How complexity does a number achieve #1

$ln(\dfrac{\omega^\omega}{\omega^{\omega^2}}$ ) $=$ $ln(\omega^\omega) - ln(\omega^{\omega^2}$ ) $=$ $(\omega- \omega^2)ln(\omega)$ .

Let $ln(\omega)=y$ so $\omega=e^{y}$ . $\omega$ is a cube root of unity also known as $\cos \theta +i\sin \theta$ or $e^{i\theta}$ where $\theta = \dfrac{2\pi}{3}$ .

So $y=\dfrac {2i\pi}{3}$ . Since $\omega^2+\omega+1=0.$ Then $\omega-\omega^2 =2\omega+1$ .

$(2\omega+1)(\dfrac{2i\pi}{3})=(2(\dfrac{-1 +i\sqrt3}{2}) +1)(\dfrac {2i\pi}{3} )= i^2 \dfrac{2\pi\sqrt3}{3} = \boxed{\frac{-2\pi}{\sqrt3}}$ .

We have $\omega$ as a complex root. and so according to condition we shift the denominator part to the numerator as follows:- $\omega^{\omega-\omega^2}=e^{-2\pi/\sqrt3}$ ..(it is a property). and thus we find out logarithm of this:- $ln(\omega)^{\omega-\omega^2}=\boxed{\dfrac{-2\pi}{\sqrt3}}$