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:

$\begin{align*} \Im(f(1)) & = i+i\Im(\alpha)+i\Im(\gamma) \\ \Im(f(i)) & = -i+i\Re(\alpha)+i\Im(\gamma) \end{align*}$

Let $p=\Im(\gamma)$ and $q=\Re{(\gamma)},$ then we know $\Im(\alpha)=-p-1$ and $\Re(\alpha)=1-p.$ Therefore $|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^2}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},$ which reaches its minimum $\sqrt 2$ when $p=q=0$ by the Trivial Inequality. Thus, the answer is $\boxed B.$

Solution 2:

$f(1)=4+i+\alpha+\gamma$ $f(i)=-4-i+\alpha \cdot i +\gamma$

Since $f(1)$ and $f(i)$ are both real we get, $\alpha+\gamma=-i$ $\alpha \cdot i+\gamma=i$

Solving, we get $\alpha=1-i$ , $\gamma$ can be anything, to minimize the value we set $\gamma=0$ , so then the answer is $\sqrt{1^2+1^2}=\sqrt{2}$ . Thus, the answer is $\boxed{B}$

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

Page

Toolbox

Search

2008 AMC 12B Problems/Problem 19

Problem 19

Solution

See Also