Show that every complex number $c$ with $|c|\leq n$ can be written as $c=a_1+a_2+\cdots + a_n$ where $|a_j|=1$ for every $j$. 
Let $n\ge 2$ be a positive integer. Show that every complex number $c$ with $|c|\leq n$ can be written as $c=a_1+a_2+\cdots + a_n$ where $|a_j|=1$ for every $j$.

I think one can come up with a continuous complex-valued function that is the sum of n terms all of which have absolute value 1 so that it contains all the complex numbers from 0 to n in its range. Consider $f(t) = \sum_{j=0}^{n-1} \mathrm{sign}((-1)^j t)e^{ijt}$, where $sign(0):=1$ and $sign(x) = \dfrac{x}{|x|}$ for any other x. $f(0) = n,$ but even though f is continuous it might not have all the numbers from 0 to n in its range. Maybe if I could write a sum of trig functions as a sum of complex exponentials it might be useful? For instance, the Dirichlet kernel satisfies this property. Every $|c|\leq n$ can be written as $x+iy$ for some $x^2 + y^2 \leq n$ and in particular if $r = |c|$ then $c = re^{i\theta}$ for some $\theta \in [0,2\pi)$.
 A: With no loss of generality we may assume that $c\in \mathbb{R}.$

*

*$n=2m\ge 2$
Let $d=\sqrt{1-{c^2\over n^2}}.$ Then the numbers
$u_{k}={c\over n}+(-1)^k di$ satisfy $|u_k|=1.$
We have
$$c=u_1+u_2+\ldots +u_n$$


*$n=2m+1\ge 3$
We may assume that $c\ge 0,$ multiplying by $-1$ if necessary. Then $c=1+(c-1)$ and $|c-1|\le 2m.$
Now we apply the case 1. to represent $c-1$ as a sum of $2m$ terms.
In both cases the summands are given explicitly.
A: We can assume that $c $ is real and non negative without loss of generality.
Since $e^{it}+ e^{-it} = 2 \cos t$ we see that the result is true for $n=2$. Suppose it is true for $n$ and choose $x \in [0,n+1]$. It is clear that the circle $|z-x| = 1$ intersects the closed ball $\overline{B(0,n)}$, let $p$ be a point of intersection, then there are $a_1,...,a_n$ such that $a_1+\cdots+a_n = p$ and we let $a_{n+1} = x-p$, hence the result is true for $n+1$.
A: Along the lines you suggest, you can prove this using connectedness and continuity.  Let $C$ denote the unit circle and define $f:C^n\to\mathbb{R}$ by $f(a_1,\dots,a_n)=|a_1+\dots+a_n|$.  Note that $f(1,1,\dots,1)=n$ and $f(1,e^{2\pi i/n},e^{4\pi i/n},\dots,e^{(2n-2)\pi i/n})=0$ (here we use the assumption that $n\geq 2$).  So, since $C^n$ is connected, $f$ must take all values between $0$ and $n$.  So for any $c$ with $|c|\leq n$, we can find $a_1,\dots,a_n\in C$ with $|a_1+\dots+a_n|=|c|$.  We can then multiply each $a_k$ by a complex number of absolute value $1$ to make their sum equal to $c$ itself.
