How to prove this graphically on a complex plane?

Question

If |z1 + z2| = |z1| + |z2|, then prove that arg(z1) = arg(z2)

If |z1 - z2| = |z1| + |z2|, then prove that arg(z1) - arg(z2) = pie

Answer 1

Consider a paralelogram $ABCD$ on the picture. Points $A,B,C,D$ correspond to compleks numbers $0,z_1,z_2,z_1+z_2$. If $|z_1+z_2|=|z_1|+|z_2|$ then triangles $ABD$ and $ACD$ are degnerate so $B$ and $C$ are on $AD$ and thus $z_1$ and $z_2$ have the same argument.

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Answer 2

a) Let $w=z_2$ and $z=z_1$ then we have $$|z+w|=|z|+|w|$$ so after squaring we get $$zz'+zw'+wz'+ww'= zz'+2|zw|+ ww'$$ Let $u=zw'$, then $${u+u'\over 2 } = |u|$$

so $$R(u) = |u| \implies Im(u) = 0 \implies zw'\in \mathbb{R}$$

So there is a real number $k$ so that $zw' =k$ so $z|w|^2 = kw$ so $z$ is real multiple of $w$ and that is iff $z$ and $w$ are collinear with $0$ in complex plane.

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I assume the second one goes similary.