Difference between revisions of "2021 AMC 12B Problems/Problem 18"
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+ | ==Solution 5== | ||
+ | Observe that all the answer choices are real. Therefore, <math>z</math> and <math>\frac{6}{z}</math> must be complex conjugates as this is the only way for both their sum (one of the answer choices) and their product (<math>6</math>) to be real. Thus <math>|z|=|\tfrac{6}{z}|=\sqrt{6}</math>. We will test all the answer choices, starting with <math>\textbf{(A)}</math>. Suppose the answer is <math>\textbf{(A)}</math>. If <math>z+\tfrac{6}{z}=-2</math> then <math>z^{2}+2z+6=0</math> and <math>z=\frac{-2\pm\sqrt{4-24}}{2}=-1\pm\sqrt{5}i</math>. Note that if <math>z=-1+\sqrt{5}i</math> works, then so does <math>-1-\sqrt{5}i</math>. It is relatively easy to see that if <math>z=-1+\sqrt{5}i</math>, then <math>12|z|^{2}=12\cdot 6=72, 2|z+2|^{2}=2|1+\sqrt{5}i|=2\cdot 6=12, |z^{2}+1|^{2}=|-3-2\sqrt{5}i|^{2}=29,</math> and <math>72=12+29+31</math>. Thus the condition <cmath>12|z|^{2}=2|z+2|^{2}+|z^{2}+1|^{2}+31</cmath> is satisfied for <math>z+\tfrac{6}{z}=-2</math>, and the answer is <math>\boxed{\textbf{(A)} ~-2}</math>. | ||
==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 14:46, 15 February 2021
Contents
[hide]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 that is real and , so . Then we have Plugging the above back to the original equation, we have So .
~Sequoia
Solution 4 (funny observations)
There are actually several ways to see that I present two troll ways of seeing it, and a legitimate way of checking.
Rewrite using
Symmetric in and so if is a sol, then so is
TROLL OBSERVATION #1: ALL THE ANSWERS ARE REAL. THUS, which means they must be conjugates and so
TROLL OBSERVATION #2: Note that because either solution must give the same answer! which means that
Alternatively, you can check: Let and Thus, we have and the discriminant of this must be nonnegative as is real. Thus, or which forces as claimed.
Thus, we plug in and get: ie. or which means and that's our answer since we know
- ccx09
Solution 5
Observe that all the answer choices are real. Therefore, and must be complex conjugates as this is the only way for both their sum (one of the answer choices) and their product () to be real. Thus . We will test all the answer choices, starting with . Suppose the answer is . If then and . Note that if works, then so does . It is relatively easy to see that if , then and . Thus the condition is satisfied for , and the answer is .
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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