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Difference between revisions of "2008 AMC 12B Problems/Problem 19"
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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 two 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>\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 two 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>.
 +
  
 
<math>| \alpha | + |\gamma | = |1-i| + |0| = \sqrt{2}</math>, answer choice B.
 
<math>| \alpha | + |\gamma | = |1-i| + |0| = \sqrt{2}</math>, answer choice B.
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==See Also==
 +
{{AMC12 box|year=2008|ab=B|num-b=18|num-a=20}}

Revision as of 22:24, 1 March 2008

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:

$1) f(1) = i+\alpha_{imaginary}+\gamma_{imaginary}$

$2) f(i) = -i+i\alpha_{real}+\gamma_{imaginary}$

Since $\gamma_{imaginary}$ 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 two tells us that the real part of $\alpha$ must be $i/i = 1$. Therefore, $\alpha = 1-i$. There are no restrictions on $\gamma_{real}$, so to minimize $\gamma$'s absolute value, we let $\gamma_{real} = 0$.


$| \alpha | + |\gamma | = |1-i| + |0| = \sqrt{2}$, answer choice 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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