2012 AIME II Problem 8

Question

The complex numbers

z

and

w

satisfy the system

z + \frac{20\mathrm{i}}{w} = 5 + \mathrm{i},

w + \frac{12\mathrm{i}}{z} = -4 + 10\mathrm{i}.

Find the smallest possible value of

|zw|^2.

Answer

Multiply the two equations: the cross terms become the constants

12\mathrm{i}

and

20\mathrm{i},

leaving an equation in the single unknown

zw

The quadratic in

zw

needs

\sqrt{416 - 210\mathrm{i}};

find it by setting

(a + b\mathrm{i})^2 = 416 - 210\mathrm{i}

with integers

a

and

b

Multiplying the two equations gives

\begin{aligned} &zw + 12\mathrm{i} + 20\mathrm{i} \\ &\quad {}- \frac{240}{zw} = (5 + \mathrm{i})(-4 + 10\mathrm{i}) \\ &\quad = -30 + 46\mathrm{i}, \end{aligned}

so

zw - \frac{240}{zw} = -30 + 14\mathrm{i}.

Setting

v = zw

yields

v^2 + (30 - 14\mathrm{i})v - 240 = 0.

By the quadratic formula,

v = -15 + 7\mathrm{i}

\pm \sqrt{(15 - 7\mathrm{i})^2 + 240}

= -15 + 7\mathrm{i}

\pm \sqrt{416 - 210\mathrm{i}}.

Writing

(a + b\mathrm{i})^2 = 416 - 210\mathrm{i}

requires

a^2 - b^2 = 416

and

ab = -105,

which gives

a + b\mathrm{i} = \pm(21 - 5\mathrm{i}).

Hence

v = 6 + 2\mathrm{i}

or

v = -36 + 12\mathrm{i},

with

|v|^2 = 40

or

1440.

The smaller value is attained:

z = 1 - \mathrm{i},

w = 2 + 4\mathrm{i}

satisfies both equations with

zw = 6 + 2\mathrm{i}.

So the smallest possible value of

|zw|^2

is

40.

Problem 8 in other years