# Given any cubic polynomial with real coefficients, find the argument of the complex roots

I have the cubic equation $$y = 2x^3 - 2x^2 - 6x - 3 = 0$$, which I am given has one real root and two complex conjugate roots. Is it possible, without explicitly finding these roots, to find the argument (angle with the real axis in the complex plane) of these complex conjugates?

This question was inspired a problem of the opposite kind, "form a cubic polynomial with real coefficients whose roots $$\alpha, \beta$$ and $$\gamma$$ satisfy $$\text{Im}(\alpha) = 0$$, $$\text{arg}(\beta) = \pi/6$$, and $$|\gamma| = \sqrt{3}$$ " (The solution is not the above polynomial)

• Do you know the formulas by Cardano? – Dr. Sonnhard Graubner Apr 25 at 14:20
• I have, although I only understand how to use them to solve cubics to find their roots, which is what I wanted to avoid. – Nick_2440 Apr 25 at 14:28
• Using Newton’s identities? – mathcounterexamples.net Apr 25 at 14:50
• I used those equations (I didn't know they were called Newton's identities) for the inverse problem. I cannot see how to answer my question using those alone. – Nick_2440 Apr 25 at 14:58
• Suppose that the roots are $re^{i\theta}, re^{-i\theta}, x$ where $x$ is the real root. You can write 3 equations using Newton’s identities ($r^2x = 3/2$ is one of those). And your goal is to find $\theta$. – mathcounterexamples.net Apr 25 at 15:03