Difference between revisions of "1983 AHSME Problems/Problem 17"
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\textbf{(E)} \ E </math> | \textbf{(E)} \ E </math> | ||
| − | ==Solution== | + | ==Solution 1== |
Write <math>F</math> as <math>a + bi</math>, where we see from the diagram that <math>a, b > 0</math> and <math>a^2+b^2>1</math> (as <math>F</math> is outside the unit circle). We have <math>\frac{1}{a+bi} = \frac{a-bi}{a^2+b^2} = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}i</math>, so, since <math>a, b > 0</math>, the reciprocal of <math>F</math> has a positive real part and negative imaginary part. Also, the reciprocal has magnitude equal to the reciprocal of <math>F</math>'s magnitude (since <math>|a||b| = |ab|</math>); as <math>F</math>'s magnitude is greater than <math>1</math>, its reciprocal's magnitude will thus be between <math>0</math> and <math>1</math>, so its reciprocal will be inside the unit circle. Therefore, the only point shown which could be the reciprocal of <math>F</math> is point <math>\boxed{\textbf{C}}</math>. | Write <math>F</math> as <math>a + bi</math>, where we see from the diagram that <math>a, b > 0</math> and <math>a^2+b^2>1</math> (as <math>F</math> is outside the unit circle). We have <math>\frac{1}{a+bi} = \frac{a-bi}{a^2+b^2} = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}i</math>, so, since <math>a, b > 0</math>, the reciprocal of <math>F</math> has a positive real part and negative imaginary part. Also, the reciprocal has magnitude equal to the reciprocal of <math>F</math>'s magnitude (since <math>|a||b| = |ab|</math>); as <math>F</math>'s magnitude is greater than <math>1</math>, its reciprocal's magnitude will thus be between <math>0</math> and <math>1</math>, so its reciprocal will be inside the unit circle. Therefore, the only point shown which could be the reciprocal of <math>F</math> is point <math>\boxed{\textbf{C}}</math>. | ||
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| + | ==Solution 2 (Polar Form)== | ||
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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>. | ||
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| + | -j314andrews | ||
==See Also== | ==See Also== | ||
Latest revision as of 15:53, 2 July 2025
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 
. Which one?
Solution 1
Write as 
, where we see from the diagram that 
and 
(as is outside the unit circle). We have 
, so, since , the reciprocal of has a positive real part and negative imaginary part. Also, the reciprocal has magnitude equal to the reciprocal of 's magnitude (since 
); as 's magnitude is greater than 
, its reciprocal's magnitude will thus be between 
and , so its reciprocal will be inside the unit circle. Therefore, the only point shown which could be the reciprocal of is point 
.
Solution 2 (Polar Form)
Let 
be the complex number at in polar form. Then 
. Since is outside the circle, 
, and therefore 
. So 
must be inside the circle and on the line which is 
clockwise from the real axis, that is 
.
-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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
