2008 AMC 12B Problems/Problem 19

Problem 19

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

$\textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad$

Solution

We need only concern ourselves with the imaginary portions of $f(1)$ and $f(i)$ (both of which must be 0). These are:

$1) f(1) = i+i\textrm{Im}(\alpha)+i\textrm{Im}(\gamma)$

$2) f(i) = -i+i\textrm{Re}(\alpha)+i\textrm{Im}(\gamma)$

Since $\textrm{Im}(\gamma)$ appears in both equations, we let it equal 0 to simplify the equations. This yields two single-variable equations. Equation 1 tells us that the imaginary part of $\alpha$ must be $-1$ , and equation 2 tells us that the real part of $\alpha$ must be $i/i = 1$ . Therefore, $\alpha = 1-i$ . There are no restrictions on $\textrm{Re}(\gamma)$ , so to minimize $\gamma$ 's absolute value, we let $\textrm{Re}(\gamma) = 0$ .

$| \alpha | + |\gamma | = |1-i| + |0| = \sqrt{2} \Rightarrow \boxed{B}$ .

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

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2008 AMC 12B Problems/Problem 19

Problem 19

Solution

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