2008 AMC 12B Problems/Problem 19

Problem

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 1:

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 $\boxed{B}$

By: Quaratinium

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
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

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