Assume $|z_1+z_2|=|z_1|+|z_2|$. Show that this implies $\arg(z_1)-\arg(z_2)=2n\pi$ The hint I am given is that the relationship of $|z1||z2|$ implies $\arg(z1)-\arg(z2)=2n\pi$ is to be used somewhere.  I think the only way this can be done is to square it but after that I'm not getting anywhere further.  Need some help.  
 A: If $|z_1 + z_2| = |z_1| + |z_2|$, then taking square on both sides we have
$$|z_1 + z_2|^2 = (|z_1| + |z_2|)^2$$
$$|z_1|^2 + 2\Re(z_1 \bar{z_2}) + |z_2|^2 = |z_1|^2 + 2|z_1||z_2| + |z_2|^2 $$
$$\Re(z_1 \bar{z_2}) = |z_1\bar{z_2}|$$
Hence 
$$z_1 \bar{z_2} \in \mathbb{R}$$
and
$$arg(z_1) - arg(z_2) = arg(z_1 \bar{z_2}) = 2k\pi$$
A: Writing the complex numbers using the exponential representation
$$ z_1 = r_1 e^{i \arg(z_1)}, z_2 = r_2 e^{i \arg(z_2)} $$
one sees right away that
$$ | z_1 | + | z_2 | = r_1 + r_2 $$
and
$$ | z_1 + z_2 | = | r_1 e^{i \arg(z_1)} + r_2 e^{i \arg(z_2)} |. $$
For these to be equal the angular parts of the representations need to be equal
$$ | z_1 + z_2 | = | z_1 | + | z_2 | \Rightarrow e^{i \arg(z_1)} = e^{i \arg(z_2)} $$
which implies
$$ \arg(z_1) = \arg(z_2) + 2 n \pi $$
or written in the form of the title
$$ \arg(z_1) - \arg(z_2) = 2 n \pi .$$
A: why not use polar form:
$z_1=r_1e^{i\theta_1}$,$z_2=r_2e^{i\theta_2}$
Now putting it in the given relation:$|r_1e^{i\theta_1}+r_2e^{i\theta_2}|=r_1+r_2$
$\implies |r_1e^{i\theta_1}+r_2e^{i\theta_2}|^2=(r_1+r_2)^2$
Simplifying and after cancellation we are left with
$(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta _2)=1$
$\cos(\theta_1-\theta_2)=1$
$\implies \theta_1-\theta_2=2n\pi$
