Dados os números complexos z1 = 1, z2 = -i e z3 = z1 + z2, a forma trigonométrica de (z3)2 é?
Gabarito:
2(cos 3pi/2 + i sen 3pi/2)
Ensino Médio ⇒ Complexo (fórmula trigonométrica) Tópico resolvido
Moderador: [ Moderadores TTB ]
Mai 2018
29
08:19
Re: Complexo (fórmula trigonométrica)
Desenvolvendo a forma algébrica:
[tex3]z_3=z_1+z_2=1-i[/tex3]
[tex3]{z_3}^2=(1-i)^2=1-2i+i^2=1-2i-1=-2i[/tex3]
[tex3]|{z_3}^2|=\sqrt{0^2+(-2)^2}=2[/tex3]
[tex3]\begin{cases}\cos\theta=\frac{0}{2}=0\\\sin\theta=\frac{-2}{2}=-1\end{cases}\rightarrow\theta=\frac{3\pi}{2}[/tex3]
[tex3]{z_3}^2=2\(\cos\frac{3\pi}{2}+i\cdot\sin\frac{3\pi}{2}\)[/tex3]
De outra forma:
[tex3]z_3=z_1+z_2=1-i[/tex3]
[tex3]|z_3|=\sqrt{1^2+(-1)^2}=\sqrt{2}[/tex3]
[tex3]\begin{cases}\cos\theta=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\\\sin\theta=\frac{-1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\end{cases}\rightarrow\theta=\frac{7\pi}{4}[/tex3]
[tex3]z_3=\sqrt{2}\(\cos\frac{7\pi}{4}+i\cdot\sin\frac{7\pi}{4}\)[/tex3]
[tex3]{z_3}^2=\sqrt{2}^2\(\cos\frac{2\cdot7\pi}{4}+i\cdot\sin\frac{2\cdot7\pi}{4}\)=2\(\cos\frac{7\pi}{2}+i\cdot\sin\frac{7\pi}{2}\)=2\[\cos\(2\pi+\frac{3\pi}{2}\)+i\cdot\sin\(2\pi+\frac{3\pi}{2}\)\]=2\(\cos\frac{3\pi}{2}+i\cdot\sin\frac{3\pi}{2}\)[/tex3]
[tex3]z_3=z_1+z_2=1-i[/tex3]
[tex3]{z_3}^2=(1-i)^2=1-2i+i^2=1-2i-1=-2i[/tex3]
[tex3]|{z_3}^2|=\sqrt{0^2+(-2)^2}=2[/tex3]
[tex3]\begin{cases}\cos\theta=\frac{0}{2}=0\\\sin\theta=\frac{-2}{2}=-1\end{cases}\rightarrow\theta=\frac{3\pi}{2}[/tex3]
[tex3]{z_3}^2=2\(\cos\frac{3\pi}{2}+i\cdot\sin\frac{3\pi}{2}\)[/tex3]
De outra forma:
[tex3]z_3=z_1+z_2=1-i[/tex3]
[tex3]|z_3|=\sqrt{1^2+(-1)^2}=\sqrt{2}[/tex3]
[tex3]\begin{cases}\cos\theta=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\\\sin\theta=\frac{-1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\end{cases}\rightarrow\theta=\frac{7\pi}{4}[/tex3]
[tex3]z_3=\sqrt{2}\(\cos\frac{7\pi}{4}+i\cdot\sin\frac{7\pi}{4}\)[/tex3]
[tex3]{z_3}^2=\sqrt{2}^2\(\cos\frac{2\cdot7\pi}{4}+i\cdot\sin\frac{2\cdot7\pi}{4}\)=2\(\cos\frac{7\pi}{2}+i\cdot\sin\frac{7\pi}{2}\)=2\[\cos\(2\pi+\frac{3\pi}{2}\)+i\cdot\sin\(2\pi+\frac{3\pi}{2}\)\]=2\(\cos\frac{3\pi}{2}+i\cdot\sin\frac{3\pi}{2}\)[/tex3]
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