Let $x-\frac 1x=\sqrt 2 i$. Then find the value of $x^{2187}-\frac{1}{x^{2187}}$

Question

Solving the quadratic equation, we find x to be $$x=\frac{i \pm 1}{\sqrt 2}$$ Moving back to the original expression $$\frac{(x^2)^{2187}-1}{x^{2187}}$$ $$=\frac{-i-1}{x^{2187}}$$ I don’t know how to solve further

The answer is $\sqrt 2 i$

Answer 1

Rewrite the given as $x^2-\sqrt 2 i x - 1=0$, which has the solution

$$ x = \frac{i\pm 1}{\sqrt2} = ie^{\pm i\frac\pi4}\implies x^{\pm 2187}=x^{\pm 3}$$

Thus,

$$x^{2187}-\frac1{x^{2187}}=x^{3}-\frac1{x^{3}}=\left(x-\frac1x\right)^3+3\left(x-\frac1x\right)=(\sqrt2 i)^3+3\sqrt2 i=\sqrt2 i$$

Answer 2

Hint: $\left( \dfrac{i\pm1}{\sqrt2}\right)^8=1$

Answer 3

Since $x=\pm\exp\frac{\pm\pi i}{4}$, $x^8=1$ and$$x^{2187}-x^{-2187}=x^3-x^{-3}=\pm\left(\exp\frac{\pm3\pi i}{4}-\exp\frac{\mp3\pi i}{4}\right)=2i\Im\exp\frac{3\pi i}{4}=i\sqrt{2}.$$

Answer 4

Let $x^{3^n}-x^{-3^n}=A_n.$ Then $(A_n)^3=A_{n+1}-3A_n.$ So if $A_n=i\sqrt 2$ then $A_{n+1}=i\sqrt 2=A_n.$ We are given $A_0=i\sqrt 2.$ Therefore $$i\sqrt 2=A_0=A_1=A_2=A_3=A_4=A_5=A_6=A_7=x^{2187}-x^{-2187}.$$

Answer 5

Hint

$$(x-1/x)^2=-2$$

$$\implies x^4+1=0,x^4=?$$

As $2187=594(4)-1,$

$x^{2187}=(x^4)^{594}x^{-1}=\cdots=\dfrac1x$