2

Consider the cubic equation $$az^3-bz^2+\bar{b}z-\bar{a}=0$$ where $a$ and $b$ are non-zero complex numbers. Suppose $z_1, z_2$ and $z_3$ are the roots.

Question: Which $a$ and $b$ gives $|z_1|=|z_2|=|z_3|=1$ ?

We can easily prove $|z_1z_2z_3|=1$ and if $z$ is a root, then $\dfrac{1}{\bar{z}}$ is also a root. This implies that there is a root with absolute value $1.$
But after that how can we continue ?

Bumblebee
  • 18,220
  • 5
  • 47
  • 87
  • 1
    One of the roots must be real if all the coefficients are real. In this case, that is not true, so all the roots might be complex. – marty cohen Apr 28 '16 at 03:05
  • @martycohen: $z^3-z^2+z-1=(z-1)(z-i)(z+i)$ – Bumblebee Apr 28 '16 at 04:00
  • Your example has all the coefficients real, so it must have a real root. OPs equation can have all non-real coefficients, so it can have all complex roots. – marty cohen Apr 28 '16 at 04:10
  • How does this imply that tthere is a root with absolute value 1? – miracle173 Apr 28 '16 at 06:31
  • @miracle173: Suppose $z_3=\dfrac{1}{\bar{z_1}},$ Then $|z_1z_3|=1.$ Hence $|z_2|=1.$ – Bumblebee Apr 28 '16 at 06:44
  • And why can I suppose that $z_3=\dfrac{1}{\bar{z_1}}$? – miracle173 Apr 28 '16 at 06:49
  • @miracle173: If all are on the unit circle, we have the result. Suppose at least one of them ($z_1$) is not. Then $\dfrac{1}{\bar{z_1}}(\not=z_1)$ is also satisfies the given cubic and it is not on the unit circle. – Bumblebee Apr 28 '16 at 06:53
  • 1
    this is related: http://math.stackexchange.com/questions/866081/what-is-the-condition-for-roots-of-conjugate-reciprocal-polynomials-to-be-on-the – Chip Apr 29 '16 at 02:54

1 Answers1

3

For $z^{3}+w z^{2}+\bar{w} z+1=0$, there's a paper discussed. Still need to check for your case.

Ng Chung Tak
  • 18,990