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

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

Solution 1

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:

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

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 $\boxed{\textbf{(B)}}$ .

Solution 2

Notice that $w = \cos40^\circ+i\sin40^\circ = \cos\left(\frac{2\pi}{9}\right)+i\sin\left(\frac{2\pi}{9}\right)$ is the primitive root of the equation $x^9 = 1$ .

Consider the simple geometric progression,

$f(x) = 1 + x + x^2 +\cdots + x^9 = \frac{x^{10} - 1}{x - 1}$

Differentiating this equation (using the Quotient Rule), we obtain the following equation,

$f'(x) = 1 + 2x + 3x^2 + \cdots + 9x^8 = \frac{9x^{10} - 10x^9 + 1}{(x - 1)^2}$

Now, we can evaluate the given expression,

$w + 2w^2 + 3w^3 + \cdots + 9w^9 = w \cdot (1+2w+3w^2+...+9w^8)$ $= w\cdot f'(w)$ $= w\cdot\frac{9w^{10} - 10w^9 + 1}{(w - 1)^2}$ $= \frac{9w}{w-1} \; (\because w^9 = 1)$

Finally, we can calculate the given modulus,

$\lvert w+2w^2+3w^3+...+9w^9 \rvert^{-1} = \lvert \frac{1}{9}\cdot\left(1 - w^8) \right) \rvert$ $= \lvert \frac{1}{9}\cdot\left(1 - \cos\left(320^\circ\right)+i\sin\left(320^\circ\right)) \right) \rvert$ $= \frac{1}{9}\sqrt{\left(1 - \cos\left(320^\circ\right)\right)^2+\left(\sin\left(320^\circ\right)\right)^2}$ $= \frac{1}{9}\sqrt{2 - 2\cos\left(320^\circ\right)}$ $= \frac{2}{9}\sin\left(160^\circ\right)$ $= \boxed{\frac{2}{9}\sin\left(20^\circ\right)}$

~ plusone

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1984 AHSME Problems/Problem 30

Contents

Problem

Solution 1

Solution 2

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