1
$\begingroup$

Let $y=\frac{x}{1+x}$, where $$x={{\omega^{2009}}^{2009}}^{. ...\text{upto}\;2009\;\text{times}}$$ and $\omega$ is a complex cube root of $1$. Then $y$ is

$a) \omega;\quad b)-\omega;\quad c)\omega^2; \quad d)-\omega^2.$

My attempt:

We have $\omega^{2009}=\omega^2$. Now taking $2009$ power in both side gives ${(\omega^{2009}})^{2009}={(\omega^2)}^{2009}={(\omega^{2009})}^{2}=\omega.$

But ${\omega^{2009}}^{2009}\neq{(\omega^{2009}})^{2009}$, since ${a}^{m^n}\neq a^{mn}$.

I am stuck here. So give me only hints. (If it is required to change the tags of the problem then please feel free to change it)

$\endgroup$
  • 2
    $\begingroup$ The value of $\omega^n$ depends only on the residue of $n$ modulo 3. The value of $2009^m$ modulo 3 depends only on the residue of $m$ modulo 2. $\endgroup$ – Gerry Myerson Apr 11 '18 at 9:43
  • 2
    $\begingroup$ related math.stackexchange.com/questions/2251030/… $\endgroup$ – David Quinn Apr 11 '18 at 10:01
  • $\begingroup$ Ok.. I understood. Thanks $\endgroup$ – SAHEB PAL Apr 11 '18 at 10:08
  • $\begingroup$ If you understand, Saheb, let me encourage you to write it up and post it as an answer. $\endgroup$ – Gerry Myerson Apr 12 '18 at 10:04
  • $\begingroup$ Yes. It will be greatful if you post as an answer. $\endgroup$ – SAHEB PAL Apr 12 '18 at 12:11
1
$\begingroup$

We have $2009 \equiv - 1(mod\;3)$. Also ${{{2009}^{2009}}^{2009}}^{...}$ is odd.

Therefore $${{2009}^{2009}}^{. …\text{upto}\;2009\;\text{times}}\equiv - 1(mod\;3).$$ $\therefore{{2009}^{2009}}^{. …\text{upto}\;2009\;\text{times}}=3k-1$ for some integer $k$.

Thus $$x={{\omega^{2009}}^{2009}}^{. …\text{upto}\;2009\;\text{times}}=\omega^{3k-1}=\omega^{-1}=\omega^2.$$

$$\therefore y=\frac{x}{1+x}=\frac{\omega^2}{1+\omega^2}=\frac{\omega^2}{-\omega}=-\omega.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.