Question

Let the complex numbers $z_1,z_2,z_3$ represent the vertices of an equilateral triangle. Then prove that $\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+\frac{1}{z_3-z_1}=0$.

Answer 1

Since $z_1,z_2,z_3$ represent the vertices of an equilateral triangle .

Then $|z_1-z_2|=|z_2-z_3|=|z_3-z_1|$.......(1).

Now $\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+\frac{1}{z_3-z_1}$

$=\frac{\overline{(z_1-z_2)}}{|z_1-z_2{|}^2}+\frac{\overline{(z_2-z_3)}}{|z_2-z_3{|}^2}+\frac{\overline{(z_3-z_1)}}{|z_3-z_1{|}^2}$

$=\frac{\overline{\left(z_1\right)}-\overline{\left(z_2\right)}}{|z_1-z_2{|}^2}+\frac{\overline{\left(z_2\right)}-\overline{\left(z_3\right)}}{|z_1-z_2{|}^2}+\frac{\overline{\left(z_3\right)}-\overline{\left(z_1\right)}}{|z_1-z_2{|}^2}$ [Using (1) and for complex number $a$ and $b$,$\overline{a+b}=\overline{a}+\overline{b}$]

$=0$.