Polinomio del grado dos mil.

Notice the golden ratio. When applicable, geometry is very powerful in solving algebra problems. Consider a triangle $ABC$ with $\angle B = \angle C = \frac{2\pi}{5}$ and $\angle A = \frac{\pi}{5}$ .

In $\triangle ABC$ notice that $AB/BC$ is the golden ratio. Using the law of sines $\frac{AB}{BC} = \frac{\sin\frac{2\pi}{5}}{\sin\frac{\pi}{5}} = \frac{2\sin\frac{\pi}{5}\cos\frac{{\pi}}{5}}{\sin\frac{\pi}{5}} = 2\cos\frac{\pi}{5}$

let $x = e^{i\theta}, \frac{1}{x} = e^{-i\theta}$ , $x+\frac{1}{x} = 2\cos\frac{\pi}{5}$

$x^{2000}+\frac{1}{x^{2000}} = 2\cos\frac{2000\pi}{5}$ $= 2\cos400\pi = \boxed{2}$

Let $\phi=\frac{1+\sqrt{5}}2$ . One can easily check that this number, also known as the golden ratio, satisfies $\phi^2-\phi-1=0$ . Substituting $\phi=x+\frac1x$ we obtain:

$0 =x^2-x+1-\frac1x+\frac1{x^2} =\frac1{x^2}(x^4-x^3+x^2-x+1) =\frac{x^5+1}{x^2(x+1)}.$

We see that $x^5=-1$ , meaning that

$x^{2000}+\frac1{x^{2000}} =(-1)^{400}+\frac1{(-1)^{400}} =1+1=\boxed{2}$ .

$x+\frac { 1 }{ x } =\frac { 1+\sqrt { 5 } }{ 2 } \\ multiply\quad by\quad x.\\ { x }^{ 2 }-(\frac { 1+\sqrt { 5 } }{ 2 } )x+1=0\\ \therefore \quad x=\frac { \frac { 1+\sqrt { 5 } }{ 2 } \pm \sqrt { { (\frac { 1+\sqrt { 5 } }{ 2 } ) }^{ 2 }-4 } }{ 2 } \\ x=\frac { \frac { 1+\sqrt { 5 } }{ 2 } \pm \sqrt { \frac { 6+2\sqrt { 5 } }{ 4 } -4 } }{ 2 } ,\quad multiply\quad by\quad \frac { 2 }{ 2 } \\ x=\frac { 1+\sqrt { 5 } }{ 4 } \pm \quad \frac { \sqrt { 2\sqrt { 5 } -10 } }{ 4 } \\ note\quad that\quad 2\sqrt { 5 } -10\quad <0\\ so\quad x\quad should\quad be\quad written\quad as:\\ x=\frac { 1+\sqrt { 5 } }{ 4 } \pm \quad \frac { i\sqrt { 10-2\sqrt { 5 } } }{ 4 } \\ we\quad obtain\quad the\quad modulus\quad and\quad argument\quad of\quad this\quad complex\quad number.\\ r=\sqrt { { (\frac { 1+\sqrt { 5 } }{ 4 } ) }^{ 2 }+{ (\quad \frac { \sqrt { 10-2\sqrt { 5 } } }{ 4 } ) }^{ 2 } } \\ \quad =\sqrt { \frac { 6+2\sqrt { 5 } }{ 16 } +\frac { 10-2\sqrt { 5 } }{ 16 } } =1\\ \theta =\tan ^{ -1 }{ \frac { \sqrt { 10-2\sqrt { 5 } } }{ 1+\sqrt { 5 } } } =36\\ \therefore \quad x=cos(36)\pm i\quad sin(36).\\ according\quad to\quad de-moivre's\quad theorem:\\ { x }^{ 2000 }={ 1 }^{ 2000 }(cos(36\times 2000)\pm i\quad sin(36\times 2000))\\ \quad \quad \quad =cos(200\times 2\pi )\pm i\quad sin(200\times 2\pi )\\ \quad \quad \quad =cos(2\pi )\pm i\quad sin(2\pi )\\ \quad \quad \quad =1+0=1\\ \\ \therefore \quad { x }^{ 2000 }+\frac { 1 }{ { x }^{ 2000 } } =1+\frac { 1 }{ 1 } =\boxed { \quad 2\quad }$

Solving the given equation as a quadratic equation you find that the solutions of $x$ is complex. Hence ${z}^{n}+\frac{1}{{z}^{n}}=2\cos n\theta$ is applicable (which is a formula derived from de Moirve's theorem). Using the given equation you can solve this for $n=1$ and find $\theta$ . $2\cos \theta=\frac{1+\sqrt{5}}{2}\therefore\theta=\frac{\pi}{5}$ . Hence you can plug this value back into ${z}^{n}+\frac{1}{{z}^{n}}=2\cos n\theta$ for $n=2000$ to get ${x}^{2000}+\frac{1}{{x}^{2000}}=2\cos 2000\frac{\pi}{5}=2$ .

simple ,, let x=cos a + i sin a and then u have 1/x= cos a - i sin a (i am assuming modulus of x as 1 here ) then we have 2 cos a = 1+root5/2 so which clearly tells a is 36 degrees, now write down x in complex number form and use demouvires theorem to get the value of Re(e^(i2000a))= 1 ,,, hence adding the real parts of x and 1/x ,, we get 2

Let x=cos∆+isin∆ =>∆ is 36° so GE =2cos2000∆ which is nothing but 2