2008 AMC 12B Problem 19

Attempt Problem 19 of the 2008 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12B solutions, or check the answer key.

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19.

A function ff is defined by f(z)=(4+i)z2+αz+γf(z) = (4 + i)z^2 + \alpha z + \gamma for all complex numbers z,z, where α\alpha and γ\gamma are complex numbers and i2=1.i^2 = -1. Suppose that f(1)f(1) and f(i)f(i) are both real. What is the smallest possible value of α+γ?|\alpha| + |\gamma|?

11

2\sqrt{2}

22

222\sqrt{2}

44

Answer: B
Concepts:complex numberoptimization
Difficulty rating: 1990
Solution:

Let α=a+bi\alpha = a + bi and γ=c+di.\gamma = c + di. Then f(1)=(4+a+c)f(1) = (4 + a + c) +(1+b+d)i+ (1 + b + d)i and f(i)=(4b+c)f(i) = (-4 - b + c) +(1+a+d)i.+ (-1 + a + d)i.

Both being real forces 1+b+d=01 + b + d = 0 and 1+a+d=0,-1 + a + d = 0, i.e. a=1da = 1 - d and b=1d.b = -1 - d.

Hence α+γ=(1d)2+(1+d)2+c2+d2=2+2d2+c2+d2, \begin{aligned} &|\alpha| + |\gamma| \\ &= \sqrt{(1 - d)^2 + (1 + d)^2} \\ &\quad {}+ \sqrt{c^2 + d^2} \\ &= \sqrt{2 + 2d^2} \\ &\quad {}+ \sqrt{c^2 + d^2}, \end{aligned} which is smallest when c=d=0,c = d = 0, giving 2.\sqrt{2}.

Thus, the correct answer is B.

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