2012 AIME II Problem 8

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8.

The complex numbers zz and ww satisfy the system z+20iw=5+i,z + \frac{20i}{w} = 5 + i, w+12iz=4+10i.w + \frac{12i}{z} = -4 + 10i.

Find the smallest possible value of zw2.|zw|^2.

Answer: 40
Concepts:complex numbersystem of equationsquadratic
Difficulty rating: 2840
Solution:

Multiplying the two equations gives zw+12i+20i240zw=(5+i)(4+10i)=30+46i, \begin{aligned} &zw + 12i + 20i \\ &\quad {}- \frac{240}{zw} = (5 + i)(-4 + 10i) \\ &\quad = -30 + 46i, \end{aligned} so zw240zw=30+14i.zw - \frac{240}{zw} = -30 + 14i. Setting v=zwv = zw yields v2+(3014i)v240=0.v^2 + (30 - 14i)v - 240 = 0.

By the quadratic formula, v=15+7iv = -15 + 7i ±(157i)2+240\pm \sqrt{(15 - 7i)^2 + 240} =15+7i= -15 + 7i ±416210i.\pm \sqrt{416 - 210i}. Writing (a+bi)2=416210i(a + bi)^2 = 416 - 210i requires a2b2=416a^2 - b^2 = 416 and ab=105,ab = -105, which gives a+bi=±(215i).a + bi = \pm(21 - 5i). Hence v=6+2iv = 6 + 2i or v=36+12i,v = -36 + 12i, with v2=40|v|^2 = 40 or 1440.1440.

The smaller value is attained: z=1i,z = 1 - i, w=2+4iw = 2 + 4i satisfies both equations with zw=6+2i.zw = 6 + 2i. So the smallest possible value of zw2|zw|^2 is 40.40.

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