Limit of $\lim _{ n->\infty }{ ({ z+{ z }^{ -1 }) }^{ n } } $ I'd like to compute the limit of $\lim _{ n->\infty  }{ ({ z+{ z }^{ -1 }) }^{ n } } $ for $z\neq 0$
My attempt was to use the Polar-Cor. representation for complexe numbers.So
${ (z+{ z }^{ -1 })^{n} }$=${ { (r }_{ 1 }(\cos { \theta  } +i\sin { \theta )+\frac { 1 }{ { r }_{ 2 }(\cos { \phi +i } \sin { \phi ) }  }  } ) }^{ n }$
and using Moivre's Formula we get
${ { r }_{ 1 } }^{ n }{ e }^{ i*n*\theta  }+\frac { 1 }{ { { { r }_{ 2 } }^{ n } }{ e }^{ i*n*\phi  } } $. How do I continue from here?
 A: Let 
$$z+\frac1z=w.$$
Obviously, $w^n$ converges to zero if $|w|<1$ or to one if $w=1$.
By solving the above identity for $z$,
$$z=\frac{w\pm\sqrt{w^2-4}}2=\frac{re^{i\theta}\pm\sqrt{r^2e^{i2\theta}-4}}2$$ with $r<1$, or
$$z=\frac{1\pm i\sqrt3}2.$$
A: Hint. Let $w=z+1/z$ and consider three cases: 1) $|w|<1$, 2) $|w|=1$, and $|w|>1$. Moreover note that for $z=re^{i\theta}$ with $r>0$,  $$w=(r+\frac{1}{r})\cos(\theta)+i(r-\frac{1}{r})\sin(\theta)$$
and
$$|w|^2=(r+\frac{1}{r})^2\cos^2(\theta)+(r-\frac{1}{r})^2\sin^2(\theta)=
r^2+\frac{1}{r^2}+2\cos(2\theta).$$
P.S. $w$ is on ellipse 
$$\frac{x^2}{(r+1/r)^2} + \frac{y^2}{(r-1/r)^2}=1$$.
A: As pointed out in the other answer the behavior is dependent of $|z+z^{-1}|$.
With $z=re^{i \theta}$ you have:
$$ \left|z +\frac{1}{z} \right|^2=\left|r e^{i \theta} +\frac{1}{r}e^{-i \theta} \right|^2=r^2+\frac{1}{r^2}+2 \cos(2 \theta)$$
to compare this expression with $1$  you can notice that:
$$ \left|z +\frac{1}{z} \right|^2-1=\frac{1}{r^2} \left[(r^2)^2+(2\cos(2 \theta)-1) r^2+1 \right]$$
so you only have to study the sign of a quadratic function ($X^2+(2\cos(2\theta)-1)X+1$).
