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2008 AMC 12B Problems/Problem 19: Difference between revisions

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==Problem 19==
==Problem 19==
A function <math>f</math> is defined by <math>f(z) = (4 + i) z^2 + \alpha z + \gamma</math> for all complex numbers <math>z</math>, where <math>\alpha</math> and <math>\gamma</math> are complex numbers and <math>i^2 = - 1</math>. Suppose that <math>f(1)</math> and <math>f(i)</math> are both real. What is the smallest possible value of <math>| \alpha | + |\gamma |</math>
A function <math>f</math> is defined by <math>f(z) = (4 + i) z^2 + \alpha z + \gamma</math> for all complex numbers <math>z</math>, where <math>\alpha</math> and <math>\gamma</math> are complex numbers and <math>i^2 = - 1</math>. Suppose that <math>f(1)</math> and <math>f(i)</math> are both real. What is the smallest possible value of <math>| \alpha | + |\gamma |</math> ?


<math>\textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad</math>
<math>\textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad</math>
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We need only concern ourselves with the imaginary portions of <math>f(1)</math> and <math>f(i)</math> (both of which must be 0). These are:
We need only concern ourselves with the imaginary portions of <math>f(1)</math> and <math>f(i)</math> (both of which must be 0). These are:


<math>1) f(1) = i+\alpha_{imaginary}+\gamma_{imaginary}</math>
<math>1) f(1) = i+\textrm{Im}(\alpha)+\textrm{Im}(\gamma)</math>


<math>2) f(i) = -i+i\alpha_{real}+\gamma_{imaginary}</math>
<math>2) f(i) = -i+i\textrm{Re}(\alpha)+\textrm{Im}(\gamma)</math>
 
Since <math>\gamma_{imaginary}</math> 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 <math>\alpha</math> must be <math>-i</math>, and equation 2 tells us that the real part of <math>\alpha</math> must be <math>i/i = 1</math>. Therefore, <math>\alpha = 1-i</math>. There are no restrictions on <math>\gamma_{real}</math>, so to minimize <math>\gamma</math>'s absolute value, we let <math>\gamma_{real} = 0</math>.


Since <math>\textrm{Im}(\gamma)</math> 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 <math>\alpha</math> must be <math>-i</math>, and equation 2 tells us that the real part of <math>\alpha</math> must be <math>i/i = 1</math>. Therefore, <math>\alpha = 1-i</math>. There are no restrictions on <math>\textrm{Re}(\gamma)</math>, so to minimize <math>\gamma</math>'s absolute value, we let <math>\textrm{Re}(\gamma) = 0</math>.


<math>| \alpha | + |\gamma | = |1-i| + |0| = \sqrt{2} \Rightarrow \boxed{B}</math>.
<math>| \alpha | + |\gamma | = |1-i| + |0| = \sqrt{2} \Rightarrow \boxed{B}</math>.

Revision as of 17:06, 1 June 2011

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 $\textrm{Re}(\gamma) = 0$.

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

See Also

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
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