Question

If $z_{1},z_{2}$ and $z_{3}$ are $3$ distinct complex numbers such that $\frac{3}{\left|z_{1} - z_{2}\right|}=\frac{5}{\left|z_{2} - z_{3}\right|}=\frac{7}{\left|z_{3} - z_{1}\right|}$ , then the value of $\frac{9}{z_{1} - z_{2}}+\frac{25}{z_{2} - z_{3}}+\frac{49}{z_{3} - z_{1}}$ is equal to

• A. $0$
• B. $1$
• C. $-1$
• D. $15$

Answer 1

Answer: (A)

Let, $\frac{3}{\left|z_{1}-z_{2}\right|}=\frac{5}{\left|z_{2}-z_{3}\right|}=\frac{7}{\left|z_{3}-z_{1}\right|}=k$
$\frac{9}{\left|z_{1}-z_{2}\right|^{2}}=\frac{25}{\left|z_{2}-z_{3}\right|^{2}}=\frac{49}{\left|z_{3}-z_{1}\right|^{2}}= k ^{2}$
$\Rightarrow \frac{9}{\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)}=k^{2} \Rightarrow \frac{9}{z_{1}-z_{2}}=k^{2}\left(\bar{z}_{1}-\bar{z}_{2}\right)$
similarly, $\frac{25}{\left(z_{2}-z_{3}\right)}=k^{2}\left(\bar{z}_{2}-\bar{z}_{3}\right)$
and $\frac{49}{\left(z_{3}-z_{1}\right)}=k^{2}\left(\bar{z}_{3}-\bar{z}_{1}\right)$
$\Rightarrow \frac{9}{z_{1}-z_{2}}+\frac{25}{z_{2}-z_{3}}+\frac{49}{z_{3}-z_{1}}$
$=k^{2}\left(\bar{z}_{1}-\bar{z}_{2}+\bar{z}_{2}-\bar{z}_{3}+\bar{z}_{3}-\bar{z}_{1}\right)=0$