# If $\omega$ is a primitive cube root of unity, simplify ${\omega}^{{2009}^{{2009}^{{2009}^{\cdots 2009}}}}$

Let $$\omega$$ be a primitive cube root of unity. Let $$x = {\omega}^{{2009}^{{2009}^{{2009}^{\cdots 2009}}}}$$ (up to $$2009$$ times). Simplify the value of $$x$$.

My attempt: Let $$m = {{2009}^{{2009}^{{2009}^{\cdots 2009}}}}$$ (up to $$2007$$ times). Then since $$2009$$ is odd so $$2009^m$$ is also odd. Let $$k = 2009^m$$. Now since $$k$$ is an odd integer so $$2^k \equiv 2\ (\text {mod}\ 3)$$. Also $$2009 \equiv 2\ (\text {mod}\ 3)$$. Therefore, $$2009^k \equiv 2\ (\text {mod}\ 3)$$. Let $$n = 2009^k$$. Then $$n = 3k' + 2$$ for some $$k' \in \Bbb N$$. Therefore $$x = {\omega}^n = {\omega}^2$$

Am I right? Please verify it.

• You did not say what $\omega$ is – DIdier_ May 20 at 7:10
• @Didlier $\omega$ is a primitive cube root of unity. – math maniac. May 20 at 7:11
• Yep, you're right – Shashwat1337 May 20 at 7:15
• As which primitive root $\omega$ is is not specified and the answer seems to be unambiguous, then it must be $1$. – Yves Daoust May 20 at 7:17
• @Yves Daoust what must be $1$? If $\omega$ is a primitive cube root of unity then how can $\omega^2 = 1$? Usually $\frac {-1 + \sqrt 3 i} {2}$ is considered as $\omega.$ But then $\omega^2 = \frac {-1 - \sqrt 3 i} {2} \neq 1.$ – math maniac. May 20 at 7:20

The answer is $$\omega^2$$ and your method is right and check my method $$2009^{odd}$$can be written as $${(2010-1)}^{odd}$$. That is $$2010 \times m - 1$$($$m$$ is some integer), as $$2010$$ is divisible by $$3$$ it is $$\frac1{\omega} = \omega^2$$
$$2009$$ can be expressed as $$6m-1$$ where $$m$$ is any integer
Now observe that $$(6m-1)^{6m-1}\equiv-1\pmod6,$$ so again of the form $$6m'-1$$
If $$w$$ is a root of unity,
$$w^{6n-1}=(w^3)^{2n-1}w^2=?$$