Prove that complex numbers $z_1$, $z_2$ and $\frac{z_1-iz_2}{1-i}$ form a right triangle

Question

For reference this should be able to be solved using division in the complex plane, but I have been trying for a few days with no luck.

Any help would be greatly appreciated.

Answer 1

Let the three vertexes be $A=z_1$, $B=z_2$ and $C= \frac{z_1-iz_2}{1-i}$. Then

$$BC = \frac{z_1-iz_2}{1-i} - z_2 = \frac{z_1-z_2}{1-i} ,\>\>\>\>\> CA = z_1-\frac{z_1-iz_2}{1-i} = \frac{-i(z_1-z_2)}{1-i} $$

Evaluate

$$\frac{BC}{CA} = i = e^{i\frac\pi2}\implies \text{Arg} (BC) - \text{Arg} (CA) = \frac\pi2$$

i.e. the sides $BC$ and $CA$ of the triangle are at a right angle.