If we have two angles $$\phi_1,\phi_2\in[0,2\pi]$$ such that $$\phi_1\le\phi_2$$ and we perform standard addition on complex numbers from the subset of $$S=\{ z\in \mathbb{C} : arg(z) \in[\phi_1 ,\phi_2]\}$$ do we get a closed algebraic structure $(S,+)$? In other words, for every $z_1,z_2 \in S$, is $z_1+z_2 \in S$ ?
Intuitively it makes sense to me that this should be closed, since if we add 2D geometrical vectors (represented by the complex numbers) whose arguments are bounded, using the geometrical rules for vector addition (eg. the parallelogram rule) the resultant vector (complex number) has a bounded argument to the same boundaries as the initial two vectors.
I can't find any counterexamples, but I also cannot figure out how to show closedness. I've expressed the arguments of $z_1=x_1+iy_1$ and$z_2=x_2+iy_2$ as $\theta_1 = \arctan{(y_1/x_1)}$ and $\theta_2 = \arctan{(y_2/x_2})$, but I don't see how to express the argument of their sum $arg(z_1+z_2)=\theta_{12} = \tan(\frac{y_1+y_2}{x_1+x_2})$ in terms of their arguments or how to show that $\theta_{12}\in[\phi_1 ,\phi_2]$.
I know that I can use the identity for the tan of an angle sum (which we would have if we were multiplying the complex numbers $z_1$ and $z_2$) to express it in terms of the tans of the $z_1$ and $z_2$, but I don't see that as being the case here.
EDIT: Note to self - when looking for counter-examples, look more at the extremes.
If $\phi_2-\phi_1<\pi$ you get a sector in the complex plane between two half-lines. That's closed under addition. When $\phi_2-\phi_1=\pi$ you get a half-plane, but when $\phi_1-\phi_0>\pi$ you get a sector with a reflex angle, and that's not closed under addition.
As a concrete example, take $\phi_1=0$, $\phi_2=3\pi/2$, $z_1=1$ and $z_2=-i$. Then $z_1+z_2=1-i\notin S$.