Question

If $\frac{a}{|z_1-z_2|}=\frac{b}{|z_2-z_3|}=\frac{c}{|z_3-z_1|}\text{where}(a,b,c\in R),$ then value of $\frac{a^2}{z_1-z_2}+\frac{b^2}{z_2-z_3}+\frac{c^2}{z_3-z_1}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is A $0$

Given$\frac{a}{|z_1-z_2|}=\frac{b}{|z_2-z_3|}=\frac{c}{|z_3-z_1|}$
Let $\frac{a}{|z_1-z_2|}=\frac{b}{|z_2-z_3|}=\frac{c}{|z_3-z_1|}={\lambda }^{\prime }$
On squaring both sides, we get $\frac{a^2}{|z_1-z_2{|}^2}=\frac{b}{|z_2-z_3{|}^2}=\frac{c^2}{|z_3-z_1{|}^2}={\lambda }^{\prime 2}=\lambda \cdots \left(i\right)$
We know $|z_1-z_2{|}^2=(z_1-z_2)({\overline{z}}_1-{\overline{z}}_2)$
$\therefore a^2=\lambda (z_1-z_2)({\overline{z}}_1-{\overline{z}}_2)\cdots \left(ii\right)$
$b^2=\lambda (z_2-z_3)({\overline{z}}_2-{\overline{z}}_3)\cdots \left(iii\right)$
$c^2=\lambda (z_3-z_1)({\overline{z}}_3-{\overline{z}}_1)\cdots \left(iv\right)$ for $\lambda \in R.$
By putting this values of $a^2,b^2,c^2$ in $\frac{a^2}{z_1-z_2}+\frac{b^2}{z_2-z_3}+\frac{c^2}{z_3-z_1}$, we get
$\sum \frac{a^2}{z_1-z_2}=0$