1984 AHSME Problems/Problem 30

Problem

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

$\mathrm{(A) \ }\frac{1}{9}\sin40^\circ \qquad \mathrm{(B) \ }\frac{2}{9}\sin20^\circ \qquad \mathrm{(C) \ } \frac{1}{9}\cos40^\circ \qquad \mathrm{(D) \ }\frac{1}{18}\cos20^\circ \qquad \mathrm{(E) \ } \text{None of these}$

Solution

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 $1-w$ :

$S(1-w)=\sum_{i=1}^{9}\sum_{j=i}^{9} w^j(1-w)$

However, $\sum_{j=i}^{9} w^j(1-w)$ 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 $|S|^{-1}=|S^{-1}|=|-S^{-1}|$ , so the number we seek is equal to $|\frac{1-w}{9w}|$ .

Now we plug $w$ into the fraction:

$\frac{1-w}{9w}=\frac{(1-\cos{40})-i\sin{40}}{9\cos{40}+9i\sin{40}}$

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

$\frac{1-w}{9w}=\frac{(\cos{40}-i\sin{40})((1-\cos{40})-i\sin{40})}{9}$

$=\frac{\cos{40}-\cos^2{40}-\sin^2{40}+i(-\sin{40}+\sin{40}\cos{40}-\sin{40}\cos{40})}{9}$

$=\frac{(\cos{40}-1)-i\sin{40}}{9}$

The absolute value of this is

$|\frac{1-w}{9w}|=\frac{1}{9}\sqrt{(\cos{40}-1)^2+\sin^2{40}}=\frac{1}{9}\sqrt{1-2\cos{40}+\cos^2{40}+\sin^2{40}}=\frac{\sqrt{2}}{9}\sqrt{1-\cos{40}}$

Note that, from double angle formulas, $\cos{40}=\cos^2{20}-\sin^2{20}$ , so $1-\cos{40}=\cos^2{20}+\sin^2{20}-(\cos^2{20}-\sin^2{20})=2\sin^2{20}$ . Therefore

$|\frac{1-w}{9w}|=\frac{\sqrt{2}}{9} \sqrt{2\sin^2{20}}$

$=\frac{2}{9}\sin{20}$

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

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1984 AHSME Problems/Problem 30

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