Difference between revisions of "1991 AHSME Problems/Problem 18"

Latest revision as of 08:00, 8 October 2024

Problem

If $S$ is the set of points $z$ in the complex plane such that $(3+4i)z$ is a real number, then $S$ is a

(A) right triangle (B) circle (C) hyperbola (D) line (E) parabola

Solution

$\fbox{D}$

Solution 1

We want $(3+4i)z$ a real number, so we want the $4i$ term to be canceled out. Then, we can make $z$ be in the form $(n-\frac{4}{3}ni)$ to make sure the imaginary terms cancel out when it's multiplied together. $(n-\frac{4}{3}ni)$ is a line, so the answer is $\textbf{(D) } \text{line}$

Solution 2

Let $z = a + bi$ . Then, we have $(3 + 4i)(a + bi)$ which simplifies to $3 - 4b + (4a + 3b)i$ . So whenever $4a + 3b = 0$ , the value is real and thus, it produces a line. $\textbf{(D) }$ ~elpianista227

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.

@@ Line 11: / Line 11: @@
 == Solution ==
 <math>\fbox{D}</math>
+==Solution 1==
+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>
+==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 ==

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