1984 AHSME Problems/Problem 10
Problem
Four complex numbers lie at the vertices of a square in the complex plane. Three of the numbers are , and . The fourth number is
Solution
Perhaps the easiest way to attack this is to transfer this to the Cartesian plane. The points then would be and , assuming the real axis was horizontal. Let these points be and , respectively. The remaining point is then the intersection of the following perpendicular lines: The one perpenicular to and passing through and the one perpendicular to and passing through . The slope of the first line is the negative reciprocal of the slope of the line through , which, using the slope formula, is , so the slope of the perpendicular line is . It passes through , so the equation of the line in point slope form is , or . Similarly, the slope of the second line is , and, since it passes through , its equation is , or . To find the intersection, we have , and solving for yields . Plugging this back into the equation yields , so the remaining point in the Cartesian plane is , and in the complex plane is .
NICE! You can also just graph and use pythag.
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 11 | |
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