Difference between revisions of "1991 AHSME Problems/Problem 18"
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− | We want <math>(3+4i)z</math> a real number, so we want the <math>4i</math> term to be canceled out. Then, we can make <math>z</math> be in the form <math>(n-\frac{4}{3}ni)</math> to make sure the imaginary terms cancel out when it's multiplied together. <math>(n-\frac{4}{3}ni)</math> is a line | + | We want <math>(3+4i)z</math> a real number, so we want the <math>4i</math> term to be canceled out. Then, we can make <math>z</math> be in the form <math>(n-\frac{4}{3}ni)</math> to make sure the imaginary terms cancel out when it's multiplied together. <math>(n-\frac{4}{3}ni)</math> is a line, so the answer is <math>\textbf{(D) } \text{line}</math> |
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+ | ==Solution 2== | ||
+ | Let <math>z = a + bi</math>. Then, we have <math>(3 + 4i)(a + bi)</math> which simplifies to <math>3 - 4b + (4a + 3b)i</math>. So whenever <math>4a + 3b = 0</math>, the value is real and thus, it produces a line. <math>\textbf{(D) }</math> | ||
+ | ~elpianista227 | ||
== See also == | == See also == |
Latest revision as of 08:00, 8 October 2024
Problem
If is the set of points in the complex plane such that is a real number, then is a
(A) right triangle (B) circle (C) hyperbola (D) line (E) parabola
Solution
Solution 1
We want a real number, so we want the term to be canceled out. Then, we can make be in the form to make sure the imaginary terms cancel out when it's multiplied together. is a line, so the answer is
Solution 2
Let . Then, we have which simplifies to . So whenever , the value is real and thus, it produces a line. ~elpianista227
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.