Case of near equality in triangle inequality Let $a_i$, $1 \leq i \leq n$ be complex numbers such that $\sum a_i=1$, and $\sum |a_i| \leq 1+\epsilon$, where $\epsilon>0$ is small.
This corresponds to a case of "near equality" in the triangle inequality. Therefore it is strongly expected that all the $a_i$ have small imaginary part (controlled by $\epsilon$) and real or "smally negative" real part. 
I'm looking for a simple algebraic proof of this fact, which has been eluding me so far. The best I have for now is a somewhat complicated geometric proof using relationships between area and perimeter of a triangle. 
Thanks in advance !
 A: For convenience of notation, let's write $a_j = x_j + i y_j$ with $x_j,\, y_j \in \mathbb{R}$. We note that we have $\lvert x_j\rvert \leqslant \lvert a_j\rvert$ for all $j$.
Concerning the real parts, we see
$$1 + \varepsilon \geqslant \sum \lvert x_j\rvert = \sum_{x_j \geqslant 0} x_j - \sum_{x_j < 0} x_j$$
and hence
$$\varepsilon = (1 + \varepsilon) - 1 \geqslant \sum_{x_j \geqslant 0} x_j - \sum_{x_j < 0} x_j - \sum_{j=1}^n x_j = -2\sum_{x_j < 0} x_j = 2\sum_{x_j < 0} \lvert x_j\rvert,$$
thus $\sum\limits_{x_j < 0} \lvert x_j\rvert \leqslant \frac{\varepsilon}{2}$.
For the imaginary parts it's not as easy, but for any $1 \leqslant k \leqslant n$, we have
$$1 + \varepsilon \geqslant \sum_{j=1}^n \lvert a_j\rvert \geqslant \sum_{j=1}^n \lvert x_j\rvert + (\lvert a_k\rvert - \lvert x_k\rvert).$$
The sum of the $\lvert x_j\rvert$ is $\geqslant 1$, hence $\lvert a_k\rvert - \lvert x_k\rvert \leqslant \varepsilon$. From that we obtain
$$\begin{gather}\lvert a_k\rvert^2 = x_k^2 + y_k^2 \leqslant (\lvert x_k\rvert + \varepsilon)^2 = x_k^2 + 2\lvert x_k\rvert \varepsilon + \varepsilon^2\\
y_k^2 \leqslant 2\lvert x_k\rvert \varepsilon + \varepsilon^2.
\end{gather}$$
For $\lvert x_k\rvert \leqslant 1$, that immediately yields $\lvert y_k\rvert \leqslant \sqrt{\varepsilon(2+\varepsilon)}$ as a crude estimate (rather crude, you have essentially a factor of $\sqrt{\lvert x_k\rvert}$ that makes it smaller until $x_k$ is close to $0$, when you have essentially a bound $\lvert y_k\rvert \leqslant \varepsilon$, and the imaginary parts must sum to $0$, so you must have at least one of the opposite sign, that reduces the bound too).
For $\lvert x_k\rvert > 1$, we use $\lvert a_k\rvert \leqslant 1 + \varepsilon$, which tells us
$$x_k^2 + y_k^2 \leqslant (1+\varepsilon)^2 = 1 + 2\varepsilon + \varepsilon^2,$$
whence $y_k^2 \leqslant (1 - x_k^2) + 2\varepsilon + \varepsilon^2 < \varepsilon(2+\varepsilon).$
If we consider not only a single $a_k$, we find that
$$\sum_{j=1}^n (\lvert a_j\rvert - \lvert x_j\rvert) \leqslant \varepsilon,$$
which shows that the above bound is too large, but doesn't easily lead to a better bound.
