Difference between revisions of "2008 AMC 12B Problems/Problem 19"

Revision as of 09:54, 4 July 2013

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:

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

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

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 $-i$ , 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
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.

@@ Line 17: / Line 17: @@
 ==See Also==
 {{AMC12 box|year=2008|ab=B|num-b=18|num-a=20}}
+{{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "2008 AMC 12B Problems/Problem 19"

Revision as of 09:54, 4 July 2013

Problem 19

Solution

See Also