Prove It!

Given complex number u, v and their product uv:

Show that the triangle formed by uv, v and uv-v is similar to that formed by u, 1 and u-1.

Here is my proof:

From Fig. 1, we see that

\angle COB = \theta_{uv},  \angle DOB = \theta_v, \angle AOB = \theta_u.

It follows that

\angle COD = \angle COB - \angle DOB = \theta_{uv}-\theta_v \overset{\theta_{uv} = \theta_u + \theta_v}{=} \theta_u + \theta_v - \theta_v = \theta_u = \angle AOB.


\frac{|u\cdot v|}{|u|} \overset{|u\cdot v|=|u|\cdot|v|}{=} \frac{|u|\cdot|v|}{|u|} = |v| =\frac{|v|}{1} = \frac{|v|}{|1|}\implies \frac{OC}{OA}=\frac{OD}{OB}.


\left(\angle COD = \angle AOB, \frac{OC}{OA} = \frac{OD}{OB}\right) \implies \triangle OCD \sim \triangle OAB.