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?  

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.
