Difference between revisions of "2021 AMC 12B Problems/Problem 18"
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==Solution 2== | ==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 | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let <math>x = z + \frac{6}{z}</math>. Then <math>z = \frac{x \pm \sqrt{x^2-24}}{2}</math>. From the answer choices we know <math>x</math> is real and <math>|x|<24</math>, so <math>z = \frac{x \pm i\sqrt{24-x^2}}{2}</math>. We'll take the plus sign for now since we know the answer is unique. Then we have | ||
+ | <cmath> |z|^2 = 6</cmath> | ||
+ | <cmath> |z+2|^2 = (\frac{x}{2} + 2)^2 + \frac{24-x^2}{4} = 2x+10</cmath> | ||
+ | <cmath> |z^2+1|^2 = |xz -6 +1|^2 = (\frac{x^2}{2}-5)^2 + \frac{x^2(24-x^2)}{4} = x^2 +25</cmath> | ||
+ | Plug the above back to the original equation, we have | ||
+ | <cmath> 12*6 = 2(2x+10) + x^2 + 25 + 31</cmath> | ||
+ | <cmath> (x+2)^2 = 0</cmath> | ||
+ | So <math>x = -2</math> <math>\boxed{\textbf{(A) }-2}</math>. | ||
+ | |||
+ | ~Sequoia | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2021|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:17, 13 February 2021
Problem
Let be a complex number satisfying What is the value of
Solution 1
Using the fact , the equation rewrites itself as
As the two quantities in the parentheses are real, both quantities must equal so
Solution 2
The answer being in the form means that there are two solutions, some complex number and its complex conjugate. 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
Solution 3
Let . Then . From the answer choices we know is real and , so . We'll take the plus sign for now since we know the answer is unique. Then we have Plug the above back to the original equation, we have So .
~Sequoia
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 |
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