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Difference between revisions of "2012 AIME II Problems/Problem 8"
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== Problem 8 ==
 
== Problem 8 ==
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The complex numbers <math>z</math> and <math>w</math> satisfy the system <cmath> z + \frac{20i}w = 5+i \\ \\
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The complex numbers <math>z</math> and <math>w</math> satisfy the system <cmath> z + \frac{20i}w = 5+i </cmath>
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w+\frac{12i}z = -4+10i </cmath> Find the smallest possible value of <math>\vert zw\vert^2</math>.
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<cmath> w+\frac{12i}z = -4+10i </cmath> Find the smallest possible value of <math>\vert zw\vert^2</math>.
  
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== Solution ==
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Multiplying the two equations together gives us <cmath>zw + 32i - \frac{240}{zw} = -30 + 46i</cmath> and multiplying by <math>zw</math> then gives us a quadratic in <math>zw</math>: <cmath>(zw)^2 + (30-14i)zw - 240 =0.</cmath> Using the quadratic formula, we find the two possible values of <math>zw</math> to be <math>7i-15 \pm \sqrt{(15-7i)^2 + 240}</math> = <math>6+2i,</math> <math>12i - 36.</math> The smallest possible value of <math>\vert zw\vert^2</math> is then obviously <math>6^2 + 2^2 = \boxed{040}</math>.
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===Note===
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A key thing to note here is that <math>|zw|^2=|z|^2\cdot|w|^2,</math> which can be proved as follows:
  
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== Solution ==
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Proof: Using the values for <math>z</math> and <math>w</math> that we used above, we get:
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 +
\begin{align*}
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|zw|^2&=|(ac-bd)+i(bc+ad)|^2\\
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&=(ac-bd)^2+(bc+ad)^2\\
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&=a^2c^2+b^2d^2+b^2c^2+a^2d^2-2abcd+2abc\\
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&=a^2c^2+b^2d^2+b^2c^2+a^2d^2
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\end{align*}
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Also, <math>|z|^2=a^2+b^2</math> and <math>|w|^2=c^2+d^2</math>. Therefore:
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<cmath>|z|^2\cdot|w|^2=(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2</cmath>
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and our proof is complete.
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 +
Now, also note that we found <math>\sqrt{416-210i}</math> by letting <math>416-210i=(a-bi)^2</math> and solving for <math>a</math> and <math>b</math> by considering real and imaginary parts. Then, we substitute that into <math>a-bi</math> which is the value of <math>\sqrt{416-210i}</math> and continue from there.
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'''mathboy282'''
  
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== See also ==
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== See Also ==
 
{{AIME box|year=2012|n=II|num-b=7|num-a=9}}
 
{{AIME box|year=2012|n=II|num-b=7|num-a=9}}
 +
{{MAA Notice}}

Latest revision as of 20:54, 14 January 2024

Problem 8

The complex numbers $z$ and $w$ satisfy the system \[z + \frac{20i}w = 5+i\] \[w+\frac{12i}z = -4+10i\] Find the smallest possible value of $\vert zw\vert^2$.

Solution

Multiplying the two equations together gives us \[zw + 32i - \frac{240}{zw} = -30 + 46i\] and multiplying by $zw$ then gives us a quadratic in $zw$: \[(zw)^2 + (30-14i)zw - 240 =0.\] Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \pm \sqrt{(15-7i)^2 + 240}$ = $6+2i,$ $12i - 36.$ The smallest possible value of $\vert zw\vert^2$ is then obviously $6^2 + 2^2 = \boxed{040}$.

Note

A key thing to note here is that $|zw|^2=|z|^2\cdot|w|^2,$ which can be proved as follows:

Proof: Using the values for $z$ and $w$ that we used above, we get:

\begin{align*} |zw|^2&=|(ac-bd)+i(bc+ad)|^2\\ &=(ac-bd)^2+(bc+ad)^2\\ &=a^2c^2+b^2d^2+b^2c^2+a^2d^2-2abcd+2abc\\ &=a^2c^2+b^2d^2+b^2c^2+a^2d^2 \end{align*} Also, $|z|^2=a^2+b^2$ and $|w|^2=c^2+d^2$. Therefore: \[|z|^2\cdot|w|^2=(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\] and our proof is complete.

Now, also note that we found $\sqrt{416-210i}$ by letting $416-210i=(a-bi)^2$ and solving for $a$ and $b$ by considering real and imaginary parts. Then, we substitute that into $a-bi$ which is the value of $\sqrt{416-210i}$ and continue from there.

mathboy282

See Also

2012 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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