If $\text{ }\big(x-\frac{1}x\big)=i\sqrt{2}$. Then compute $\bigg(x^{2187}-\frac{1}{x^{2187}}\bigg)$. Here $i=\sqrt{-1}$ 
QUESTION: If $\text{ }\big(x-\frac{1}x\big)=i\sqrt{2}$ , $\text{ }$then compute $$\bigg(x^{2187}-\frac{1}{x^{2187}}\bigg)$$ Here $i=\sqrt{-1}$ .


MY ANSWER: I have done it using the Quadratic formula and De Moivre's Theorem. Let me write down my working before I propose my doubt.. Here's how I did it..
Solving the equation we get $$x^2-(i\sqrt{2})x-1=0$$ $$\implies x=\frac{i\sqrt{2} \pm \sqrt{(i\sqrt{2})^2+4}}{2} $$ $$\implies x=\frac{i\sqrt{2}\pm\sqrt{2}}{2}$$ Take $x=(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)=e^{\frac{i\pi}4}$
Now we know that $2187=(273\times8)+3$
$$\therefore x^{2187}=e^{2187\times \frac{i\pi}4}=e^{(273\times 2\pi + \frac{3\pi}4)i}=e^{\frac{{3\pi}}{4}i}=\frac{i-1}{\sqrt{2}}$$
$$\therefore x^{2187}-\frac{1}{x^{2187}}= \frac{i-1}{\sqrt{2}}-\frac{\sqrt{2}}{i-1}$$ $$=\frac{(i-1)^2-2}{(i-1)\sqrt{2}}$$ $$=\frac{2}{\sqrt{2}}\frac{(1+i)}{(1-i)}$$ $$=\frac{\sqrt{2}}{2} (1+i)^2$$ $$=\boxed{\sqrt{2}i}$$


Now my first question is that, the quadratic relation gave us two different values for $x$. One with which I have worked out to reach the answer of $\sqrt {2}i$ and the other, $\big(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\big)$ which I had left behind. Now working with that I find that the angle turns out to be $\frac{\pi}{10}$ and stuff becomes much more complicated after that. The official answer to this one is $\sqrt{2}i$ (which matches with what I have found out).




My doubt is why do we not consider the other value of $x$ ?




And is there any alternative (preferably simpler) method(s) to solve this one?


Thank You so much for your help and support.. :)

In fact, it is easy to verify that both values of $x$ yield the same result. For the whole problem, you just need De Moivre's formula twice (two lines of paper without explanation).
For $x=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$, you have shown the answer is $i\sqrt 2$.
Now, let $x=-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i=\cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4}$. Using the De Moivre's formula and the fact that
$$z-\frac{1}{z}=2i\sin(\arg(z))$$
you get
$$x^{2187}-\frac{1}{x^{2187}} = x^3-\frac{1}{x^3} = 2i\sin\frac{9\pi}{4}=i\sqrt 2$$ Done!
