1983 AHSME Problems/Problem 17

Problem

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 1

Write $F$ as $a + bi$ , where we see from the diagram that $a, b > 0$ and $a^2+b^2>1$ (as $F$ 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 $\boxed{\textbf{C}}$ .

Solution 2 (Polar Form)

Let $z = r(\cos\theta + i\sin\theta)$ be the complex number at $F$ in polar form. Then $\frac{1}{z} = \frac{1}{r}(\cos(-\theta) + i\sin(-\theta))$ . Since $F$ is outside the circle, $r > 1$ , and therefore $\frac{1}{r} < 1$ . So $\frac{1}{z}$ must be inside the circle and on the line which is $\theta$ clockwise from the real axis, that is $\boxed{(\mathbf{C})\ C}$ .

-j314andrews

1983 AHSME (Problems • Answer Key • Resources)
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1983 AHSME Problems/Problem 17

Contents

Problem

Solution 1

Solution 2 (Polar Form)

See Also