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

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<math>\fbox{D}</math>
 
<math>\fbox{D}</math>
  
Solution 1 (maybe wrong):
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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 (I think), so the answer is <math>\textbf{(D) } line</math>
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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, so the answer is <math>\textbf{(D) } line</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 18:53, 11 October 2016

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 $\textbf{(D) } line$

See also

1991 AHSME (ProblemsAnswer KeyResources)
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

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