Question

Sea $z$ un número complejo que satisface $12|z|^2$ $=2|z+2|^2+|z^2+1|^2+31.$ ¿Cuál es el valor de $z+\dfrac{6}{z}$ ?

Let $z$ be a complex number satisfying $12|z|^2$ $=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\dfrac{6}{z}?$

$-2$

$-1$

$\dfrac{1}{2}$

$1$

$4$

Sea $p=|z|^2=z\bar z$ y $s=z+\bar z.$ Entonces $|z+2|^2=p+2s+4,$ y $|z^2+1|^2$ $=p^2+(z^2+\bar z^2)+1$ $=p^2+(s^2-2p)+1.$

Sustituyendo, $12p=2(p+2s+4)$ $+p^2+s^2-2p+1+31,$ lo que se simplifica a $p^2-12p+s^2+4s+40=0.$

Completando el cuadrado se obtiene $(p-6)^2+(s+2)^2=0,$ así que $p=6$ y $s=-2.$

Entonces $z+\dfrac{6}{z}=z+\dfrac{6\bar z}{|z|^2}=z+\bar z=-2.$

Por lo tanto, la respuesta correcta es A.

Let $p=|z|^2=z\bar z$ and $s=z+\bar z.$ Then $|z+2|^2=p+2s+4,$ and $|z^2+1|^2$ $=p^2+(z^2+\bar z^2)+1$ $=p^2+(s^2-2p)+1.$

Substituting, $12p=2(p+2s+4)$ $+p^2+s^2-2p+1+31,$ which simplifies to $p^2-12p+s^2+4s+40=0.$

Completing the square gives $(p-6)^2+(s+2)^2=0,$ so $p=6$ and $s=-2.$

Then $z+\dfrac{6}{z}=z+\dfrac{6\bar z}{|z|^2}=z+\bar z=-2.$

Thus, the correct answer is A.