Pré-Vestibular ⇒ (UNICAMP - 2003) Números Complexos: Forma Algébrica

[b4]

Considere a função quadrática [tex3]f(x)=x^2+x cos \alpha+ \text{sen} \alpha[/tex3].

a) Resolva a equação [tex3]f(x)=0[/tex3] para [tex3]\alpha=\frac{3\pi}{2}.[/tex3]
b) Encontre os valores de [tex3]\alpha[/tex3] para os quais o número complexo [tex3]\frac{1}{2} + \frac{\sqrt{3}}{2}i[/tex3] é raiz da equação [tex3]f(x)+1=0.[/tex3]

Respostas:

---

a) [tex3]S= \{-1; 1\}[/tex3]
b) [tex3]\alpha= (2k+1)\cdot \pi ,\, k \in \mathbb{Z}[/tex3]

[matbatrobin]

a)

• [tex3]x^2+x\cdot \cos\left(\frac{3\pi}{2}\right)+\text{sen}\left(\frac{3\pi}{2}\right)=0[/tex3]
  
  [tex3]x^2-1=0\Longrightarrow x^2=1\Longrightarrow x=1 \text{ ou } x=-1[/tex3]

Resposta: [tex3]S=\{1;-1\}[/tex3]

b)

• [tex3]\begin{array}{rl}f\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)&= \left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)^2+\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\cdot \cos\alpha+\text{sen}\alpha\\
  &=\frac{1}{4}+\frac{\sqrt{3}}{2}i-\frac{3}{4}+\frac{\cos\alpha}{2}+\frac{\sqrt{3}}{2}\cos\alpha i+\text{sen}\alpha \\
  &=\text{sen}\alpha + \frac{\cos\alpha}{2}-\frac{1}{2}+\frac{\sqrt{3}}{2}\left(\cos\alpha +1\right)i
  \end{array}[/tex3]

• [tex3]f\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)+1=0\Longrightarrow \text{sen}\alpha + \frac{\cos\alpha}{2}+\frac{1}{2}+\frac{\sqrt{3}}{2}\left(\cos\alpha +1\right)i=0[/tex3]

Para que [tex3]\frac{1}{2} + \frac{\sqrt{3}}{2}i[/tex3] seja raiz, devemos ter

• [tex3]\begin{cases}\cos\alpha +1=0\\\text{ e}\\\text{sen}\alpha + \frac{\cos\alpha}{2}+\frac{1}{2}=0 \end{cases} \Longrightarrow \begin{cases}\cos\alpha =-1\\\text{ e}\\\text{sen}\alpha =0 \end{cases} \Longrightarrow \alpha=\pi+2k\pi,\,k \in \mathbb{Z}[/tex3]

Resposta: [tex3]\alpha=(2k+1)\pi,\,k \in \mathbb{Z}[/tex3]