Complex Enough Yet?

Same method with Ruslan Abdulgani . I am putting it in proper LaTex.

$x+\dfrac{1}{x} = \sqrt{3}\quad \Rightarrow x^2-\sqrt{3}x +1 = 0 \\ \Rightarrow x = \dfrac {\sqrt{3}\pm\sqrt{3-4}}{2} = \frac{\sqrt3}{2} \pm \frac{1}{2}i = \cos {\frac{\pi}{6}} \pm \sin{\frac{\pi}{6}}i = e^{\pm \frac{\pi}{6}i}$

$\begin{aligned} \Rightarrow x^{200}+\dfrac{1}{x^{200}} & = e^{\frac{200\pi}{6}i}+e^{-\frac{200\pi}{6}i} \\ & = e^{\frac{4\pi}{3}i}+e^{-\frac{4\pi}{3}i} \\ & = -\frac{1}{2} + \frac{\sqrt3}{2}i -\frac{1}{2} - \frac{\sqrt3}{2}i \\ & = \boxed{-1} \end{aligned}$

Another way to solve this is using Newton's Sums method. We know that:

$\alpha + \beta = \sqrt{3}\quad \Rightarrow \alpha \beta = 1$

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta = 3 - 2 = 1$

For $\begin{aligned} n > 2\quad \Rightarrow \alpha^n + \beta^n & = (\alpha + \beta)(\alpha^{n-1} + \beta^{n-1}) - \alpha \beta (\alpha^{n-2} + \beta^{n-2} ) \\ & = \sqrt{3}(\alpha^{n-1} + \beta^{n-1}) - (\alpha^{n-2} + \beta^{n-2} ) \end{aligned}$

Using an Excel spreadsheet, we can find out the values of $\alpha^{n-2} + \beta^{n-2}$ see below:

It is noted that the values of $\alpha^{n} + \beta^{n}$ repeat in a cycle of $12$ . For $200 \text{ mod } 12 = 8$ we have $\alpha^{200} + \beta^{200} = \alpha^8 + \beta^8 = \boxed{-1}$

$x^2 - \sqrt 3 x + 1 = 0 \\ \Rightarrow x= \dfrac{\sqrt 3 \pm i}{2}=e^{\frac{i \pi}{6}}$ .

So $x^{200} + \dfrac{1}{x^{200}} = e^{\frac{i100 \pi}{3}}+e^{\frac{-i100 \pi}{3}}=-1.\square$

Let's us define a sequence $s_{n}=x^n+(1/x)^n$ where $s_{0}=2$ and $s_{1}=x+1/x =\sqrt{3}$ . It is easy to prove that that sequence satisfies the following recursion $s_{n+1}=\sqrt{3} s_{n} -s_{n-1}$ and then by mathematical induction one can prove that $s_{n} =2 \cos( n*\pi/6)$ . Therefore $s_{200}=2 \cos(200*\pi/6)= 2 \cos(32\pi+4\pi/3)=2 \cos(4\pi/3)=2(-1/2)=-1$ .

Just to show another approach:

We use the following identities, remembering that $x^n*\dfrac 1 {x^n}=1$ .

If $x^n+\dfrac 1 {x^n}=y \\ x^{2n}+\dfrac 1 {x^{2n}} =y^2-2, \qquad (1) \\ x^{3n}+\dfrac 1 {x^{3n}} =y^3-3y. \qquad (2 \\{x^a+\dfrac 1 {x^a}\}*\{x^b+\dfrac 1 {x^b}\}\\ = x^{a+b}+\dfrac 1 {x^{a+b}}+x^{a-b}+\dfrac 1 {x^{a-b}}, \qquad (3)}\\ x+\dfrac 1 {x}=\sqrt 3. \\ x^2+\dfrac 1 {x^2}=1 using \ (1) \qquad (4) \\ x^3+\dfrac 1 {x^3}= 0 \ using \ (2) \qquad (5) \\\displaystyle x^5+\dfrac 1 {x^5}=-\sqrt 3 ~~from~(4)*(5)~~using~(3) \qquad (6) \\\displaystyle x^{10}+\dfrac 1 {x^{10}} =1~~from~(6)~~using~(1) \qquad (7) \\\displaystyle x^{20}+\dfrac 1 {x^{20}} =-1~~from~(7)~~using(1) \qquad (8) \\\displaystyle x^{40}+\dfrac 1 {x^{40}}~=-1~~from~(8)~~using(1) \qquad (9) \\\displaystyle x^{80}+\dfrac 1 {x^{80}}~=-1~~from~(9)~~using(1) \qquad (10) \\\displaystyle x^{120}+\dfrac 1 {x^{120}}~=2~~from~(9)*(10)~~using(3) \qquad (11) \\\displaystyle x^{160}+\dfrac 1 {x^{160}}=-1~~from~(11)~~using(1) \qquad (12) \\\displaystyle {\Large \color{#D61F06}{x^{200}+\dfrac 1 {x^{200}}=-1} }~~from~(9)*(12)~~using(3)$

Note:

For integer

$n>0,x^{10*2^n}+\dfrac 1 {x^{10*2^n}} =-1 \qquad (8) \text{ to } (10) \\ x^{3^n}+\dfrac 1 {x^{3^n}} =0$

and such other.