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?  
 A: Squaring both sides, we see this is equivalent to:
$$(z_1+z_2)(\overline z_1+\overline z_2)=z_1\overline z_1 + 2|z_1||z_2| + z_2\overline z_2$$
which is equivalent to:
$$\mathrm{Re}(z_1\overline z_2)=\frac{1}{2}\left(z_1\overline z_2+z_2\overline z_1\right)=|z_1 \overline z_2|$$
So $z_1\overline z_2=\alpha$ must be  non-negative real. Then you get:
$$z_1|z_2|^2 = \alpha z_2$$
So either $z_2=0$ or $z_1=\frac{\alpha}{|z_2|^2}z_2$.
A: 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 $$u+u^{-1}=2\iff u=1\iff\theta_1-\theta_2\equiv 0\mod 2\pi\iff\theta_1\equiv\theta_2\equiv0\mod 2\pi.$$
A: Here's an alternate solution using the Law of Cosines. If we say $z_1=\lvert z_1\rvert\text{cis} \ \theta_1$ and $z_2=\lvert z_2\rvert\text{cis} \ \theta_2$, then:
$$\lvert z_1+z_2 \rvert^2=\lvert z_1\rvert^2+\lvert z_2\rvert^2-2\lvert z_1 \rvert\lvert z_2\rvert\cos(\pi-(\theta_2-\theta_1))$$
Thus, by squaring both sides of $\lvert z_1+z_2 \rvert=\lvert z_1\rvert+\lvert z_2\rvert$, we get:
$$\lvert z_1+z_2 \rvert^2=\lvert z_1\rvert^2+\lvert z_2\rvert^2+2\lvert z_1\rvert\lvert z_2\rvert$$
$$\lvert z_1\rvert^2+\lvert z_2\rvert^2-2\lvert z_1 \rvert\lvert z_2\rvert\cos(\pi-(\theta_2-\theta_1))=\lvert z_1\rvert^2+\lvert z_2\rvert^2+2\lvert z_1\rvert\lvert z_2\rvert$$
Subtract both sides by $\lvert z_1\rvert^2+\lvert z_2\rvert^2$:
$$-2\lvert z_1 \rvert\lvert z_2\rvert\cos(\pi-(\theta_2-\theta_1))=2\lvert z_1\rvert\lvert z_2\rvert$$
Divde both sides by $2\lvert z_1 \rvert\lvert z_2\rvert$:
$$-\cos(\pi-(\theta_2-\theta_1))=1$$
Multiply both sides by $-1$:
$$\cos(\pi-(\theta_2-\theta_1))=-1$$
Take the $\arccos$ of both sides:
$$\pi-(\theta_2-\theta_1)=\pi+2\pi n$$
Solve for $\theta_2-\theta_1$:
$$\theta_2-\theta_1=2\pi n$$
Thus, we have:
$$\theta_2=\theta_1+2\pi n$$
This means $z_2=\lvert z_2 \rvert\text{cis} \ (\theta_1+2\pi n)=\lvert z_2 \rvert \text{cis} \ \theta_1$. Thus: $$z_2=\frac{\lvert z_2 \rvert}{\lvert z_1 \rvert}z_1$$
Clearly, $\frac{\lvert z_2 \rvert}{\lvert z_1 \rvert} \in \Bbb{R}^{\geq 0}$, so we are done.
