1991 AHSME Problems/Problem 18

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.

Page

Toolbox

Search

1991 AHSME Problems/Problem 18

Contents

Problem

Solution

Solution 1

Solution 2

See also