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

Question

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

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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!!

Answer 1

Hint: $\omega + 2\omega ^2 + 3\omega ^3 +\cdots+9\omega ^9 = g(\omega)$, where $g(x)=xf'(x)$ and $f(x)=1+x+\cdots+x^9=\dfrac{x^{10}-1}{x-1}$.

Answer 2

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 $x=20^\circ$

Answer 3

Denote $w = e^{i \alpha}$ with $ \alpha = 40°$

$$\frac{w}{1- w} = \frac{e^{i \alpha}}{1- e^{i \alpha}}=\frac{e^{i \alpha/2}}{e^{-i \alpha/2}- e^{i \alpha/2}} = \frac{e^{i \alpha/2}}{-2i\sin\left(\alpha/2)\right)}$$

Hence

$$\left|\frac{9w}{1-w}\right|^{-1} = \frac{2\sin\left(20°\right)}{9}$$