1984 AMC 12 Problem 30

Attempt Problem 30 of the 1984 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1984 AMC 12 solutions, or check the answer key.

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30.

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+\cdots+9w^9|^{-1}

equals

\frac19\sin40^\circ

\frac29\sin20^\circ

\frac19\cos40^\circ

\frac1{18}\cos20^\circ

none of these

Answer: B

Concepts:complex number roots of unity arithmetico-geometric series

Difficulty rating: 2460

Small hint:

Let

S=w+2w^2+\cdots+9w^9

and subtract

wS

from

S

Big hint:

Use

w^9=1

and

|1-e^{i\theta}|=2\sin(\frac{\theta}{2})

Solution:

Let

S=\sum_{k=1}^9kw^k.

Since

w^9=1

and

w^{10}=w,

\begin{aligned} (1-w)S &=(w+w^2+\cdots+w^9)\\ &\quad-9w^{10}\\ &=-9w. \end{aligned}

Therefore

|S|=\frac{9}{|1-w|},

because

|w|=1.

Thus

|S|^{-1}=\frac{|1-w|}{9} =\frac{2}{9}\sin20^\circ.

Therefore, the correct answer is B.