<s>[INST] Calculate $\lim_{n\to\infty} ((a+b+c)^n+(a+\epsilon b+\epsilon^2c)^n+(a+\epsilon^2b+\epsilon c)^n)$ Calculate $\lim_{n\to\infty} ((a+b+c)^n+(a+\epsilon b+\epsilon^2c)^n+(a+\epsilon^2b+\epsilon c)^n)$ with $a,b,c \in \Bbb R$ and $\epsilon \in \Bbb C \setminus \Bbb R, \epsilon^3=1$.
Since $a+\epsilon b + \epsilon^2c=\overline {a+\epsilon^2b+\epsilon c}$, the expression above should be real, but I don't know how to deal with the $\epsilon$ in order to bring it to a form where I can calculate its limit. 
Also, for this problem, $|a+b+c|<1$ and $ab+bc+ac=0$.
 [/INST] Let
$$ z_1 = a + b + c \\
z_2 = a + \epsilon b + \epsilon^2 c \\\
z_3 = a + \epsilon^2 b + \epsilon c $$
Observe that $z_1 \in \Bbb{R}$, and $z_2 = \overline {z_3}$, so there is some $r$ and $\theta$ such that $z_2 = re^{i \theta}$ and $z_3 = re^{-i\theta}$.
$$r^2 = z_2 z_3 = (a+\epsilon b+\epsilon^2 c)(a + \epsilon^2 b+\epsilon c) = a^2 + b^2 + c^2 + \epsilon (ab + ac + bc) + \epsilon^2 (ab + ac + bc)$$
Plugging in $ab + bc + ac = 0$ twice we get
$$r^2 = a^2 + b^2 + c^2$$
Adding twice again $ab + bc + ac = 0$, we get
$$r^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = (a+b+c)^2$$
leading to the conclusion
$$|a+b+c| = r = |z_2| =  |z_3| = |z_1| < 1$$
which means the limit must be 0.
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