If $\omega =\cos 40 + i\sin 40$ then $|\omega + 2\omega ^2 + 3\omega ^3....9\omega ^9|^{-1}$ If $\omega =\cos 40 + i\sin 40$ then $|\omega + 2\omega ^2 + 3\omega ^3....9\omega ^9|^{-1}$

All angles are in degrees
I will be writing $\omega$ as w to make it easier to type
From the value of $w$ it is clear that $w^9=1$
Solving the arithmetic geometric progression we get the the sum inside the modulus as 
$$\frac{9w}{1-w}$$
Taking the inverse of the modulus ie.
$$\left|\frac{9w}{1-w}\right|^{-1}$$
I don’t know how to proceed. Please help.
The answer is $\frac 29 \sin 20$
Thanks!! 

If $S=\omega + 2\omega ^2 + 3\omega ^3+\cdots+9\omega ^9$
$(1-\omega)S=\omega + \omega ^2 + \omega ^3+\cdots+\omega ^9-9\omega ^{10}$
$=1+\omega + \omega ^2 + \omega ^3+\cdots+\omega^8-9w$ as $\omega ^9=1$
$=\dfrac{1-\omega^9}{1-w}-9w=-9w$
$$S=\dfrac{-9\omega}{1-\omega}=\dfrac9{1-\omega^{-1}}$$
$1-\omega^{-1}=1-(\cos2x-i\sin2x)=2\sin x(\cos x+i\sin x)$
$\dfrac1{1-\omega^{-1}}=\dfrac1{2\sin x(\cos x+i\sin x)}=\dfrac{\cos x-i\sin x}{2\sin x}$
Here $x=20^\circ$
