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

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

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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.. :)

Answer 1

$2187=3^7$. This is a clue. Powers of $3$ are significant. Now $$\left(x-\frac1x\right)^3=(i\sqrt2)^3=-2i\sqrt2$$ and $$\left(x-\frac1x\right)^3=x^3-\frac1{x^3}-3\left(x-\frac1x\right) =x^3-\frac1{x^3}-3i\sqrt2.$$ So $$x^3-\frac1{x^3}=i\sqrt2.$$ Repeating this, $$x^9-\frac1{x^9}=i\sqrt2,$$ $$x^{27}-\frac1{x^{27}}=i\sqrt2$$ etc. Eventually, $$x^{2187}-\frac1{x^{2187}}=i\sqrt2.$$

Answer 2

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!

Answer 3

Your result from the quadratic formula can also be handled in this fashion. Since $$ \ z \ = \ \frac{i\sqrt{2}\pm\sqrt{2}}{2} \ \ \Rightarrow \ \ z_1 \ = \ \frac{1}{\sqrt{2}} · (1 + i) \ \ = \ \ e^{\pi / 4} \ \ , \ \ z_2 \ = \ \frac{1}{\sqrt{2}} · (-1 + i) \ \ = \ \ e^{3\pi / 4} \ \ , $$ we have $ \ \frac{1}{z_{1}} \ = \ e^{-\pi / 4} \ = \ \overline{z_1} \ \ $ and, similarly, $ \ \frac{1}{z_{2}} \ = \ \overline{z_2} \ \ . $ We then confirm that $$ z_{1,2} \ - \ \frac{1}{z_{1,2}} \ \ = \ \ z_{1,2} \ - \ \overline{z_{1,2}} \ \ = \ \ 2 · \mathfrak{Im}(z_{1,2}) \ \ = \ \ 2 · \left(\frac{i}{\sqrt2} \right) \ \ = \ \ i·\sqrt2 \ \ . $$

We note the following properties:

• $ \ \ -z_1 \ = \ \frac{1}{\sqrt{2}} · (-1 - i) \ \ = \ \ e^{-3\pi / 4} \ = \ \overline{z_2} \ \ $ and $ \ -z_2 \ = \ \frac{1}{\sqrt{2}} · (1 - i) \ \ = \ \ e^{-\pi / 4} \ = \ \overline{z_1} \ \ ; $

• $ \ \ \ z_1^4 \ = \ e^{4 \ · \ (\pi / 4)} \ = \ -1 \ = \ z_2^4 \ = \ e^{4 \ · \ (3\pi / 4)} \ \ \Rightarrow \ \ z_{1,2}^8 \ = \ 1 \ \ . $

From the foregoing, we obtain $$ z_{1,2}^3 \ = \ -\frac{1}{z_{1,2}} \ \ = \ \ -\overline{z_{1,2}} \ \ . $$

We may then calculate that $$ z^{2187} \ - \ \frac{1}{z^{2187}} \ \ = \ \ z^{(8·273) \ + \ 3} \ - \ \frac{1}{z^{(8·273) \ + \ 3}} \ \ = \ \ (z^8)^{273} · z^3 \ - \ \frac{1}{(z^8)^{273} · z^3} $$ $$ = \ \ 1 · z^3_{1,2} \ - \ \left(\frac{1}{1· z_{1,2}} \right)^3 \ \ = \ \ z^3_{1,2} \ - \ (\overline{z_{1,2}})^3 \ \ = \ \ -\overline{z_{1,2}} \ - \ \left( \ -\overline{\overline{z_{1,2}}} \ \right) \ \ = \ \ z_{2,1} \ - \ \overline{z_{2,1}} \ \ , $$ which we have seen earlier is equal to $ \ i·\sqrt2 \ \ . $

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ADDENDUM (12/14) --

Another approach makes use of the factorization of a "difference of odd powers": we may write $$ z^{2187} \ - \ z^{-2187} $$ $$ = \ \ (z \ - \ z^{-1}) \ · \ (z^{2186} \ + \ z^{2185} · z^{-1} \ + \ z^{2184} · z^{-2} \ + \ \ldots \ + \ z^{2} · z^{-2184} \ + \ z^{1} · z^{-2185} \ + \ z^{-2186}) $$ $$ = \ \ (z \ - \ z^{-1}) \ · \ (z^{2186} \ + \ z^{2184} \ + \ z^{2182} \ + \ \ldots \ + \ z^2 \ + \ z^0 \ + \ z^{-2} \ + \ \ldots \ + \ z^{-2182} \ + \ z^{-2184} \ + \ z^{-2186}) \ \ . $$

From the results you derived, we obtain $$ z_1^2 \ = \ e^{\pi / 2} \ = \ i \ \ \ , \ \ \ z_2^2 \ = \ e^{3\pi / 2} \ = \ -i \ \ , $$

from which it follows that $$ z_1^2 \ + \ z_1^4 \ + \ z_1^6 \ + \ z_1^8 \ \ = \ \ i \ + \ (-1) \ + \ (-i) \ + \ 1 \ \ = \ \ 0 \ \ , $$ $$ z_2^2 \ + \ z_2^4 \ + \ z_2^6 \ + \ z_2^8 \ \ = \ \ (-i) \ + \ (-1) \ + \ i \ + \ 1 \ \ = \ \ 0 \ \ , $$ and, since $ \ \frac{1}{z_{1,2}} \ = \ \overline{z_{1,2}} \ \ , $ we also have $$ z_{1,2}^{-2} \ + \ z_{1,2}^{-4} \ + \ z_{1,2}^{-6} \ + \ z_{1,2}^{-8} \ \ = \ \ \overline{0} \ \ = \ \ 0 \ \ . $$

The "central terms" of the summation-factor are $$ \ldots \ + \ (z^8 \ + \ z^6 \ + \ z^4 \ + \ z^2) \ + \ z^0 \ + \ (z^{-2} \ + \ z^{-4} \ + \ z^{-6} \ + \ z^{-8}) \ + \ \ldots $$ and with $ \ z_{1,2}^8 \ = \ 1 \ \ , $ we can write, for instance, $$ z^{10} \ + \ z^{12} \ + \ z^{14} \ + \ z^{16} \ \ = \ \ z^8 \ · \ (z^2 \ + \ z^4 \ + \ z^6 \ + \ z^8) \ \ = \ \ 0 $$ and $$ z^{-10} \ + \ z^{-12} \ + \ z^{-14} \ + \ z^{-16} \ \ = \ \ z^{-8} \ · \ (z^{-2} \ + \ z^{-4} \ + \ z^{-6} \ + \ z^{-8}) \ \ = \ \ 0 \ \ . $$

Since $ \ 2184 \ = \ 273·8 \ \ , $ our factorization reduces to just $$ z_{1,2}^{2187} \ - \ z_{1,2}^{-2187} \ \ = \ \ (z_{1,2} \ - \ z_{1,2}^{-1}) \ · \ (z_{1,2}^{2186} \ + \ z_{1,2}^0 \ + \ z_{1,2}^{-2186}) \ \ . $$

It was given that $ z_{1,2} - \frac{1}{z_{1,2}} \ = \ i\sqrt2 \ $ and we have said that $ \ \frac{1}{z_{1,2}} \ = \ \overline{z_{1,2}} \ \ . $ Hence, $$ z_{1,2}^{2187} \ - \ z_{1,2}^{-2187} \ \ = \ \ (i·\sqrt2) \ · \ (z_{1,2}^{2186} \ + \ 1 \ + \ \overline{z_{1,2}^{2186}}) \ \ . $$

Finally, $ \ z_{1,2}^{2186} \ = \ z_{1,2}^{273·8 \ + \ 2} \ = \ z_{1,2}^2 \ \ , $ leading to $$ z_1^{2187} \ - \ z_1^{-2187} \ \ = \ \ (i·\sqrt2) \ · \ ( \ i \ + \ 1 \ + \ \overline{i} \ ) \ \ = \ \ (i·\sqrt2) \ · \ 1 $$ and $$ z_2^{2187} \ - \ z_2^{-2187} \ \ = \ \ (i·\sqrt2) \ · \ ( \ -i \ + \ 1 \ + \ \overline{-i} \ ) \ \ = \ \ (i·\sqrt2) \ · \ 1 \ \ . $$