1983 AHSME Problems/Problem 17

Revision as of 21:01, 26 January 2019 by Sevenoptimus (talk | contribs) (Added a solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Pdfresizer.com-pdf-convert-q17.png

The diagram above shows several numbers in the complex plane. The circle is the unit circle centered at the origin. One of these numbers is the reciprocal of $F$. Which one?

$\textbf{(A)} \ A \qquad  \textbf{(B)} \ B \qquad  \textbf{(C)} \ C \qquad  \textbf{(D)} \ D \qquad  \textbf{(E)} \ E$

Solution

Write $F$ as $a + bi$, where we see from the diagram that $a, b > 0$ and $a^2+b^2>1$ ($F$ is outside the unit circle). We have $\frac{1}{a+bi} = \frac{a-bi}{a^2+b^2} = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}i$, so, since $a, b > 0$, the reciprocal of $F$ has a positive real part and negative imaginary part. Also, the reciprocal has magnitude equal to the reciprocal of $F$'s magnitude (since $|a||b| = |ab|$); as $F$'s magnitude is greater than $1$, its reciprocal's magnitude will thus be between $0$ and $1$, so its reciprocal will be inside the unit circle. Therefore, the only point shown which could be the reciprocal of $F$ is point $\boxed{\textbf{C}}$.