I am recently studying complex numbers. If there is given 6 points, $a,b,c,d,e,f$ and I want to show that $\angle ABC = \angle DEF$. My approach: First, translate a and c by minus b. Since, $$arg z_{1}z_{2} = arg z_{1} + arg z_{2}$$ And we get $$arg(\frac{a-b}{c-b})= \angle ABC$$ We can repeat the same thing and get $$arg(\frac{d-e}{f-e}) = \angle DEF$$ Thus, $$arg(\frac{a-b}{c-b}) = arg(\frac{d-e}{f-e})$$ $$\frac{a-b}{c-b} = \frac{d-e}{f-e}$$ Is my approach correct? I am not very sure with my answer.
Edit: What if all the points lie on the unit circle?
If you can show that $$ \arg\left(\frac{a-b}{c-b}\right) = \pm\arg\left(\frac{d-e}{f-e}\right) \tag1$$ then you have shown the angles are equal in magnitude. If you also care about the orientation of the angles, delete the $\pm$ sign.
But consider the following list of six complex numbers that all lie on the unit circle:
\begin{align} a&=1, &b&=\exp(-i\pi/6),& c&=\exp(i\pi/6),\\ d&=1,& e&=-1,& f&=\exp(i\pi/6). \end{align}
Then $$ \frac{a-b}{c-b} = \frac12 - i\left(\frac{1 - \sqrt3}2\right) = \sqrt{2 - \sqrt3} \exp\left(-\frac\pi{12}\right) $$ whereas $$ \frac{d-e}{f-e} = 1 - i(2 - \sqrt3) = 2\sqrt{2 - \sqrt3} \exp\left(-\frac\pi{12}\right). $$
That is, $$ \arg\left(\frac{a-b}{c-b}\right) = \arg\left(\frac{d-e}{f-e}\right) = -\frac\pi{12} $$ but $$ \left(\frac{a-b}{c-b}\right) = \frac12\left(\frac{d-e}{f-e}\right) \neq \left(\frac{d-e}{f-e}\right). $$
So it is correct to compare the arguments of the two ratios, but not to compare the two ratios themselves.
Knowing that all six of $a, b, c, d, e, f$ lie on the unit circle, however, you might be able to use the inscribed angle theorem. The thing to watch out for there is that you then need either that $b$ and $e$ are both on the smaller arc between the endpoints of their angles, or both on the larger arc between the endpoints of their angles, and this might be tedious to test.
Another alternative (which works for any six points, like Equation $(1)$) is to compute $$ \left(\frac{a-b}{c-b}\right) \left(\frac{f-e}{d-e}\right). $$ If the result is a positive real number then the angles $\angle ABC$ and $\angle DEF$ are equal and oriented in the same direction. On the other hand, if $$ \left(\frac{a-b}{c-b}\right) \left(\frac{f-e}{d-e}\right) $$ is a positive real number then the angles $\angle ABC$ and $\angle DEF$ are equal in magnitude and oriented in opposite directions.