Question

Una función $f$ se define por $f(z) = (4 + i)z^2 + \alpha z + \gamma$ para todos los números complejos $z,$ donde $\alpha$ y $\gamma$ son números complejos e $i^2 = -1.$ Supón que $f(1)$ y $f(i)$ son ambos reales. ¿Cuál es el menor valor posible de $|\alpha| + |\gamma|$ ?

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 $|\alpha| + |\gamma|?$

$1$

$\sqrt{2}$

$2$

$2\sqrt{2}$

$4$

Solución:

Sea $\alpha = a + bi$ y $\gamma = c + di.$ Entonces $f(1) = (4 + a + c)$ $+ (1 + b + d)i$ y $f(i) = (-4 - b + c)$ $+ (-1 + a + d)i.$

Que ambos sean reales obliga a $1 + b + d = 0$ y $-1 + a + d = 0,$ es decir $a = 1 - d$ y $b = -1 - d.$

Por lo tanto $\begin{aligned} &|\alpha| + |\gamma| \\ &= \sqrt{(1 - d)^2 + (1 + d)^2} \\ &\quad {}+ \sqrt{c^2 + d^2} \\ &= \sqrt{2 + 2d^2} \\ &\quad {}+ \sqrt{c^2 + d^2}, \end{aligned}$ que es mínimo cuando $c = d = 0,$ dando $\sqrt{2}.$

Por lo tanto, la respuesta correcta es B.

Let $\alpha = a + bi$ and $\gamma = c + di.$ Then $f(1) = (4 + a + c)$ $+ (1 + b + d)i$ and $f(i) = (-4 - b + c)$ $+ (-1 + a + d)i.$

Both being real forces $1 + b + d = 0$ and $-1 + a + d = 0,$ i.e. $a = 1 - d$ and $b = -1 - d.$

Hence $\begin{aligned} &|\alpha| + |\gamma| \\ &= \sqrt{(1 - d)^2 + (1 + d)^2} \\ &\quad {}+ \sqrt{c^2 + d^2} \\ &= \sqrt{2 + 2d^2} \\ &\quad {}+ \sqrt{c^2 + d^2}, \end{aligned}$ which is smallest when $c = d = 0,$ giving $\sqrt{2}.$

Thus, the correct answer is B.

2008 AMC 12B Problema 19

Solución:

El Problema 19 en otros años