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

(New page: ==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>\gamm...)
 
(Solution)
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==Solution==
 
==Solution==
 
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</math>
 
<math>1) f(1) = i+\alpha</math>
  

Revision as of 23:19, 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:

$1) f(1) = i+\alpha$

$2) f(i) = -i+i\alpha$

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, so to minimize it's absolute value, we let $\gamma = 0$.

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