Showing that equality of complex numbers implies that they lie on the same ray

[kalish1]

This problem has been on my mind for a while.

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**Problem:**

Show that **if**
\begin{equation}
|z_1+z_2+\dots+z_n| = |z_1| + |z_2| + \dots + |z_n|
\end{equation}
**then** $z_k/z_{\ell} \ge 0$ for any integers $k$ and $\ell$, $1 \leq k, \ell \leq n,$ for which $z_{\ell} \ne 0.$

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**Proposed solution:**
For the base case $n = 2,$ we want to find a condition for which $|z_1+z_2| = |z_1|+|z_2|.$ From the book and class discussions, we see that equality occurs if $z_1$ and $z_2$ are collinear. Now assume the statement is true for $n=m.$ That is, assume:
\begin{equation}
|z_1+z_2+\dots+z_m| = |z_1| + |z_2| + \dots + |z_m|
\end{equation}
if and only if $z_k/z_{\ell} \ge 0$ for all $1 \le k,\ell \le m$ when $z_{\ell} \neq 0.$

Let $m \ge 2.$ We need to show:
\begin{equation}
|z_1+z_2+\dots+z_m+z_{m+1}| = |z_1| + |z_2| + \dots + |z_m| + |z_{m+1}|
\end{equation}
if and only if $z_k/z_{\ell} \ge 0$ for all $1 \le k,\ell \le m+1$ when $z_{\ell} \neq 0.$

Let $z=z_1+\ldots+z_m.$ We have $$|z_1 + \ldots + z_{m+1}| = |z + z_{m+1}| \overset{(1)}{\leq} |z| + |z_{m+1}| \overset{(2)}{\leq} |z_1| + \ldots + |z_m| + |z_{m+1}|$$

$(1)$: Equality holds if and only if $z/z_{m+1} \ge 0$ when $z_{m+1} \neq 0$ (according to the base case)

$(2)$: Equality holds if and only if $z_k/z_{\ell} \ge 0$ for all $1 \le k,\ell \le m$ when $z_{\ell} \neq 0$ (according to the induction hypothesis)

So what we have so far is:
$$|z_1 + \ldots + z_{m+1}| = |z_1| + \ldots + |z_{m+1}|$$ if and only if

**$\dfrac{z_1 + \ldots + z_m}{z_{m+1}} \ge 0,$ when $z_{m+1} \neq 0,$ and $z_k/z_{\ell} \ge 0$ for all $1 \le k,\ell \le m,$ when $z_{\ell} \neq 0.$**

We need to show $\mathrm{bold \ statement} \implies z_k/z_{\ell} \ge 0 \ \mathrm{for \ all} \ 1 \le k,\ell \le m+1 \ \mathrm{when} \ z_{\ell} \neq 0.$

Assume $\mathrm{bold \ statement}$ holds. We need to show $z_k/z_{\ell} \ge 0$ for $\ell=m+1$ and $1 \le k \le m+1$ (when $z_{m+1} \neq 0$).

When $k=m+1,$ we have $z_{k}/z_{m+1}=z_{m+1}/z_{m+1}=1 \ge 0.$ Assume now that $k \le m.$ Assume that $z_{k} \neq 0.$

From $\mathrm{bold \ statement}$ we know that $z_1/z_k \ge 0, \ldots, z_m/z_k \ge 0$ implies $\dfrac{z_1+ \ldots +z_m}{z_k} \ge 0.$ Note that $$\dfrac{z_1+ \ldots +z_m}{z_k} = 0 \iff z_1 = \dots = z_m = 0.$$ But since $z_k \neq 0,$ we have $\dfrac{z_1+ \ldots +z_m}{z_k} \neq 0.$

Thus
$$\dfrac{z_k}{z_{m+1}} = \underbrace{\dfrac{z_1+\ldots+z_m}{z_{m+1}}}_{\ge 0 \ \mathrm{bold \ statement}}\cdot\dfrac{z_k}{z_1+\ldots+z_m} \ge 0.$$

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I wrote this solution, but my teacher did not approve of it, saying that there is too much extraneous information that doesn't answer the problem.

## How can I solve the problem with the following information:
$z_k = t_kz_1,$ for $z_1 \neq 0,$ so that $z_k$ and $z_1$ lie on the same ray for every $k?$

Thanks for any help.

This question has been crossposted here: http://math.stackexchange.com/questions/1200965/showing-that-equality-of-complex-numbers-implies-that-they-lie-on-the-same-ray

 

[Opalg]

This problem has been on my mind for a while.

----------
**Problem:**

Show that **if**
\begin{equation}
|z_1+z_2+\dots+z_n| = |z_1| + |z_2| + \dots + |z_n|
\end{equation}
**then** $z_k/z_{\ell} \ge 0$ for any integers $k$ and $\ell$, $1 \leq k, \ell \leq n,$ for which $z_{\ell} \ne 0.$

----------

Without reading it through in fine detail, I have the impression that your argument is essentially correct. But it can certainly be considerably shortened. I think that what your teacher has in mind is something like this: $$| z_1 + z_2 + \ldots + z_n|^2 = (z_1 + z_2 + \ldots + z_n)(\bar z_1 + \bar z_2 + \ldots + \bar z_n) = \sum_{j=1}^n|z_j|^2 + \sum_{1\leqslant j<k\leqslant n}2\text{re}(z_j\bar z_k),$$ where the bars denote complex conjugates. But $\text{re}(z_j\bar z_k) \leqslant |z_j||z_k|$, with equality only if $z_j$ and $z_k$ lie on the same ray (that is essentially the result for the base case $n=2$). Therefore $$| z_1 + z_2 + \ldots + z_n|^2 \leqslant \sum_{j=1}^n|z_j|^2 + \sum_{1\leqslant j<k\leqslant n}2 |z_j||z_k| = \bigl(|z_1| + |z_2| + \ldots + |z_n|\bigr)^2,$$ with equality only if all the $z_j$ lie on the same ray. Now all you have to do is to take the square root of both sides, to get the result.