Difference between revisions of "2021 AMC 12B Problems/Problem 18"
Brendanb4321 (talk | contribs) (add solution) |
|||
| Line 4: | Line 4: | ||
<math>\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\frac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4</math> | <math>\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\frac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| + | Using the fact <math>z\bar{z}=|z|^2</math>, the equation rewrites itself as | ||
| + | |||
| + | <cmath>12z\bar{z}=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31</cmath> | ||
| + | <cmath>-12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32=0</cmath> | ||
| + | <cmath>\left((z^2+2z\bar{z}+\bar{z}^2)+4(z+\bar{z})+4\right)+\left(z^2\bar{z}^2-12z\bar{z}+36\right)=0</cmath> | ||
| + | <cmath>(z+\bar{z}+2)^2+(z\bar{z}-6)^2=0.</cmath> | ||
| + | As the two quantities in the parentheses are real, <math>z+\bar{z}=\boxed{\textbf{(A) }-2}</math>. | ||
| + | |||
| + | |||
| + | ==Solution 2== | ||
The answer being in the form <math>z+\frac 6z</math> means that there are two solutions, some complex number and its complex conjugate. <cmath>a+bi = \frac{6}{a-bi}</cmath> <cmath>a^2+b^2=6</cmath> We should then be able to test out some ordered pairs of <math>(a, b)</math>. After testing it out, we get the ordered pairs of <math>(-1, \sqrt{5})</math> and its conjugate <math>(-1, -\sqrt{5})</math>. Plugging this into answer format gives us <math>\boxed{\textbf{(A) }-2}</math> ~Lopkiloinm | The answer being in the form <math>z+\frac 6z</math> means that there are two solutions, some complex number and its complex conjugate. <cmath>a+bi = \frac{6}{a-bi}</cmath> <cmath>a^2+b^2=6</cmath> We should then be able to test out some ordered pairs of <math>(a, b)</math>. After testing it out, we get the ordered pairs of <math>(-1, \sqrt{5})</math> and its conjugate <math>(-1, -\sqrt{5})</math>. Plugging this into answer format gives us <math>\boxed{\textbf{(A) }-2}</math> ~Lopkiloinm | ||
Revision as of 14:34, 12 February 2021
Contents
Problem
Let 
be a complex number satisfying 
What is the value of 
Solution 1
Using the fact 
, the equation rewrites itself as
![\[12z\bar{z}=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31\]](http://latex.artofproblemsolving.com/6/6/f/66fefa3a0b5e00167c2cbd0db94f835fa2978c4a.png)
![\[-12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32=0\]](http://latex.artofproblemsolving.com/b/f/c/bfc565e2044ed9a2c9cc6ed8682b7b1438283d64.png)
![\[\left((z^2+2z\bar{z}+\bar{z}^2)+4(z+\bar{z})+4\right)+\left(z^2\bar{z}^2-12z\bar{z}+36\right)=0\]](http://latex.artofproblemsolving.com/3/f/9/3f99d6913a2dce611593a3ff5cdf5603a54b801d.png)
![\[(z+\bar{z}+2)^2+(z\bar{z}-6)^2=0.\]](http://latex.artofproblemsolving.com/a/a/8/aa857e478c17e61f7a409a9043247d493a1b5b76.png)
As the two quantities in the parentheses are real, 
.
Solution 2
The answer being in the form 
means that there are two solutions, some complex number and its complex conjugate. ![\[a+bi = \frac{6}{a-bi}\]](http://latex.artofproblemsolving.com/5/f/3/5f333ff36bfce8cc1cbef23b56cdc66c678671eb.png)
![\[a^2+b^2=6\]](http://latex.artofproblemsolving.com/a/0/0/a006fc3bc9d7d0863ef53f8dd41755cca1848ac8.png)
We should then be able to test out some ordered pairs of 
. After testing it out, we get the ordered pairs of 
and its conjugate 
. Plugging this into answer format gives us 
~Lopkiloinm
See Also
| 2021 AMC 12B (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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
