Find $x_1^n+x_2^n$ on any quadratic equation, general case. I have a simple quadratic (with $x^2$) equation, x can Be complex too:
$$x^2+x+1=0$$
But it could be any equation, the equation above is just an example. I need to compute $x_1^{10}+x_2^{10}$, but it could have another exponents (ex: $x_1^{50}+x_2^{50}$).
I need to know, on a general case, how to find $x_1^n+x_2^n,\ n\in\mathbb{N}\ ax^2+bx+c=0,\ a\ is\ not\ 0$?
I ask this because I have to create a software which computes this (user writes the equation and the number n = exponent) and I can't find the roots always, because sometimes are complex. I think I should make use of Viete, but I don't know how to compute $x_1^n+x_2^n$.
Thank you very much!!
 A: One approach is to represent the roots in polar form, and take advantage of their symmetry.  For the sake of ease of exposition, let the quadratic be monic; that is, $a = 1$.  If it's not already in that form, it's trivial to convert it to a monic form.
If the roots are real—if $b^2-4c \geq 0$—I assume you know how to handle that.  So we'll just consider the case where they're not real.  In that case, $b^2-4c < 0$, and the roots are given by
$$
x_{1, 2} = \frac{-b \pm \sqrt{b^2-4c}}{2}
         = \frac{-b}{2} \pm \frac{\sqrt{4c-b^2}}{2}i
$$
Note that $c > 0$ necessarily if the roots are not real.  We can see, using the Pythagorean theorem, that $|x_1| = |x_2| = \sqrt{c}$.  Now, find $\theta, 0 \leq \theta \leq \pi$ such that
$$
\cos\theta = -\frac{b}{2\sqrt{c}}
$$
$$
\sin\theta = \sqrt{1-\frac{b^2}{4c}}
$$
This can be done using the atan2 function in many programming languages.  Then $x_{1, 2}$ can be represented as $\sqrt{c} \text{ cis } (\pm\theta) \equiv \sqrt{c} \left[\cos(\pm\theta)+i\sin(\pm\theta)\right]$.  Then
$$
x_1^n = \sqrt{c^n} \text{ cis }(n \theta)
$$
$$
x_2^n = \sqrt{c^n} \text{ cis } (-n\theta)
$$
Since $x_1^n$ and $x_2^n$ form a conjugate pair (just as $x_1$ and $x_2$ do), we therefore have
$$
x_1^n+x_2^n = 2\sqrt{c^n}\cos (n\theta)
$$
A: Newton-Girard's relations are very general, so we can redo the computations in the case of two roots.
Set $x+y= s; \enspace xy=p$. We want to compute recursively the sums of powers 
$$P_n=x^n+y^n$$
as polynomials in $s$ and $p$. 
Initialisation:
$$P_0=2,\quad P_1=s=-1, \quad P_2=x^2+y^2=s^2-2p=-1.$$
Recursion relation:
\begin{align*}
P_{n+1}&=(x^n+y^n)(x+y)-x^ny-xy^n=(x^n+y^n)(x+y)-xy(x^{n-1}+y^{n-1})\\
&=sP_n-pP_{n-1}.
\end{align*}
Thus,
\begin{align*}
P_3&=sP_2-pP_1=s^3-3ps\\
P_4&=sP_3-pP_2=s^4-4ps^2+2p^2\\
P_5&=sP_4-sP_3=s^4-5ps^3+5p^2s\\
\&c.&\;\&c.
\end{align*}
However, for this precise equation, we have $s=-1,\enspace p=1$ and the recursion relation becomes
$$P_{n+1}=-(P_n+P_{n-1}),$$
so the computation is very simple:
\begin{align*}
P_0&=2,&P_1&=-1,&P_2&=-1,& P_3&=2,&P_4&=-1,&P_5&=-1.\end{align*}
This is more than enough to see the sequence  $(P_n)$ is periodic and that
$$P_n=\begin{cases}2&\text{if}\enspace n\equiv 0\mod 3,\\
-1&\text{if}\enspace n\not\equiv 0\mod 3.\end{cases} $$
So the sums are
$$x_1^{10}+x_2^{10}=x_1^{50}+x_2^{50}=-1.$$
A: \begin{align*}
  \alpha+\beta &= -\frac{b}{a} \\
  \alpha \beta &=  \frac{c}{a} \\
  \alpha^n+\beta^n  &=
  (\alpha+\beta)^{n}+\sum_{k=1}^{\left \lfloor \frac{n}{2} \right \rfloor}
  \binom{n-k}{k} \frac{n(-\alpha \beta)^{k} (\alpha+\beta)^{n-2k}}{n-k} \\
  &=
  (\alpha+\beta)^{n}-n\alpha \beta (\alpha+\beta)^{n-2}+
  \frac{n(n-3)}{2!} \alpha^2 \beta^2 (\alpha+\beta)^{n-4}-\ldots \\
\end{align*}
