The sum of $2N+1$ unit complex numbers in the upper half plane has modulus greater than or equal to $1$

Question

I recently came across the below property in an exam which appears to be true, but which I have been unable to prove.

Claim: Let $N$ be a positive integer and let $\theta_1, \theta_2, \ldots, \theta_{2N+1} \in [0, \pi]$ be given. Then $\left|e^{i\theta_1} + e^{i\theta_2} + \cdots + e^{i\theta_{2N+1}} \right| \geq 1$.

The exam solution relied on mathematical induction, however was flawed.

In vector terms, the claim is equivalent to the statement that the sum of an odd number of unit vectors in the same plane which lie on the same side of a line has magnitude at least $1$.

In the case where $N = 1$, the square of the magnitude reduces to \begin{align*} \left|e^{i\theta_1} + e^{i\theta_2} + e^{i\theta_3} \right|^2 &= 3 + 2 \left( \cos(\theta_1 - \theta_2) + \cos(\theta_1 - \theta_3) + \cos(\theta_2 - \theta_3)\right) \\ &= 3 + 2 \left(\cos(A-B) + \cos A + \cos B \right) \end{align*} where $A = \theta_1 - \theta_3$ and $B = \theta_2 - \theta_3$. This resembles the problem at minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$, however with the added constraint that $A, B, C \in [0, \pi]$.

Consideration of $f(A, B) = \cos(A-B) + \cos A + \cos B$ gives the minimum value over the domain $[0, \pi] \times [0, \pi]$ to be $-1$ (when $A = B = \pi$) which establishes the result for $N = 1$.

However this approach does not easily generalise. In general, we wish to minimise

$$ \left|e^{i\theta_1} + e^{i\theta_2} + \cdots + e^{i\theta_{2N+1}} \right|^{2} = 2N + 1 + 2 \left(\sum_{k=1}^{2N} \cos (X_k) + \sum_{\substack{k, \ell \\{k < \ell}}} \cos(X_k - X_\ell) \right), $$

where $X_K = \theta_K - \theta_{2N+1}$ for each $K$, over $X_{1} \times X_{2} \times \cdots \times X_{2N} \in [0, \pi]^{2N}$.

Is there a simple solution or argument (or counter-example) I am missing?

Answer 1

Strategy: Divide the angles into two groups of $N$ that flank the middle angle; then every pair of angles with one element from each group, when combined, contribute a quantity that's “enough in the direction” of the middle angle to help the sum in that direction.

To formalize this, write the angles as $0 \le \theta_{-N} \le \theta_{-(N-1)} \le \cdots \le \theta_0 \le \cdots \le \theta_{N-1} \le \theta_N \le \pi$. We will prove that $\bigl| \sum_{k=-N}^N e^{i\theta_k} \bigr| \ge 1$ by proving that $\Re\bigl( e^{-i\theta_0} \sum_{k=-N}^N e^{i\theta_k} \bigr) \ge 1$.

It suffices to show that $\Re\bigl( \sum_{k=1}^N e^{-i\theta_0} ( e^{i\theta_{-k}} + e^{i\theta_k} ) \bigr) \ge 0$, and thus suffices to show that $\Re\bigl( e^{-i\theta_0} ( e^{i\theta_{-k}} + e^{i\theta_k} ) \bigr) \ge 0$ for each $k$. More generally we show that $\Re \bigl( e^{-i\theta_0} ( e^{i\alpha} + e^{i\beta} ) \bigr) \ge 0$ whenever $0 \le \alpha \le \theta_0 \le \beta \le \pi$.

We have $e^{i\alpha} + e^{i\beta} = 2 e^{i(\alpha+\beta)/2} \cos\frac{\alpha-\beta}2$, and so the sign of $\Re \bigl( e^{-i\theta_0} ( e^{i\alpha} + e^{i\beta} ) \bigr)$ is the same as the sign of $\Re \bigl( e^{-i\theta_0} e^{i(\alpha+\beta)/2} \bigr)$. But the inequalities $0 \le \alpha \le \theta_0 \le \beta \le \pi$ imply that $$ -\frac\pi2 \le -\frac{\theta_0}2 = -\theta_0 + \frac{0+\theta_0}2 \le -\theta_0 + \frac{\alpha+\beta}2 \le -\theta_0 + \frac{\theta_0+\pi}2 = \frac{\pi-\theta_0}2 \le \frac\pi2, $$ which is enough to confirm that $\Re \bigl( e^{-i\theta_0} e^{i(\alpha+\beta)/2} \bigr) \ge 0$.

Answer 2

I think what you already did can be extended to an induction argument.

Suppose this is true for any such $2N-1$ vectors, and we move on the examine $2N+1$ such vectors. Let $U$ be the sum of the first $2N-1$ of them. So we know $U=Re^{i\theta}$ with $R\geq1$ and $\theta\in[0,\pi]$.

Now we progress to having $2N+1$ vectors $$ V=U+e^{i\theta_1}+e^{i\theta_2}=Re^{i\theta}+e^{i\theta_1}+e^{i\theta_2}$$

The square of the magnitude is

$$\begin{align*} \left|Re^{i\theta} + e^{i\theta_1} + e^{i\theta_2} \right|^2 &= R^2+2 + 2 \left( \cos(\theta_1 - \theta_2) + R\cos(\theta_1 - \theta) + R\cos(\theta_2 - \theta)\right) \\ &= R^2+2 + 2 \left(\cos(A-B) + R\cos A + R\cos B \right) \end{align*}$$

Now analyze $f(A,B,R)=\cos(A-B) + R\cos A + R\cos B$ over appropriate constraints for $A,B$ and $R\geq1$.

Note that $A=\theta_1-\theta$ can be as low as $-\pi$ up to as much as $\pi$. So $A\in[-\pi,\pi]$. Similarlly, $B\in[-\pi,\pi]$, but the two choices are not independent. You cannot simultaneously have $A=\pi$ and $B=-\pi$ or else $A-B=2\pi$, and since $A-B=\theta_1-\theta_2$, it is constrained to $[-\pi,\pi]$ as well.

$A$ and $B$ must come from the subset of $[-\pi,\pi]\times[-\pi,\pi]$ such that $A-B\in[-\pi,\pi]$. That forms a hexagon with vertices at $(-\pi,-\pi), (0,-\pi), (\pi,0), (\pi,\pi), (0,\pi), (-\pi,0)$.