1984 AHSME Problems/Problem 30

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For any complex number $w=a+bi$, $|w|$ is defined to be the 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 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:


We multiply the numerator and denominator by $9\cos{40^{\circ}}-9i\sin{40^{\circ}}$ and simplify to get




The absolute value of this is


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

\[\left|\frac{1-w}{9w}\right|=\frac{\sqrt{2}}{9} \sqrt{2\sin^2{20^{\circ}}}\]


Therefore the correct answer is $\boxed{\textbf{(B)}}$.

See Also

1984 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 29
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