Limit of $\lim _{ n->\infty }{ ({ z+{ z }^{ -1 }) }^{ n } } $

Question

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?

Answer 1

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 $$\frac{x^2}{(r+1/r)^2} + \frac{y^2}{(r-1/r)^2}=1$$.

Answer 2

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 ($X^2+(2\cos(2\theta)-1)X+1$).

Answer 3

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

$$z=\frac{1\pm i\sqrt3}2.$$