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

(Solution)
Line 7: Line 7:
 
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+\textrm{Im}(\alpha)+\textrm{Im}(\gamma)</math>
+
<math>1) f(1) = i+i\textrm{Im}(\alpha)+i\textrm{Im}(\gamma)</math>
  
<math>2) f(i) = -i+i\textrm{Re}(\alpha)+\textrm{Im}(\gamma)</math>
+
<math>2) f(i) = -i+i\textrm{Re}(\alpha)+i\textrm{Im}(\gamma)</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>.
+
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>-1</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 23:20, 30 December 2014

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:

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

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

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 $-1$, 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 (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

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