Exercício Resolvido de Números Complexos
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g) \( \displaystyle z=\frac{-3+3i}{1+i\sqrt{3}} \)
Multiplicando o numerador e o denominador pelo complexo conjugado do denominador \( \left(\overline{{z}}=1-i\sqrt{3}\right) \)
\[
\begin{gather}
z=\frac{(-3+3i)}{(1+i\sqrt{3})}.\frac{(1-i\sqrt{3})}{(1-i\sqrt{3})}\\[5pt]
z=\frac{(-3).1-3.(-i\sqrt{3})+3i.1+3i.(-i\sqrt{3})}{1.1+1.(-i\sqrt{3})+i\sqrt{3}.1+i\sqrt{3}.(-i\sqrt{3})}\\[5pt]
z=\frac{-3+i3\sqrt{3}+3i-i.i3\sqrt{3}}{1^{2}-i\sqrt{3}+i\sqrt{3}-i^{2}(\sqrt{3})^{2}}
\end{gather}
\]
sendo
\( i^{2}=-1 \)
\[
\begin{gather}
z=\frac{-3+i(3\sqrt{3}+3)-(-1).\sqrt{3}}{1-(-1).3}\\[5pt]
z=\frac{-3+i(3\sqrt{3}+3)+\sqrt{3}}{1+3}\\[5pt]
z=\frac{3\sqrt{3}-3}{4}+i\frac{3\sqrt{3}+3}{4}
\end{gather}
\]
O módulo é dado por
\[ \bbox[#99CCFF,10px]
{|z|=\sqrt{x^{2}+y^{2}}}
\]
\[
\begin{gather}
|z|=\sqrt{\left(\frac{3\sqrt{3}-3}{4}\right)^{2}+\left(\frac{3\sqrt{3}+3}{4}\right)^{2}}\\[5pt]
|z|=\sqrt{\frac{(3\sqrt{3})^{2}-2.3\sqrt{3}.3+3^{2}}{16}+\frac{(3\sqrt{3})^{2}+2.3\sqrt{3}.3+3^{2}}{16}}\\[5pt]
|z|=\sqrt{\frac{27-6\sqrt{3}+9}{16}+\frac{27+6\sqrt{3}+9}{16}}\\[5pt]
|z|=\sqrt{\frac{27-6\sqrt{3}+9+27+6\sqrt{3}+9}{16}}\\[5pt]
|z|=\sqrt{\frac{72}{16}}\\|z|=\frac{\sqrt{9.2^{3}}}{4}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{|z|=\frac{3}{2}\sqrt{2}}
\]
O argumento é dado por
\[ \bbox[#99CCFF,10px]
{\theta=\operatorname{arg}(z)=\operatorname{arctg}\left(\frac{y}{x}\right)}
\]
\[
\begin{gather}
\theta=\operatorname{arctg}\left[\frac{\dfrac{3\sqrt{3}+3}{\cancel{4}}}{\dfrac{3\sqrt{3}-3}{\cancel{4}}}\right]\\[5pt]
\theta=\operatorname{arctg}\left[\frac{3\sqrt{3}+3}{3\sqrt{3}-3}\right]\\[5pt]
\theta=\operatorname{arctg}\left[\frac{\sqrt{3}+1}{\sqrt{3}-1}\right]\\[5pt]
\theta=\operatorname{arctg}\left[\frac{1+\sqrt{3}}{-(1-\sqrt{3})}\right]\\[5pt]
\theta=\operatorname{arctg}\left[-{\frac{1+\sqrt{3}}{(1-\sqrt{3})}}\right]
\end{gather}
\]
Usando a propriedade trigonométrica
\[
\operatorname{arctg}(-x)=-\operatorname{arctg}(x)
\]
\[
\theta=-\operatorname{arctg}\left[\frac{1+\sqrt{3}}{1-\sqrt{3}}\right]
\]
Usando a propriedade trigonométrica
\[
\operatorname{arctg}\left(\frac{x+y}{1-xy}\right)=\operatorname{arctg}(x)+\operatorname{arctg}(y)
\]
definindo x = 1 e
\( y=\sqrt{3} \),
temos
\[
\operatorname{arctg}\left(\frac{1+\sqrt{3}}{1-1.\sqrt{3}}\right)=\operatorname{arctg}(1)+\operatorname{arctg}\left(\sqrt{3}\right)
\]
\[
\begin{gather}
\theta=-\left[\operatorname{arctg}(1)+\operatorname{arctg}\left(\sqrt{3}\right)\right]\\[5pt]
\theta=-\left[\frac{\pi}{4}+\frac{\pi}{3}\right]\\[5pt]
\theta =-{\frac{7\pi}{12}}=\frac{5\pi}{12}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{\theta =\frac{5\pi}{12}}
\]
Escrevendo z na forma polar
\[ \bbox[#99CCFF,10px]
{z=r(\cos \theta +i\operatorname{sen}\theta )\quad \text{,}\quad r=|z|}
\]
\[ \bbox[#FFCCCC,10px]
{z=\frac{3}{2}\sqrt{2}\;\left(\cos \frac{5\pi}{12}+i\operatorname{sen}\frac{5\pi}{12}\right)}
\]
Gráfico 1
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Fisicaexe - Exercícios Resolvidos de Física de Elcio Brandani Mondadori está licenciado com uma Licença Creative Commons - Atribuição-NãoComercial-Compartilha Igual 4.0 Internacional .