The value of $y=\frac{x}{1+x}$, where $x={{\omega^{2009}}^{2009}}^{. …\text{upto}\;2009\;\text{times}}$

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)

• 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. – Gerry Myerson Apr 11 '18 at 9:43
• – David Quinn Apr 11 '18 at 10:01
• Ok.. I understood. Thanks – SAHEB PAL Apr 11 '18 at 10:08
• If you understand, Saheb, let me encourage you to write it up and post it as an answer. – Gerry Myerson Apr 12 '18 at 10:04
• Yes. It will be greatful if you post as an answer. – SAHEB PAL Apr 12 '18 at 12:11

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