Proving an inequality related to complex numbers.

Question

If $$\sum_{i=1}^4b_iz_i=0,\sum_{i=1}^4b_i=0,|z_i|=r$$ How can one prove that $$b_1b_2|z_1-z_2|^2=b_3b_4|z_3-z_4|$$

---

I tried LHS: $$b_1b_2|z_1-z_2|^2=b_1b_2(z_1-z_2)(\bar z_1-\bar z_2)=b_1b_2(2r-z_1\bar z_2-z_2\bar z_1)$$

Answer 1

I suppose that the $b_j$ are real numbers (if not, the example in my comment show a contradiction). Wlog, we can suppose that the $z_j$ are of modulus one. Then taking the conjugate of the first equality, we get $b_1z_1^{-1}+b_2 z_2^{-1}+b_3z_3^{-1}+b_4 z_4^{-1}=0$.

Now write $b_1z_1+b_2z_2=-(b_3z_3+b_4z_4)$, $b_1z_1^{-1}+b_2z_2^{-1}=-(b_3z_3^{-1}+b_4z_4^{-1})$, and multiply these equality. We get that $$b_1^2+b_2^2+2b_1b_2-2b_1b_2+b_1b_2z_1^{-1}z_2+b_1b_2z_1z_2^{-1}$$ is equal to $$b_3^2+b_4^2+2b_3b_4-2b_3b_4+b_3b_4z_3^{-1}z_4+b_3b_4z_3z_4^{-1}$$

Using $$(b_1+b_2)^2=(b_3+b_4)^2$$ we get $$2b_1b_2-b_1b_2z_1^{-1}z_2-b_1b_2z_1z_2^{-1}=2b_3b_4-b_3b_4z_3^{-1}z_4-b_3b_4z_3z_4^{-1}$$ hence $$b_1b_2|z_1-z_2|^2=b_3b_4|z_3-z_4|^2$$ and we are done.