(IME 2023/2024) Seja i tal que i^{2}=-1. O valor do número real \alpha que satisfaz à equação
\operatorname{cis}(7 \pi / 6)-2 \operatorname{cis}(-7 \pi / 6)=\left|\begin{array}{ccc} -\sqrt{3} & -i & 0 \\\ \sqrt{3} / 2 & -\sqrt{3} & \alpha \\\ i^{5} & \sqrt{3} & -i / 2 \end{array}\right|
é:
(A) 3
(B) \sqrt{3} / 16
(B) \sqrt{3} / 2
(B) \sqrt{3} / 8
(C) \sqrt{3} / 4

Passo 1

O enunciado pede o real \alpha que satisfaz
\operatorname{cis}\Bigl(\tfrac{7\pi}{6}\Bigr) -2\,\operatorname{cis}\Bigl(-\tfrac{7\pi}{6}\Bigr) =\det\begin{pmatrix} -\sqrt{3} & -i & 0 \\\ \frac{\sqrt{3}}{2} & -\sqrt{3} & \alpha \\\ i^5 & \sqrt{3} & -\frac{i}{2} \end{pmatrix}.
Vamos primeiro simplificar o lado esquerdo usando as definições trigonométricas e em seguida calcular o determinante, igualando as partes real e imaginária.

Passo 2

Calculando as cis:
\operatorname{cis}\Bigl(\tfrac{7\pi}{6}\Bigr) =-\frac{\sqrt{3}}{2}-\frac{i}{2}, \quad \operatorname{cis}\Bigl(-\tfrac{7\pi}{6}\Bigr) =-\frac{\sqrt{3}}{2}+\frac{i}{2}.
Logo
\operatorname{cis}\Bigl(\tfrac{7\pi}{6}\Bigr) -2\,\operatorname{cis}\Bigl(-\tfrac{7\pi}{6}\Bigr) =-\frac{\sqrt{3}}{2}-\frac{i}{2}-2\Bigl(-\tfrac{\sqrt{3}}{2}+\tfrac{i}{2}\Bigr) =\frac{\sqrt{3}}{2}-\frac{3i}{2}.

Passo 3

No determinante observe que i^5=i. Expandindo pela primeira linha, só os dois primeiros cofatores são não nulos:
\det M =(-\sqrt{3}) \det\!\begin{pmatrix}-\sqrt{3}&\alpha\\sqrt{3}&-\tfrac{i}{2}\end{pmatrix} +(-1)^{1+2}(-i) \det\!\begin{pmatrix}\tfrac{\sqrt{3}}{2}&\alpha\i&-\tfrac{i}{2}\end{pmatrix}.
Calculando os menores:
\det\begin{pmatrix}-\sqrt{3}&\alpha\\sqrt{3}&-\tfrac{i}{2}\end{pmatrix} =\frac{\sqrt{3}i}{2}-\alpha\sqrt{3}, \quad \det\begin{pmatrix}\tfrac{\sqrt{3}}{2}&\alpha\i&-\tfrac{i}{2}\end{pmatrix} =-\frac{\sqrt{3}i}{4}-\alpha i.
Portanto
\det M =-\sqrt{3}\Bigl(\tfrac{\sqrt{3}i}{2}-\alpha\sqrt{3}\Bigr) + i\Bigl(-\tfrac{\sqrt{3}i}{4}-\alpha i\Bigr) =4\alpha+\frac{\sqrt{3}}{4}-\frac{3i}{2}.

Passo 4

Igualando à parte já obtida no passo 2,
4\alpha+\frac{\sqrt{3}}{4}-\frac{3i}{2} =\frac{\sqrt{3}}{2}-\frac{3i}{2},
as imaginárias conferem, e das reais
4\alpha+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2} \quad\Longrightarrow\quad 4\alpha=\frac{\sqrt{3}}{4} \quad\Longrightarrow\quad \alpha=\frac{\sqrt{3}}{16}.

Resposta

\displaystyle \alpha=\frac{\sqrt{3}}{16}.

Exercícios relacionados