For complex vectors $z_1$ and $z_2$, How do I show that if $|z_1+z_2|=|z_1|+|z_2|$ then the vectors $z_1$ and $z_2$ are parallel or antiparallel. In my complex analysis class, we went over a geometric proof of this with the triangle inequality, but I'm trying to find a more algebraic proof. I'm also trying not to use Arg because we haven't gone over it in my class yet and all I really know about an argument is that if $z_1$, $z_2$, and 0 are collinear than they'd all have the same angle off the real axis.
I haven't really gotten particularly far. I can convert the absolute values (based on the definition of modulus that, for a complex number $z=x+iy$, $|z|=sqrt{(x^2+y^2)}$). I can then square both sides of the equation twice and then simplify to get something that looks really easy to work with, but I'm not sure what the results of that could tell me about how the three points are collinear.
Any help would be much appreciated.

I will answer the question in the heading (not including $0$).
Let $z_k=x_k+iy_k$  Square both sides and remove common terms to get: $(x_1+iy_1)(x_2-iy_2)+(x_1-iy_1)(x_2+iy_2)=(2(x_1x_2+y_1y_2))=2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}$.  Now square both sides and eliminate common terms to get $2x_1x_2y_1y_2=x_1^2y_2^2+x_2^2y_1^2$   Assume none of the terms $=0$, divide by the product and get $2=\frac{x_1y_2}{x_2y_1}+\frac{x_2y_1}{x_1y_2}$  Notice that the two terms on the right are reciprocal, su that with $u=$ one ratio, we have a quadratic $u^2-2u+1=0$, so that the ratio $=1$, forcing the vectors to be scalar multiples of each other.
