Different complex numbers ${z_1,z_2,z_3,z_4}$ are roots of a polynomial ${z^4+az^3+bz^2+cz+d}$ where ${a,b,c,d}$ are real numbers. We also know that:
${|z_1|=|z_2|=|z_3|=|z_4|=1}$ and ${Im(z_1-z_2)=Im(z_3-z_4)=0}$
Prove that:
${a=c=0, d=1}$ and ${-2<b<2}$
From vieta's formulas we get:
${1) x_1+x_2+x_3+x_4=-a}$
${2) x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=b}$
${3) x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=-c}$
${4) x_1x_2x_3x_4=d}$
From the last one we have
${|x_1x_2x_3x_4|=|d|}$
${|x_1||x_2||x_3||x_4|=|d|}$
${|d|=1*1*1*1=1}$
if we divide ${3)}$ with ${4)}$
${\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+\frac{1}{z_4}=\frac{-c}{d}}$
I also tried:
${P(z_1)=z_1^4+az_1^3+bz_1^2+cz_1+d=0}$
${P(z_2)=z_2^4+az_2^3+bz_2^2+cz_2+d=0}$
${P(z_1)-P(z_2)=z_1^4-z_2^4+az_1^3-az_2^3+bz_1^2-bz_2^2+cz_1-cz_2=0}$
I end up with something like this:
${(z_1-z_2)[(z_1+z_2)(z_1^2+z_2^2)+a(z_1^2+z_1z_2+z_2^2)+b(z_1+z_2)+c]=0}$
the same can be done with ${P(z_3)}$ and ${P(z_4)}$
but I don't really see it going places, I also thought about the fact that if a polynomial has a root z then ${\overline z}$ must also be its root.
Best Answer
Hint: $\;$ $\operatorname{Im}(z_1-z_2) = 0 \iff \operatorname{Im}(z_1)=\operatorname{Im}(z_2)$. Since $|z_1|=|z_2|=1$ and $z_1 \ne z_2$ this requires $\operatorname{Re}(z_1) \ne 0$ and $z_2 = -\bar z_1$.
Similarly $\operatorname{Im}(z_3-z_4) = 0 \iff \operatorname{Im}(z_3)=\operatorname{Im}(z_4)$ and $\operatorname{Im}(z_1+z_2+z_3+z_4) = \operatorname{Im}(-a) = 0$ $\implies \operatorname{Im}(z_3)=\operatorname{Im}(z_4)$ $=-\operatorname{Im}(z_1)$. Since $|z_3|=|z_4|=1$ and $z_3 \ne z_4$ this requires $\{z_3, z_4\}=\{-z_1, \bar z_1\}$, and $z_3 \ne z_{1,2}$ requires $\operatorname{Im}(z_1) \ne 0$.
Then the polynomial must be $(z-z_1)(z+\bar z_1)(z+z_1)(z-\bar z_1) = \dots$
[ EDIT ] $\;$ The $\,b,c,d \in \mathbb R\,$ part of the premise is not needed and is not used in the above.