2012 AIME II Problems/Problem 8

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:

|zw|2=|(acbd)+i(bc+ad)|2=(acbd)2+(bc+ad)2=a2c2+b2d2+b2c2+a2d22abcd+2abc=a2c2+b2d2+b2c2+a2d2 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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png