ roots of unity having largest (positive) projection onto $v$.  Then we have 
\begin{eqnarray*}
|\sum_{j \in S} \zeta^j| &=& proj_{v}\left(\sum_{j \in S} \zeta^j\right) \\
&=& \sum_{j \in S} proj_v \left(\zeta^j\right) \\
&\leq& \sum_{j \in S'} proj_v \left(\zeta^j\right) \\
&=& proj_v \left(\sum_{j \in S'} \zeta^j \right) \\
&\leq& | \sum_{j \in S'} \zeta^j |
\end{eqnarray*}
But for any $v$ the set $S'$ consists of $q$ consecutive roots of unity, so the last expression is the right hand side of (*).  

If equality holds, then $S$ must consist of $q$ consecutive residues
modulo $r$.  In other words, the set $\{1, m, 2m, \dots, (q-1)m\}$ is
some permutation of $\{a,a+1,\dots,a+q-1\}$ (modulo $r$) for some $a$
and $q$.  Let us fix some $m,a,q$ for which this occurs.
For $1 \leq j \leq q$, let $f(j)$ be the integer between $0$ and $q-1$
such that $mj$ is congruent to $a+f(j)$ modulo $r$.  One of two things
must happen:
-If $f(j)=j$ for all $j$, then $a=m=1$, not allowed.
-Otherwise, there is some $j$ for which $f(j)>f(j+1)$.  But this would
imply $m(j+1)-mj \geq a+r-(a+q-1)$, or, rearranging, $m+q \geq r+1$,
which is also not allowed.
Essentially what the second case corresponds to is that your condition on $m$ and $q$ versus $r$ makes it impossible to "wrap around" from one end of the interval to the other between two consecutive multiples of $m$.  
