If $\vert{z_{1}+ \cdots + z_{n}}\vert$ = $\vert{z_{1}}\vert + \cdots + \vert{z_{n}}\vert$ then $z_{j} = c_{j}z_{1}$ Could someone give me just one suggestion to solve this problem?

Let $z_{1}, \ldots z_{n} \in \mathbb{C}$, with $z_{1}\neq 0$. Prove that if 
  $\vert{z_{1}+ \cdots + z_{n}}\vert$ = 
  $\vert{z_{1}}\vert + \cdots  + \vert{z_{n}}\vert$ then $z_{j} = c_{j}z_{1}$, where 
  $c_{j}\geqslant 0 $, $j=1,\ldots n $

 A: Using  $|z|^2 = z z^*$ we have 
$$
\vert{z_{1}+ \cdots + z_{n}}\vert^2 = (z_{1}+ \cdots + z_{n})(z_{1}^*+ \cdots + z_{n}^*) \\
= \sum_{i} |z_{i}|^2 +   \sum_{i > j} (z_i z_j^* +z_j z_i^* )\\
= \sum_{i} |z_{i}|^2 +  2 \sum_{i > j} \Re (z_i z_j^*) = \sum_{i} |z_{i}|^2 +  2 \sum_{i > j} |z_i||z_j| \cos(\phi_i - \phi_j)
$$
which must, according to the question, be equal to
$$
(\vert{z_{1}}\vert + \cdots + \vert{z_{n}}\vert)^2 = \sum_{i} |z_{i}|^2 +  2 \sum_{i > j} |z_i||z_j|
$$
Since all terms in the second expression are positive and since $\cos(\phi_i - \phi_j) \le 1$, we must have  $\phi_i = \phi_j$ for all $(i,j)$. Hence all $z_j$ must have the same phase angle.  This is equivalent to  saying $z_{j} = c_{j}z_{1}$ with real positive constants $c_{j}$.
A: Here's the base case.
If
$|(a+bi)+(c+di)|
=|a+bi|+|c+di|
$
then,
$\sqrt{(a+c)^2+(b+d)^2}
=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}
$.
Squaring,
$(a+c)^2+(b+d)^2
=a^2+b^2+c^2+d^2+2\sqrt{(a^2+b^2)(c^2+d^2)}
$
or
$2ac+2bd
=2\sqrt{(a^2+b^2)(c^2+d^2)}
$.
Dividing by 2
and squaring again,
$a^2c^2+2abcd+b^2d^2
=a^2c^2+a^2d^2+b^2c^2+b^2d^2
$
or
$0
=a^2d^2+b^2c^2-2abcd
=(ad-bc)^2
$
so
$ad=bc$.
Now it becomes the cases
depending on
which, if any,
of $b$ abd $d$ are zero.
If $b\ne 0, d\ne 0$,
then
$a/b = c/d$.
Letting
$r = a/b$,
then
$a = rb, c = rd$
so
$a+ib
=b(r+i)
$
and
$c+id
= d(r+i)
$
so
$c+id
= (d/b)b(r+i)
=(d/b)(a+ib)
$.
I'll let you work out the
other cases.
A: Let $P(n)$ mean that the result if true for $\le n$ variables. $P(1)$ is clear.
It is straightforward to show that $|1+w| = 1+|w|$ iff $w\ge 0$.
For $P(2)$, if $z_1 \neq 0$ then $|1+ {z_2 \over z_1}| = 1+ | {z_2 \over z_1}|$ and hence $P(2)$ is true by the preceding remark.
Suppose $P(n)$ is true and $|z_1+z_2+\cdots+ z_n + z_{n+1}| = |z_1|+ |z_2|+ \cdots + |z_n| + |z_{n+1}|$.
If $z_1 \neq 0$ we can divide through to get
$|1+ {z_2 \over z_1}+\cdots + {z_{n+1} \over z_1}| = 1+ |{z_2 \over z_1}|+\cdots + |{z_{n+1} \over z_1}|$.
To reduce clutter let $w_k = {z_k \over z_1}$.
Since
$|1+w_2+\cdots+  w_{n+1}| \le 1 + |w_2+\cdots+ w_{n+1}| \le 1 + |w_2| + \cdots + |w_{n+1}|$ we have
$|w_2+\cdots+ w_{n+1}| = |w_2| + \cdots + |w_{n+1}|$ from which we
see that $P(n+1)$ is true.
