(Complex Numbers) I've almost solved this problem but can't understand which particular case is this problem talking about.

Question

If $x,y,z$ are the roots of the equation $x^3 + px^2 + qx + p = 0$, prove that $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = n\pi$ except in one particular case.

Note: $n$ is any integer.

Answer 1

The given polynomial is $n^3+pn^2+qn+p=0$ $\qquad$

NOTE: im using $n$ as the variable to avoid confusion with it being the root

Vieta's Formula gives us that $x+y+z= -p$

$ xy+yz+zx = q$

$xyz = -p$

$\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(z) = \tan^{-1}\bigg( \frac{x+y+z -xyz}{1-(xy+yz+zx)}\bigg)$

$\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(z) = \tan^{-1}\bigg( \frac{-p+p}{1-q}\bigg)$

$\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(z) = \tan^{-1}(0)$

$\implies\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(z) = n\pi$ $\qquad ,n\in Z$

Notice that the above is not defined when $1-q = 0\implies q =1$

Answer 2

Hint:

$$\tan(\tan^{-1}x+\tan^{-1}y+\tan^{-1}z)=\dfrac{x+y+z-xyz}{1-(xy+yz+zx)}$$

Now we have $$x+y+z=xyz=-\dfrac p1$$

and $xy+yz+zx=\dfrac q1$

But $\dfrac{x+y+z-xyz}{1-(xy+yz+zx)}$ will be undefined if $xy+yz+zx=q=1$