For any [complex number](https://artofproblemsolving.com/wiki/index.php/Complex_number) $w=a+bi$, $|w|$ is defined to be the [real number](https://artofproblemsolving.com/wiki/index.php/Real_number) $\sqrt{a^2+b^2}$. If $w=\cos40^\circ+i\sin40^\circ$, then $|w+2w^2+3w^3+...+9w^9|^{-1}$ equals
$\text{(A) }\frac{1}{9}\sin40^\circ \qquad \text{(B) }\frac{2}{9}\sin20^\circ \qquad \text{(C) } \frac{1}{9}\cos40^\circ \qquad \text{(D) }\frac{1}{18}\cos20^\circ \qquad \text{(E) } \text{None of these}$
Let $S=w+2w^2+3w^3+...+9w^9$. Note that
\[S=\sum_{i=1}^{9}\sum_{j=i}^{9} w^j\]
Now we multiply $S$ by $1-w$:
\[S(1-w)=\sum_{i=1}^{9}\sum_{j=i}^{9} w^j(1-w)\]
By the geometric series formula, $\sum_{j=i}^{9} w^j(1-w)$ is simply $w^i-w^{10}$. Therefore
\[S(1-w)=\sum_{i=1}^{9} w^i-w^{10}=(w+w^2+\cdots +w^9)-9w^{10}\]
A simple application of [De Moivre's Theorem](https://artofproblemsolving.com/wiki/index.php/De_Moivre%27s_Theorem) shows that $w$ is a ninth root of unity ($w^9=1$), so
\[(w+w^2+\cdots +w^9)(1-w)=w-w^{10}=w(1-w^{9})=0\]
This shows that $S=\frac{-9w^{10}}{1-w}$. Note that $w^{10}=w\cdot w^9=w$, so $S=\frac{-9w}{1-w}$. 
It's not hard to show that $\left|S\right|^{-1}=\left|S^{-1}\right|=\left|-S^{-1}\right|$, so the number we seek is equal to $\left|\frac{1-w}{9w}\right|$.
Now we plug $w$ into the fraction:
\[\frac{1-w}{9w}=\frac{(1-\cos{40^{\circ}})-i\sin{40^{\circ}}}{9\cos{40}+9i\sin{40^{\circ}}}\]
We multiply the numerator and denominator by $9\cos{40^{\circ}}-9i\sin{40^{\circ}}$ and simplify to get
\[\frac{1-w}{9w}=\frac{(\cos{40^{\circ}}-i\sin{40^{\circ}})((1-\cos{40^{\circ}})-i\sin{40^{\circ}})}{9}\]
\[=\frac{\cos{40^{\circ}}-\cos^2{40^{\circ}}-\sin^2{40^{\circ}}+i(-\sin{40^{\circ}}+\sin{40^{\circ}}\cos{40^{\circ}}-\sin{40^{\circ}}\cos{40^{\circ}})}{9}\]
\[=\frac{(\cos{40^{\circ}}-1)-i\sin{40^{\circ}}}{9}\]
The absolute value of this is
\[\left|\frac{1-w}{9w}\right|=\frac{1}{9}\sqrt{(\cos{40^{\circ}}-1)^2+\sin^2{40^{\circ}}}=\frac{1}{9}\sqrt{1-2\cos{40^{\circ}}+\cos^2{40^{\circ}}+\sin^2{40^{\circ}}}=\frac{\sqrt{2}}{9}\sqrt{1-\cos{40^{\circ}}}\]
Note that, from double angle formulas, $\cos{40^{\circ}}=\cos^2{20^{\circ}}-\sin^2{20^{\circ}}$, so $1-\cos{40^{\circ}}=\cos^2{20^{\circ}}+\sin^2{20^{\circ}}-(\cos^2{20^{\circ}}-\sin^2{20^{\circ}})=2\sin^2{20^{\circ}}$. Therefore
\[\left|\frac{1-w}{9w}\right|=\frac{\sqrt{2}}{9} \sqrt{2\sin^2{20^{\circ}}}\]
\[=\frac{2}{9}\sin{20^{\circ}}\]
Therefore the correct answer is $\boxed{\textbf{(B)}}$.