I have a following polynomial. (See WolframAlpha ):

$$x^9-6x^8+14x^7-16x^6+36x^5-56x^4+ 24x^3-320x+\frac{640}{9}=0 \tag{1}$$

Wolfram says that $(1)$ has three real roots and three pairs of complex conjugate roots.

I know one of the real roots of this polynomial is equal to:

$$x_1=\tan \left(\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right) =0.22267663636945$$

I know this is a root, because I derived the polynomial from this expression.


By @IvanNeretin's comment we also have two other real roots:

$$x_2=\tan \left(\frac{\pi}{3} +\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$$

$$x_3=\tan \left(\frac{2\pi}{3} +\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$$

A very surprising (for me) discovery: all six complex roots have the same absolute value of the imaginary part:

$$b=Im(x)= \pm \frac{\sqrt[3]{2}}{\sqrt{3}} (1+\sqrt[3]{2})= \pm 1.643902181980216,~~~~\text{for all } x \notin \mathbb{R}$$

I found it using ISC, but hadn't found closed forms for the real parts.

How can we find explicitly the real parts of the complex roots?

My thoughts - if we substitute in $(1)$:

$$x=a \pm i b$$

then we obtain two equations for one real varibale $a$ (for each three pairs of complex roots of course), and by adding and subtracting these equations we can lower their order.

Since there should be three distinct real parts $a$, we could probably bring it down to a cubic equation.

But I need a CAS for that and I'm not sure it will lead to a solution.

  • $\begingroup$ @IvanNeretin, could you please explain this? My algebra probably stayed at school level. This seems connected to the three cubic roots of unity $\endgroup$
    – Yuriy S
    Jun 23 '16 at 11:45

The other two real roots, quite expectedly, are $\tan \left({\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$ and $\tan \left({2\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$. The complex roots are produced in a similar manner, remembering that $\tanh$ is a periodic function with imaginary period.

Long story short, all nine roots are $$\tan \left({\pi\over3}\cdot m+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left({\pi i\over3}\cdot n+\frac{1}{3} \text{arctanh} \frac{1}{3} \right),\;n,m\in\{0,1,2\}$$ A rigorous explanation is beyond my expertise. Informally speaking, inverse trigonometric and inverse hyperbolic functions are multivalued, but the polynomial "knows nothing" of this; to it, they are all the same, so if one value is a root, so must be the others.

  • $\begingroup$ This is great, thank you. $\endgroup$
    – Yuriy S
    Jun 23 '16 at 11:54
  • $\begingroup$ I figured out how to get the real parts explicitly, so I'm accepting your answer, thanks again $\endgroup$
    – Yuriy S
    Jun 23 '16 at 11:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.