If $\frac{a}{|z_2-z_3|}=\frac{b}{|z_3-z_1|}=\frac{c}{|z_1-z_2|}$, then why is $\frac{a^2}{z_2-z_3}+\frac{b^2}{z_3-z_1}+\frac{c^2}{z_1-z_2}=0$?

Question

If $\dfrac{a}{|z_2-z_3|}=\dfrac{b}{|z_3-z_1|}=\dfrac{c}{|z_1-z_2|}$, then why is $\dfrac{a^2}{z_2-z_3}+\dfrac{b^2}{z_3-z_1}+\dfrac{c^2}{z_1-z_2}=0$?

This is a complex number property given in my book but the derivation is not given. How do I get this result? I refer to the chapter complex numbers of the book 39 years chapterwise topicwise solved papers by Arihant.

Answer 1

Let $\frac{a}{|z_2-z_3|}=\frac{b}{|z_3-z_1|}=\frac{c}{|z_1-z_2|}=k$, then $a=k|z_2-z_3|,b=k|z_3-z_1|,c=k|z_1-z_2|$. So $$\frac{a^2}{z_2-z_3}=k^2(\bar{z_2}-\bar{z_3});$$ $$\frac{b^2}{z_3-z_1}=k^2(\bar{z_3}-\bar{z_1});$$ $$\frac{c^2}{z_1-z_2}=k^2(\bar{z_1}-\bar{z_2}).$$ This implies $$\frac{a^2}{z_2-z_3}+\frac{b^2}{z_3-z_1}+\frac{c^2}{z_1-z_2}=0.$$

Answer 2

Assume $a,b,c\in \mathbb{R}$.

Let $z_2-z_3=ka e^{i\alpha}$, $z_3-z_1=kbe^{i\beta}$ and $z_1-z_2=kce^{i\gamma}$ where $k\ne 0$

\begin{align} ka e^{i\alpha}+kbe^{i\beta}+kce^{i\gamma} &= (z_2-z_3)+(z_3-z_2)+(z_1-z_2) \\ a e^{i\alpha}+be^{i\beta}+ce^{i\gamma} &= 0 \tag{1} \\ \frac{a^2}{z_2-z_3}+\frac{b^2}{z_3-z_1}+\frac{c^2}{z_1-z_2} &= \frac{a}{ke^{i\alpha}}+\frac{b}{ke^{i\beta}}+\frac{c}{ke^{i\gamma}} \\ &= \frac{a e^{-i\alpha}+be^{-i\beta}+ce^{-i\gamma}}{k} \\ \end{align}

Equating real and imaginary parts for $(1)$, the result follows.