Question

If $z_1,z_2,z_3$ are three distinct complex numbers such that $\frac{3}{|z_2-z_3|}=\frac{4}{|z_3-z_1|}=\frac{5}{|z_1-z_2|},$ then $\frac{9}{z_2-z_3}+\frac{16}{z_3-z_1}+\frac{25}{z_1-z_2}$ equals

• A. 0.00
• B. 0
• C. 0.0
• D. 00

Answer 1

Answer: (A), (B), (C), (D)

$\frac{3}{|z_2-z_3|}=\frac{4}{|z_3-z_1|}=\frac{5}{|z_1-z_2|}=k$ (say)
$\Rightarrow \frac{9}{|z_2-z_3{|}^2}=\frac{16}{|z_3-z_1{|}^2}=\frac{25}{|z_1-z_2{|}^2}=k^2$

$\frac{9}{|z_2-z_3{|}^2}=k^2$
$\Rightarrow \frac{9}{z_2-z_3}=k^2(\overline{z_2}-\overline{z_3})\dots \left(1\right)$ $[\because |z{|}^2=z\cdot \overline{z}]$

Similarly, $\frac{16}{(z_3-z_1)}=k^2(\overline{z_3}-\overline{z_1})\dots \left(2\right)$
and $\frac{25}{(z_1-z_2)}=k^2(\overline{z_1}-\overline{z_2})\dots \left(3\right)$
Adding $\left(1\right),\left(2\right),\left(3\right),$
$\frac{9}{z_2-z_3}+\frac{16}{(z_3-z_1)}+\frac{25}{(z_1-z_2)}=k^2(\overline{z_2}-\overline{z_3}+\overline{z_3}-\overline{z_1}+\overline{z_1}-\overline{z_2})=0$
Hence, the value of given expression is $0$