Lower bound on the argument of difference of two complex number?

Question

Suppose I have two complex numbers $z_{1} = a_{1}+ ib_{1}$ and $z_2 = a_2+ ib_2$. Also $\arg(z_1),\arg(z_2)$ $\in$ $(\pi/2,3\pi/2)$ with $\arg(z_2) > \arg (z_1)$. Can I get a lower bound on $\arg(z_2-z_1)$ in terms of $\arg(z_2)$ and $\arg(z_1)$. Geometrically using parallelogram law it seems that $\arg(z_2-z_1)$ > $\arg(z_2)$. How can I prove it? [Convention used here is angle measured in counter-clockwise direction is positive]