Difference between revisions of "1991 AHSME Problems/Problem 18"
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==Solution 2== | ==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 4a + 3b = 0, the value is real and thus, it produces a line. <math>\textbf{(D) }</math> | 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 4a + 3b = 0, the value is real and thus, it produces a line. <math>\textbf{(D) }</math> | ||
| + | ~elpianista227 | ||
== See also == | == See also == | ||
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 4a + 3b = 0, 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. 
