Minimum of $|z_1| + |z_2|+ \dots + |z_n|$ when $z_1 + \dots + z_n = 1$ and all the arguments of $z_m$ are prescribed. Let $-\frac{\pi}{2} \le \phi_1 < \dots < \phi_k < 0 < \phi_{k+1} < \dots < \phi_N < \frac{\pi}{2}$ be $N$ prescribed phases. Let $a_1, \dots, a_N$ be $N$ real numbers such that
$$
a_1 e^{i \phi_1} + \dots + a_N e^{i \phi_N} = 1.
$$
What is the minimum of $|a_1|+\dots+|a_N|$?
The minimum is achieved when $a_m=0$ for $m \ne k, k+1$. The proof can be easily understood in a geometrically point of view, as kindly suggested in the following answer. Many thanks to all the answers and comments.
 A: A proof by functional analysis:
Let $a=\{a_j\}_j$ be such that $$1=\sum_j{a_je^{\phi_ji}}$$
Let $f^*=\begin{bmatrix}\cos{\phi_1}&\cdots&\cos{\phi_N}\end{bmatrix}$, $g^*=\begin{bmatrix}\sin{\phi_1}&\cdots&\sin{\phi_N}\end{bmatrix}$.  Then, taking real and imaginary parts, we find that we must solve \begin{gather*}
1=f^*(a) \\
0=g^*(a)
\end{gather*} which I will abbreviate as $(1,0)=(f,g)^*(a)$.
Note that this equation is homogenous: $(\alpha,0)=(f,g)^*(a)$.  But so is the $1$-norm $\|a\|_1=\sum_j{|a_j|}$ (well, for nonnegative $\alpha$, anyways).  Thus our minimum is $$m=\min_{(1,0)=(f,g)^*(a)}{\|a\|_1}=\min_{(\alpha,0)=(f,g)^*(\alpha a)}{\frac{\|a\|_1}{|\alpha|}}$$ where $\alpha$ is any number, possibly dependent on $a$.  Changing variables $\alpha a\mapsto a$, we have $$m=\min_{\substack{0=g^*(a)\\0\neq f^*(a)}}{\frac{\|a\|_1}{|f^*(a)|}}=\left(\max_{0=g^*(a)}{\frac{|f^*(a)|}{\|a\|_1}}\right)^{-1}$$
That maximum is a well-understood quantity in functional analysis.  To see this, first recall that any normed vector space $(V,\|\cdot\|_V)$ has a dual space $V^*$ consisting of the (continuous) linear functionals $V\to\mathbb{R}$.  On this dual space, the quantity $\sup_{v\neq0}{\frac{|f(v)|}{\|v\|_V}}$ is a norm; moreover, for many different normed vector spaces it has already been computed.  For example, the dual space of $(\mathbb{R}^n,\|\cdot\|_1)$ is $(\mathbb{R}^n,\|\cdot\|_{\infty})$, where $\|x\|_{\infty}=\max_j{|x_j|}$.
So what is our vector space?  Well, $\ker{(g^*)}$ inherits a norm from the inclusion $\ker{(g^*)}\subseteq(\mathbb{R}^n,\|\cdot\|_1)$.  We want to compute the dual norm $\|\cdot\|_d$ and then $\|f^*|_{\ker{(g^*)}}\|_d$.
Well, $g^*$ defines the short exact sequence $$0\to\ker{(g^*)}\to(\mathbb{R}^n,\|\cdot\|_1)\to\mathbb{R}1\to0$$  Taking duals reverses the arrows, so that $$0\to\mathbb{R}g^*\to(\mathbb{R}^n,\|\cdot\|_{\infty})\to\ker{(g^*)}^*\to0$$  By the First Isomorphism Theorem, as Banach spaces, $\ker{(g^*)}^*=(\mathbb{R}^n,\|\cdot\|_{\infty})/(\mathbb{R}g^*)$.  What is the norm of this quotient?  For any $x^*\in(\mathbb{R}^n,\|\cdot\|_{\infty})$, $$\|x^*\|_d=\min_{\lambda}{\|x^*+\lambda g^*\|_{\infty}}$$
So now we know: we need to minimize $$\|f^*+\lambda g^*\|_{\infty}=\max_j{\left|\cos(\phi_j)+\lambda\sin(\phi_j)\right|}=\max_j{\left|\cos(\phi_j)(1+\lambda\tan(\phi_j))\right|}$$  Call the $j$th term in this maximum $l_j(\lambda)$, and the maximum as a whole $M(\lambda)$.  Each $l_j$ is the absolute value of a line making angle $\phi_j$ with the $x$-axis and intersecting said axis at $-\cot{\phi_j}$.
$M$ is convex, so it has a unique minimum.  That minimum must be the intersection of an increasing line (to the right) and a decreasing one (to the left).  We claim that it is the intersection of $l_k$ and $l_{k+1}$.
To see this, first calculate: $$M(0)=\max_j{|\cos(\phi_j)|}\in\{\cos{\phi_k},\cos{\phi_{k+1}}\}$$  Now, if we swap $\phi_j\mapsto-\phi_j$, we send $M(\lambda)\mapsto M(-\lambda)$.  Thus we may assume $M(0)=\cos{\phi_{k+1}}$ without loss of generality and $M=l_{k+1}$ on a neighborhood of $0$.
Now, $l_k$ and $l_{k+1}$ intersect at $$\lambda_0=-\frac{\cos(\phi_{k+1})-\cos(\phi_k)}{\sin(\phi_{k+1})-\sin(\phi_k)}$$  Since $l_k$ is decreasing, the minimum of $M$ is attained somewhere in $[\lambda_0,0]$.
Suppose for contradiction the minimum were attained not at $\lambda_0$, but rather at some $\lambda_1\in(\lambda_0,0]$.  Then some other, decreasing, $l_j$ intersects $l_{k+1}$ at $\lambda_1$.  Since $l_j$ and $l_k$ are decreasing but $l_{k+1}$ increasing, the relative order is then preserved as we increase $\lambda$: for $\lambda\in[\lambda_1,0]$, we have $$l_k<l_j<l_{k+1}$$  But: how do I know that $l_j$ and $l_k$ do not "bounce" off the $x$-axis on this region?  First, suppose $j<k$.  Then $l_j$ and $l_k$ both intersect the $x$-axis at $-\cot{(\phi_j)}$ and $-\cot{(\phi_k)}$, respectively...which are both positive numbers.  Second, suppose $j>k+1$.  Then $-\cot{\phi_j}$ is negative; it must be a small negative number.  But then $\cos{\phi_j}=\cot(\phi_j)\sin(\phi_j)$ is even smaller, and so $l_j$ cannot dominate $l_k$ either.
So $\lambda\to0^+$, $$l_k<l_j<l_{k+1}$$  But this is impossible: the $\cos{\phi_j}$ terms dominate the $\lambda\sin{\phi_j}$ terms, so that we must have $l_j<l_k$.  This is a contradiction, and $\lambda_0$ is the minimum.
So $M$ attains its minimum at $\lambda_0$; by algebra, $$M(\lambda_0)=\frac{\sin(\phi_{k+1}-\phi_k)}{\sin(\phi_{k+1})-\sin(\phi_k)}$$  Thus $$m=\frac{\sin(\phi_{k+1})-\sin(\phi_k)}{\sin(\phi_{k+1}-\phi_k)}$$ and it is easy to verify that this is attained when $a_j=0$ for $j\notin\{k,k+1\}$.
A: A proof with painting a drawing with words. Not very rigorous. :)
Let's look at the problem geometrically. Suppose you chose three values of $\phi$ and built a three-segment path with vertices $0, P, Q, 1$ on the complex plane. Then the middle segment $PQ$ has a slope somewhere between the other two segments'.
Suppose then $P$ is closer to the real axis than $Q$. Let's replace segments $0P$ and $PQ$ with a segment $0Q$. This makes the path shorter at a cost of loosing the slope.
However, we can restore the slope of $0Q$ to the original one of $PQ$ by shitfting $Q$ towards $1$ to some appropriate $R$ – and at the same time we shorten the path again.
This applies to any sequence of $\phi$ arguments: if you choose three or more, you can get rid of the one (or more) initial or terminal values and improve the result, as long as you keep at least one positive and one negative term. Finally, the shortest path is reached with the two slopes closest to zero.
Edit: In the above I meant 'closest' as a neighboorhood in the $(\phi_i)$ sequence, not an arithmetic 'distance'. That is, as you proposed, $a_k<0$ and $a_{k+1}>0$.

Hopefully it can be translated from a a simple geometry to an obfuscated exponential notation of complex analysis, but I really don't feel like doing that now. :)
A: Now we have one proof very rigorous and another one more based on intuition. I'd like to post my own proof that somewhat lies between them.
Denote $R=\{ z_1, \dots, z_N \}$, $R_1 = \{ z \in R: \text{Im}  z > 0 \}$, $R_2 = \{ z \in R: \text{Im}  z < 0 \}$. Both $R_1$ and $R_2$ must be non-empty. Denote
$$
z^+ = \sum_{z \in R_1} z = |z^+| e^{i \phi_+}, \text{ and } z^- = \sum_{z \in R_2} z = |z^-| e^{i \phi_-}.
$$
There must be $\pi> \phi_+ \ge \phi_{k+1}$ and $-\pi < \phi_- \le \phi_k$.
It is easy to see that
$$
|z_1| + \dots + |z_N| \ge |z^+| + |z^-|.
$$
Because $z^+ + z^- = 1$, we have
$$
|z^+| + |z^-| = \frac{\sin \phi_+ - \sin \phi_-}{\sin (\phi_+ - \phi_-)}
$$
which is monotonically increasing with $\phi_+$ and decreasing with $\phi_-$ (just examine the sign of derivatives, or directly seen by plotting triangles with the bottom being 1). Thus, the minimum of $|z^+| + |z^-|$ is attained when $\phi_+ = \phi_{k+1}$ and $ \phi_- = \phi_k$, or $z_m=0$ for $m \ne k, k+1$.
