<s>[INST] 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 $

 [/INST] 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.
</s>