$12 \cdot 13$ For any complex number $w=a+bi, |w|$ is defined as the real number $\sqrt{a^{2}+b^{2}}$. If $w=\cos 40^{\circ}+i \sin 40^{\circ}$, then
$\left|w+2 w^{2}+3 w^{3}+\cdots+9 w^{9}\right|^{-1}$ equals
(A) $\frac{1}{9} \sin 40^{\circ}$.
(B) $\frac{2}{9} \sin 20^{\circ}$.
(C) $\frac{1}{9} \cos 40^{\circ}$.
(D) $\frac{1}{18} \cos 20^{\circ}$.
(E) None of the above.
(35th American High School Mathematics Examination, 1984)
[Solution] From the given, we know that $|w|=1$, and $40^{\circ}=\frac{360^{\circ}}{9}$, which means $w$ is a 9th root of unity. For a more general $n$th root of unity, where $n>1$, we set
$$
s=w+2 w^{2}+\cdots+n w^{n},
$$

Multiplying both sides of the equation by $w$, we get
$$
s w=w^{2}+2 w^{3}+\cdots+n w^{n+1} \text {. }
$$
Subtracting (2) from (1) gives
$$
s(1-w)=w+w^{2}+w^{3}+\cdots+w^{n}-n w^{n+1} \text {, }
$$

Since $n>1$, we have $w \neq 1$, so $s(1-w)=\frac{w^{n+1}-w}{w-1}-n w^{n+1}$. Given that $w^{n}=1$, we have $s(1-w)=-n w$, thus
$$
\frac{1}{s}=\frac{w-1}{n w} \text {, and } \quad \frac{1}{|s|}=\frac{|w-1|}{n} .
$$

Since 1 and $w$ are adjacent vertices on the unit circle, $|w-1|$ is the length of the side of a regular $n$-sided polygon inscribed in the unit circle, which is $2 \sin \frac{\pi}{n}$, so
$$
|s|^{-1}=\frac{2}{n} \sin \frac{\pi}{n} \text {. }
$$

Therefore, if $n>1$, and $w=\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}$, then
$$
\left|w^{1}+2 w^{2}+3 w^{3}+\cdots+n w^{n}\right|^{-1}=\frac{2}{n} \sin \frac{\pi}{n} \text {. }
$$

When $n=9$, $\left|w+2 w^{2}+3 w^{3}+\cdots+9 w^{9}\right|^{-1}=\frac{1}{|s|}=\frac{2}{9} \sin 20^{\circ}$. Therefore, the answer is $(B)$.