If $z_1 =2+ 3i$ ; $\;z_2=3–2i\;$ and $\;z_3= -1–2\sqrt3i,\;$ then how do you get this
$$\arg(z_3/z_2)=2\arg((z_3-z_1)/(z_2-z_1))$$ The correct answer from the test's answer key
This is from the mock test that I had last week. This is the only correct option from the options that were provided. From what I understand, the RHS is the angle between 2 lines formed by $z_3-z_1$ and $z_2-z_1$. I do not know what the LHS means. I need some help on how do i proceed with solving it. The test doesn't have the solution for this sum.
Please help me.
Since the absolute values $|z_1|=|z_2|=|z_3|=\sqrt{13},$ the points $A(2+3i), B(3-2i), C(-1-2\sqrt 3)$ lie on the circle centered at $O$ with radius $\sqrt{13}.$ Then $$\arg\left(\frac{z_3}{z_2}\right)=\arg\left(\frac{z_3-0}{z_2-0}\right)$$ is the central angle in our circle, and $$\arg\left(\frac{z_3-z_1}{z_2-z_1}\right)$$ is an inscribed angle, both related to the chord $BC.$ From this follows the relation $$\arg(z_3/z_2)=2\arg((z_3-z_1)/(z_2-z_1)).$$