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Difference between revisions of "2021 AMC 12B Problems/Problem 18"
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==Solution 1==
 
==Solution 1==
 
Using the fact <math>z\bar{z}=|z|^2</math>, the equation rewrites itself as
 
Using the fact <math>z\bar{z}=|z|^2</math>, the equation rewrites itself as
−
 
+
<cmath>\begin{align*}
−
<cmath>12z\bar{z}=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31</cmath>
+
12z\bar{z}&=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31 \\
−
<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>
+
-12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32&=0 \\
−
<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>
+
\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>(z+\bar{z}+2)^2+(z\bar{z}-6)^2=0.</cmath>
+
(z+\bar{z}+2)^2+(z\bar{z}-6)^2&=0.
 +
\end{align*}</cmath>
 
As the two quantities in the parentheses are real, both quantities must equal <math>0</math> so <cmath>z+\frac6z=z+\bar{z}=\boxed{\textbf{(A) }-2}.</cmath>
 
As the two quantities in the parentheses are real, both quantities must equal <math>0</math> so <cmath>z+\frac6z=z+\bar{z}=\boxed{\textbf{(A) }-2}.</cmath>
  
 
==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
+
 
 +
Let <math>z = a + bi</math>, <math>z^2 = a^2-b^2+2abi</math>
 +
 
 +
By the equation given in the problem
 +
 
 +
<cmath>12(a^2+b^2) = 2((a+2)^2 + b^2) + ((a^2-b^2+1)^2 + (2ab)^2) + 31</cmath>
 +
 
 +
<cmath>12a^2 + 12b^2 = 2a^2 + 8a + 8 + 2b^2 + a^4 + b^4 + 1 + 2a^2 - 2b^2 - 2a^2b^2 + 4a^2b^2 + 31</cmath>
 +
 
 +
<cmath>a^4 + b^4 - 8a^2 - 12b^2 + 2a^2b^2 + 8a + 40 = 0</cmath>
 +
 
 +
<cmath>(a^2+b^2)^2 - 12(a^2+b^2) + 4(a^2 + 2a + 1) + 36=0</cmath>
 +
 
 +
<cmath>(a^2 + b^2 - 6)^2 + 4(a+1)^2 = 0</cmath>
 +
 
 +
Therefore, <math>a^2 + b^2 - 6 = 0</math> and <math>a+1 = 0</math>
 +
 
 +
<math>a = -1</math>, <math>b^2 = 6-1 = 5</math>, <math>b = \sqrt{5}</math>
 +
 
 +
<cmath>z + \frac{6}{z} = \frac{ a^2 - b^2 + 6 + 2abi }{ a+bi } = \frac{ 1 - 5 + 6 + 2(-1)\sqrt{5} i }{ -1 + i \sqrt{5} } = \frac{ 2 - 2i \sqrt{5} }{-1 + i \sqrt{5}} = \boxed{\textbf{(A)} -2}</cmath>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==Solution 3==
 
==Solution 3==
Line 20: Line 42:
 
<cmath> |z|^2 = 6</cmath>
 
<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|^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>
+
<cmath>  |z^2+1|^2 = |xz -6 +1|^2 = \left(\frac{x^2}{2}-5\right)^2 + \frac{x^2(24-x^2)}{4} = x^2 +25</cmath>
 
Plugging the above back to the original equation, we have
 
Plugging the above back to the original equation, we have
 
<cmath> 12*6 = 2(2x+10) + x^2 + 25 + 31</cmath>
 
<cmath> 12*6 = 2(2x+10) + x^2 + 25 + 31</cmath>
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~Sequoia
 
~Sequoia
  
−
==Solution 4 (funny observations)==
+
==Solution 4 (Funny Observations)==
 
There are actually several ways to see that <math>|z|^2 = 6.</math> I present two troll ways of seeing it, and a legitimate way of checking.
 
There are actually several ways to see that <math>|z|^2 = 6.</math> I present two troll ways of seeing it, and a legitimate way of checking.
  
Line 46: Line 68:
 
Let <math>a = w + \overline{w} \in \mathbb{R},</math> and <math>r = |w|^2 \in \mathbb{R}.</math> Thus, we have <math>a^2+4a+40+r^2-12r=0,</math> and the discriminant of this must be nonnegative as <math>a</math> is real. Thus, <math>16-4(40+r^2-12r) \geq 0</math> or <math>(r-6)^2 \leq 0,</math> which forces <math>r = 6,</math> as claimed.
 
Let <math>a = w + \overline{w} \in \mathbb{R},</math> and <math>r = |w|^2 \in \mathbb{R}.</math> Thus, we have <math>a^2+4a+40+r^2-12r=0,</math> and the discriminant of this must be nonnegative as <math>a</math> is real. Thus, <math>16-4(40+r^2-12r) \geq 0</math> or <math>(r-6)^2 \leq 0,</math> which forces <math>r = 6,</math> as claimed.
  
−
 
+
Thus, we plug in <math>z \overline{z} = 6,</math> and get: <math>72 = 4(z + \overline{z}) + 76 + (z + \overline{z})^2,</math> ie. <math>(z+\overline{z})^2 + 4(z + \overline{z}) + 4 = 0,</math> or <math>(z+\overline{z} + 2)^2 = 0,</math> which means <math>z + \overline{z} = \boxed{\textbf{(A) }-2}</math> and that's our answer since we know <math>\overline{z} = 6 / z</math>
−
Thus, we plug in <math>z \overline{z} = 6,</math> and get: <math>72 = 4(z + \overline{z}) + 76 + (z + \overline{z})^2,</math> ie. <math>(z+\overline{z})^2 + 4(z + \overline{z}) + 4 = 0,</math> or <math>(z+\overline{z} + 2)^2 = 0,</math> which means <math>z + \overline{z} = \boxed{-2}</math> and that's our answer since we know <math>\overline{z} = 6 / z</math>
 
  
 
- ccx09
 
- ccx09
  
 
==Solution 5==
 
==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>.
+
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>.
 +
 
 +
==Solution 6==
 +
Using <math>z\bar{z}=|z|^2</math>, we have
 +
\begin{align*}
 +
12z\bar{z} = 2(z + 2)(\bar{z} + 2) + (z^2 + 1)(\bar{z}^2 + 1) + 31 &\implies z^2\bar{z}^2 + z^2 + \bar{z}^2 + 1 + 2z\bar{z} + 4z + 4\bar{z} + 8 + 31 - 12z\bar{z} = 0 \\
 +
&\implies  z^2\bar{z}^2 - 10z\bar{z} + (z^2 + \bar{z}^2) + 4(z + \bar{z}) + 40 = 0
 +
\end{align*}
 +
Let <math>p = z\bar{z}</math> and <math>s = z + \bar{z}</math>. Then we get
 +
\begin{align*}
 +
p^2 - 10p + s^2 - 2p + 4s + 40 = 0 \implies p^2 - 12p + s^2 + 4s + 40 = 0
 +
\end{align*}
 +
Completing the square, we get
 +
<cmath>(p-6)^2 - 36 + (s+2)^2 - 4 + 40 = (p-6)^2 + (s+2)^2 = 0</cmath>
 +
Therefore, <math>p = 6</math> and <math>s = -2</math>. So, <math>z + \bar{z} = -2</math> and <math>z\bar{z} = 6</math>. Plugging into <math>z + \frac{6}{z}</math>, we get
 +
<cmath> z + \frac{6}{z} = z + \frac{6\bar{z}}{z\bar{z}} = z + \frac{6\bar{z}}{6} = z + \bar{z} = \boxed{\textbf{(A) }-2} </cmath>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Crazyvideogamez CrazyVideoGamez]
 +
 
 +
==Video Solution by OmegaLearn (Using Complex Number Identities)==
 +
https://youtu.be/AEbMTTGEZV4
 +
~pi_is_3.14
 +
 
 +
==Video Solution== 
 +
 
 +
https://youtu.be/Yw-IJvfrT_U
 +
~MathProblemSolvingSkills.com
 +
 
 +
(includes review of complex numbers)
 +
 
 +
 
 +
 
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==

Latest revision as of 11:39, 5 November 2024

Problem

Let $z$ be a complex number satisfying $12|z|^2=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\frac 6z?$

$\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\frac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4$

Solution 1

Using the fact $z\bar{z}=|z|^2$, the equation rewrites itself as \begin{align*} 12z\bar{z}&=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31 \\ -12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32&=0 \\ \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 \\ (z+\bar{z}+2)^2+(z\bar{z}-6)^2&=0. \end{align*} As the two quantities in the parentheses are real, both quantities must equal $0$ so \[z+\frac6z=z+\bar{z}=\boxed{\textbf{(A) }-2}.\]

Solution 2

Let $z = a + bi$, $z^2 = a^2-b^2+2abi$

By the equation given in the problem

\[12(a^2+b^2) = 2((a+2)^2 + b^2) + ((a^2-b^2+1)^2 + (2ab)^2) + 31\]

\[12a^2 + 12b^2 = 2a^2 + 8a + 8 + 2b^2 + a^4 + b^4 + 1 + 2a^2 - 2b^2 - 2a^2b^2 + 4a^2b^2 + 31\]

\[a^4 + b^4 - 8a^2 - 12b^2 + 2a^2b^2 + 8a + 40 = 0\]

\[(a^2+b^2)^2 - 12(a^2+b^2) + 4(a^2 + 2a + 1) + 36=0\]

\[(a^2 + b^2 - 6)^2 + 4(a+1)^2 = 0\]

Therefore, $a^2 + b^2 - 6 = 0$ and $a+1 = 0$

$a = -1$, $b^2 = 6-1 = 5$, $b = \sqrt{5}$

\[z + \frac{6}{z} = \frac{ a^2 - b^2 + 6 + 2abi }{ a+bi } = \frac{ 1 - 5 + 6 + 2(-1)\sqrt{5} i }{ -1 + i \sqrt{5} } = \frac{ 2 - 2i \sqrt{5} }{-1 + i \sqrt{5}} = \boxed{\textbf{(A)} -2}\]

~isabelchen

Solution 3

Let $x = z + \frac{6}{z}$. Then $z = \frac{x \pm \sqrt{x^2-24}}{2}$. From the answer choices， we know that $x$ is real and $x^2<24$, so $z = \frac{x \pm i\sqrt{24-x^2}}{2}$. Then we have \[|z|^2 = 6\] \[|z+2|^2 = (\frac{x}{2} + 2)^2 + \frac{24-x^2}{4} = 2x+10\] \[|z^2+1|^2 = |xz -6 +1|^2 = \left(\frac{x^2}{2}-5\right)^2 + \frac{x^2(24-x^2)}{4} = x^2 +25\] Plugging the above back to the original equation, we have \[12*6 = 2(2x+10) + x^2 + 25 + 31\] \[(x+2)^2 = 0\] So $x = \boxed{\textbf{(A) }-2}$.

~Sequoia

Solution 4 (Funny Observations)

There are actually several ways to see that $|z|^2 = 6.$ I present two troll ways of seeing it, and a legitimate way of checking.

Rewrite using $w \overline{w} = |w|^2$

$12z \overline{z} + 2(z+2)(\overline{z} + 2) + (z^2+1)(\overline{z}^2+1)+31$ $12 z \overline{z} = 2z \overline{z} + 4z + 4 \overline{z} + 8 + z^2 \overline{z}^2+z^2+\overline{z}^2 + 1 + 31.$ $12 z \overline{z} = 4(z + \overline{z}) + (z \overline{z})^2 + (z + \overline{z})^2 + 40.$

Symmetric in $z$ and $\overline{z},$ so if $w$ is a sol, then so is $\overline{w}$

TROLL OBSERVATION #1: ALL THE ANSWERS ARE REAL. THUS, $z + \frac{6}{z} \in \mathbb{R},$ which means they must be conjugates and so $|z|^2 = 6.$

TROLL OBSERVATION #2: Note that $z+\frac{6}{z} = \overline{z} + \frac{6}{\overline{z}}$ because either solution must give the same answer! which means that $|z|^2 = 6.$

Alternatively, you can check: Let $a = w + \overline{w} \in \mathbb{R},$ and $r = |w|^2 \in \mathbb{R}.$ Thus, we have $a^2+4a+40+r^2-12r=0,$ and the discriminant of this must be nonnegative as $a$ is real. Thus, $16-4(40+r^2-12r) \geq 0$ or $(r-6)^2 \leq 0,$ which forces $r = 6,$ as claimed.

Thus, we plug in $z \overline{z} = 6,$ and get: $72 = 4(z + \overline{z}) + 76 + (z + \overline{z})^2,$ ie. $(z+\overline{z})^2 + 4(z + \overline{z}) + 4 = 0,$ or $(z+\overline{z} + 2)^2 = 0,$ which means $z + \overline{z} = \boxed{\textbf{(A) }-2}$ and that's our answer since we know $\overline{z} = 6 / z$

- ccx09

Solution 5

Observe that all the answer choices are real. Therefore, $z$ and $\frac{6}{z}$ must be complex conjugates as this is the only way for both their sum (one of the answer choices) and their product ($6$) to be real. Thus $|z|=|\tfrac{6}{z}|=\sqrt{6}$. We will test all the answer choices, starting with $\textbf{(A)}$. Suppose the answer is $\textbf{(A)}$. If $z+\tfrac{6}{z}=-2$ then $z^{2}+2z+6=0$ and $z=\frac{-2\pm\sqrt{4-24}}{2}=-1\pm\sqrt{5}i$. Note that if $z=-1+\sqrt{5}i$ works, then so does $-1-\sqrt{5}i$. It is relatively easy to see that if $z=-1+\sqrt{5}i$, then $12|z|^{2}=12\cdot 6=72, 2|z+2|^{2}=2|1+\sqrt{5}i|=2\cdot 6=12, |z^{2}+1|^{2}=|-3-2\sqrt{5}i|^{2}=29,$ and $72=12+29+31$. Thus the condition \[12|z|^{2}=2|z+2|^{2}+|z^{2}+1|^{2}+31\] is satisfied for $z+\tfrac{6}{z}=-2$, and the answer is $\boxed{\textbf{(A) }-2}$.

Solution 6

Using $z\bar{z}=|z|^2$, we have \begin{align*} 12z\bar{z} = 2(z + 2)(\bar{z} + 2) + (z^2 + 1)(\bar{z}^2 + 1) + 31 &\implies z^2\bar{z}^2 + z^2 + \bar{z}^2 + 1 + 2z\bar{z} + 4z + 4\bar{z} + 8 + 31 - 12z\bar{z} = 0 \\ &\implies z^2\bar{z}^2 - 10z\bar{z} + (z^2 + \bar{z}^2) + 4(z + \bar{z}) + 40 = 0 \end{align*} Let $p = z\bar{z}$ and $s = z + \bar{z}$. Then we get \begin{align*} p^2 - 10p + s^2 - 2p + 4s + 40 = 0 \implies p^2 - 12p + s^2 + 4s + 40 = 0 \end{align*} Completing the square, we get \[(p-6)^2 - 36 + (s+2)^2 - 4 + 40 = (p-6)^2 + (s+2)^2 = 0\] Therefore, $p = 6$ and $s = -2$. So, $z + \bar{z} = -2$ and $z\bar{z} = 6$. Plugging into $z + \frac{6}{z}$, we get \[z + \frac{6}{z} = z + \frac{6\bar{z}}{z\bar{z}} = z + \frac{6\bar{z}}{6} = z + \bar{z} = \boxed{\textbf{(A) }-2}\]

~CrazyVideoGamez

Video Solution by OmegaLearn (Using Complex Number Identities)

https://youtu.be/AEbMTTGEZV4 ~pi_is_3.14

Video Solution

https://youtu.be/Yw-IJvfrT_U ~MathProblemSolvingSkills.com

(includes review of complex numbers)



Video Solution by Punxsutawney Phil

https://youtu.be/E0HkYqZzw3s

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. AMC logo.png

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