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Difference between revisions of "1983 AHSME Problems/Problem 17"

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Let <math>z = r(\cos\theta + i\sin\theta)</math> be the complex number at <math>F</math> in polar form.  Then <math>\frac{1}{z} = \frac{1}{r}(\cos(-\theta) + i\sin(-\theta))</math>. Since <math>F</math> is outside the circle, <math>r > 1</math>, and therefore <math>\frac{1}{r} < 1</math>.  So <math>\frac{1}{z}</math> must be inside the circle and on the line which is <math>\theta</math> clockwise from the real axis, that is <math>\boxed{(\mathbf{C})\ C}</math>.
 
Let <math>z = r(\cos\theta + i\sin\theta)</math> be the complex number at <math>F</math> in polar form.  Then <math>\frac{1}{z} = \frac{1}{r}(\cos(-\theta) + i\sin(-\theta))</math>. Since <math>F</math> is outside the circle, <math>r > 1</math>, and therefore <math>\frac{1}{r} < 1</math>.  So <math>\frac{1}{z}</math> must be inside the circle and on the line which is <math>\theta</math> clockwise from the real axis, that is <math>\boxed{(\mathbf{C})\ C}</math>.
 +
-j314andrews
  
 
==See Also==
 
==See Also==

Revision as of 15:53, 2 July 2025

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 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

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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