Question

If $z_1,z_2,z_3$ represents the vertices of an equilateral trialgle taken in order, then

• A. $\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+\frac{1}{z_3-z_1}=0$
• B. $\frac{1}{z_1-z_2}-\frac{1}{z_2-z_3}-\frac{1}{z_3-z_1}=0$
• C. $\frac{1}{z_1-z_2}=\frac{1}{z_2-z_3}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+\frac{1}{z_3-z_1}=0$

$|z_1-z_2|=|z_2-z_3|=|z_3-z_1|=r$ ($\because △$ is equilateral)
Now, $\frac{1}{z_1-z_2}=\frac{\overline{z_1}-\overline{z_2}}{{|\overline{z_1}-\overline{z_2}|}^2}=\frac{\overline{z_1}-\overline{z_2}}{r^2}$ ...(1)
Similarly, $\frac{1}{z_2-z_3}=\frac{\overline{z_2}-\overline{z_3}}{r^2}$ ...(2)
And $\frac{1}{z_3-z_1}=\frac{\overline{z_3}-\overline{z_1}}{r^2}$ ...(3)
adding (1),(2) and (3), $\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+\frac{1}{z_3-z_1}=0$