Prove that $arg(z') = \frac{\pi}{2}+(\vec {BM}\;;\vec{AM}) +2k\pi$ where k is an integer. _Consider the points M of affix $z$ $(z \neq 2i)$ , and $M'$ of affix $z'$ such that $z'= \frac{i(z-2)}{z-2i}$ . A and B are the points of respective affixes 2 and 2i_. Show that $arg(z') = \frac{\pi}{2}+(\vec {BM}\;;\vec{AM}) +2k\pi$ where k is an integer. I tried to solve the question by substituting $z'$ by $\frac{i(z-2)}{z-2i}$ and finding the argument but nothing worked. How do I solve it? Thanks.

Argument of a product is equal to the sum of arguments, argument of a ratio is equal to the difference of arguments.

Angle of 2 vectors is equal to the difference of arguments of their affices: the last one minus the first one.

Therefore $$\arg(z')= \arg\left(\frac{i(z-2)}{z-2i}\right)=\arg(i)+\arg(z-2)-\arg(z-2i).$$ Since $$z-2=z_M-z_A=z_{\small{\vec{AM}}}\quad \text{and}\quad z-2i=z_M-z_B=z_{\small{\vec{BM}}},$$ we have $$\arg(z')=\frac{\pi}{2}+(\vec{BM}\;;\vec{AM}) +2k\pi,\;k\in\mathbb{Z}.$$