If $w$ is the $n$_th root of unity and $|z-w^k|\le 1$ then prove $z=0$ If $w$ is the $n$_th primitive root of unity and $|z-w^k|\le 1$ then prove $z=0$.(for all $0 \le k \le n-1$)
My attempt:We know that $w^k$ generates all roots of unity if $n$ is even by drawing the roots we can find two points that have the distance $2$ and are on diameter of a circle with radii $1$ and center $0$ so because $z$ has the distance at max $1$ from each point $z=0$.
Edit: $n>1$
 A: Put $z-\omega^k=u_k$. We have $z\omega^{-k}=1+u_k\omega^{-k}$. Adding and taking in account that $\sum_{k=0}^{n-1}\omega^{-k}=0$ we get that $-n=\sum_{k=0}^{n-1}u_k\omega^{-k}$. As $|u_k\omega^{-k}|\leq 1$, we get that $|u_k\omega^{-k}|= 1$ for all $k$, ie $|z-\omega^k|=1$ for all $k$. This imply that $|z|^2-\overline{z}\omega^k-z\omega^{-k}=0$ for all $k$. Adding for $k=0,\cdots n-1$ gives $n|z|^2=0$.   
A: Suppose $|p| = 1$. Any nonzero point $x$ in the half-plane $\{x: \; x \cdot p \le 0\}$ has $|x-p| > 1$.  So if you have points $p_j$ on the circle such that there is no $x$ with $x \cdot p_j > 0$ for all $p_j$, there is no nonzero $x$ with all $|x - p_j| \le 1$.  In particular, this is true whenever $0$ is in the convex hull of the $p_j$.  In your case with $p_j = w^j$, $0 = \sum_{j=0}^{n-1} \frac{w^j}{n}$.
A: 
For a point $M$ on the unit circle, the region $|z-z_M|\le 1$ correspond to the interior of the disk $\overline B(z_M,1)$.
Since the point $A=(1,0)$ is always a $n$-th root of unity, let's fix the blue disk.
Now, let's take two points $B$ and $C$ on the unit circle, associated to the green and pink disks.

When $C$ is located inside the dashed area [figure 1] of points that are located in the lower semi-plane delimited by $(OA)$ and the higher semi-plane delimited by $(OB)$ then the intersection of the $3$ disks is reduced to $O$.
While when $C$ is outside this region [figure 2], then the intersection in orange is not empty (i.e. not reduced to 1 point).

The case $n=2$ of square root of unity, correspond to only two disks with $B=(-1,0)$, in that case these two disks are tangent in $O$ and $z=0$ is the only solution.
For $n\ge 3$, let's associate $B$ to the primitive root of unity $w=\exp(\frac{2i\pi}n)$ and $C$ to the root $w^k$.
We shall see that there always exists $k\in[2,n-1]$ such that $C$ belongs to the dashed region.
This condition translates to : 
$\arg(z_C)\in[\pi,\pi+\arg(z_B)]\iff \pi\le\frac{2k\pi}n\le\pi+\frac{2\pi}n\iff n\le 2k\le2n+2$
$\forall n\ge 3,\ k=\lfloor\frac n2\rfloor+1$ satisfies this condition.
So $O$ is the only intersection of the three disks, meaning $z=0$ is the only solution of the $\forall k,|z-w^k|\le 1$ problem.



Edit: 16/09
When angle $\hat B$ is small, simple geometric considerations on semi-planes show why the dashed area is important. 
When $2$ centres are aligned then the corresponding $2$ disks are tangent. If you move $C$ above $(OA)$ an intersection between the two disks is created in upper semi-plane delimited by $(OA)$. If you move $C$ below $(OA)$ then an intersection is created in lower semi-plane delimited by $(OA)$. This is the same for the disks issued from $B$ and $C$ now realtively to semi-planes delimited by $(OB)$.
It is then straightforward to see that when $C$ is in the dashed area, the intersections belong to disjoint angular sectors $[\hat B,\pi]$ and $[\hat B+\pi,2\pi]$ at least for $\hat B$ small enough.
Yet for the case $n=3$, $\hat B$ is as large as $\frac {2\pi}3$ and the intersections span over wider areas, it is still possible to have a geometric argument, but in this case I'll prove it by algebra.
Let $z=re^{i\theta}$ be in the $3$ disk intersections when $C$ is in the dashed area :
$\begin{cases}
|z-a|\le 1 & a=1\\
|z-b|\le 1 & b=e^{i\beta} & 0<\beta<\pi\\
|z-c|\le 1 & c=e^{i\gamma} & \pi\le\gamma\le\pi+\beta \iff \gamma=\pi+\lambda\beta,\ \lambda\in[0,1]\\
\end{cases}$
$|z-c|^2=(z-c)(\bar z-\bar c)=\underbrace{z\bar z}_{r^2}+\underbrace{c\bar c}_1-\underbrace{(z\bar c+\bar zc)}_{2r\cos(\theta-\gamma)}\le 1$
This leads to the system
$\begin{cases}
\cos(\theta)\ge \frac r2\\
\cos(\theta-\beta)\ge \frac r2\\
\cos(\theta-\gamma)\ge \frac r2\\
\end{cases}$
$f(\lambda)=\cos(\theta-\gamma)=\cos(\theta-\pi-\lambda\beta)=-\cos(\theta-\lambda\beta)$.
$\begin{cases}
f(0)=-\cos(\theta)\le-\frac r2\le 0\\
f(1)=-\cos(\theta-\beta)\le-\frac r2\le 0\\
\exists \lambda\in]0,1[\ \mid f(\lambda)\ge\frac r2\ge 0
\end{cases}$
This condition on $\lambda$ can be realized only if $\theta$ and $\theta-\beta$ are at least $\pi$ apart, but this implies $\beta\ge\pi$ which is a contradiction.
So the only solution is $r=0$. (which is also $z=0$).
