Complex Number problem (dup.) I can’t solve this problem.

When $w=\cos 20^\circ +i\sin 20^\circ$, $\dfrac 1 {|w+2w^2+3w^3+\ldots+18w^{18}|} = {}$?

First, I solved the equation of the denominator using the De Moivre's formula. I don't know if it's right, but the value of real part came out simply. But the value of imaginary part is too complicated.
Secondly, I thought about Euler's formula, but I couldn't solve it. The answer seems to come in the form of sin10 or sin20.
 A: Let's simplify this denominator.
It is akin to a geometric series.
$(w+2w+\cdots +17w^{17}+18 w^{18})\\
\frac {(1-w)(w+2w^2 + \cdots+17w^{17}+18 w^{18})}{w-1}\\
\frac{w - w^2 + 2w^2 - 2w^3 + \cdots + 17w^{17} - 17w^{18} + 18 w^{18} - 18w^{19}}{1-w}\\
\frac {w + w^2 + \cdots + w^{18} - 18w^{19}}{1-w}$
Now we have an actual geometric series plus one more term.
$\frac {w + w^2 + \cdots + w^{18}}{1-w} - \frac {18w^{19}}{1-w}\\
\frac {(1-w)(w + w^2 + \cdots + w^{18})}{(1-w)^2} - \frac {18w^{19}}{1-w}\\
\frac {w-w^{19}}{(1-w)^2} - \frac {18w^{19}}{1-w}$
$w = e^{\frac {i\pi}{9}}\\
w^{19} = e^{\frac {i\pi}{9}+2\pi i} = w$
If this looks alien to you, you can find the same result using de Moivre's formula. $(\cos 20^\circ + i\sin 20^\circ)^{18} = 1$
So, our denominator simplifies to $-\frac {18w}{1-w}$
$\frac {|1-w|}{|18w|}$
$|w| = 1$
$|1-w|$ equals the length of the segment that is the base of an equilateral triangle with vertex angle $20^\circ$
$\frac {|1-w|}{18} = \frac {2\sin \frac {\pi}{18}}{18} = \frac {\sin 10^\circ}{9}$
