Does $|z_1+z_2|=|z_1|+|z_2|\implies z_1=kz_2$? Let $z_1,z_2\in \mathbb C$. I was wondering if 

$$|z_1+z_2|=|z_1|+|z_2|\implies z_1=k z_2\ \ or\ \ z_2=0,\quad k\in\mathbb R^+.$$

Is it true? 
I am having problems trying to show it. I'm sure it's easy to show using brute force, but if it's true, is there an elegant way to show it?  

Not exactly. It is equivalent to $z_1=\lambda z_2$ or $z_2=\lambda z_1$, $\;\lambda \in \mathbf R_+$, or, if they're both non-zero, to the fact that $z_1$ and $z_2$  have the same argument.
To prove it, we may suppose both are non-zero. So set $z_k=r_k\mathrm e^{\mathrm i \theta_k}, \enspace k=1,2$. Squaring both sides, we have to check under which conditions we have
\begin{align*}&(r_1\mathrm e^{\mathrm i \theta_1}+r_2\mathrm e^{\mathrm i \theta_2})(r_1\mathrm e^{-\mathrm i \theta_1}+r_2\mathrm e^{-\mathrm i \theta_2})=(r_1+r_2)^2\\
\iff& r_1^2+r_2^2+r_1r_2\bigl(\mathrm e^{\mathrm i(\theta_1-\theta_2)}+\mathrm e^{-\mathrm i(\theta_1-\theta_2)}\bigr)=(r_1+r_2)^2\\
\iff&\mathrm e^{\mathrm i(\theta_1-\theta_2)}+\mathrm e^{-\mathrm i(\theta_1-\theta_2)}=2.
\end{align*}
Setting $u=\mathrm e^{\mathrm i(\theta_1-\theta_2)}$, this means $$u+u^{-1}=2\iff u=1\iff\theta_1-\theta_2\equiv 0\mod 2\pi\iff\theta_1\equiv\theta_2\equiv0\mod 2\pi.$$
