Question

State True=1 and False=0

If $z_1,z_2,z_3$ are three distinct complex numbers and p, q, r are three positive real numbers such that $\frac{p}{|z_2-z_3|}=\frac{q}{|z_3-z_1|}=\frac{r}{|z_1-z_2|}$ then $\frac{p^2}{z_2-z_3}+\frac{q^2}{z_3-z_1}+\frac{r^2}{z_1-z_2}=0$.

Answer 1

$p^2=k^2{|z_2-z_3|}^2=k^2(z_2-z_3)\left(\overline{z_2-z_3}\right)$
$=k^2(z_2-z_3)(\overline{z_2}-\overline{z_3})$
$\therefore \frac{p^2}{z_2-z_3}=k^2(\overline{z_2}-\overline{z_3})$

Similarly, $\frac{q^2}{z_3-z_1}=k^2(\overline{z_3}-\overline{z_1})$ and $\frac{r^2}{z_1-z_2}=k^2(\overline{z_1}-\overline{z_2})$

$\therefore \frac{p^2}{z_2-z_3}+\frac{q^2}{z_3-z_1}+\frac{r^2}{z_1-z_2}=k^2\left[\right(\overline{z_2}-\overline{z_3})+(\overline{z_3}-\overline{z_1})+(\overline{z_1}-\overline{z_2}\left)\right]=0$
Hence, $\frac{p^2}{z_2-z_3}+\frac{q^2}{z_3-z_1}+\frac{r^2}{z_1-z_2}=0$ is true.