If $z_1$ and $z_2$ are two non-zero complex numbers such that $|z_1+z_2|=\left|z_1\right|+\left|z_2\right|,$ then arg $\left(z_1\right)-$ arg $\left(z_2\right)$ is equal to 

Let $z_1=r_1 (\cos {\theta }_1+i\sin {\theta }_1),z_2=r_2 (\cos {\theta }_2+i\sin {\theta }_2)$

$\therefore |z_1+z_2|=[{(r_1\cos {\theta }_1+r_2\cos {\theta }_2)}^2$ $+{(r_2\sin {\theta }_1+r_2\sin {\theta }_2)}^2{]}^{1/2}$

$={\left[r_1^2+r_2^2+2r_1{r}_2\cos ({\theta }_1-{\theta }_2)\right]}^{1/2}={\left[{(r_1+r_2)}^2\right]}^{1/2}$     $(\because |z_1+z_2|=\left|z_1\right|+\left|z_2\right|)$

Therefore $\cos ({\theta }_1-{\theta }_2)=1\Rightarrow {\theta }_1-{\theta }_2=0\Rightarrow {\theta }_1={\theta }_2$

Thus arg  $\left(z_1\right)-arg\left(z_2\right)=0$

Trick : $|z_1+z_2|=\left|z_1\right|+\left|z_2\right|\Rightarrow z_1,z_2$ lies on same straight line.

$\therefore arg{z}_1=arg{z}_2\Rightarrow arg{z}_1-arg{z}_2=0$