Exercício Resolvido de Números Complexos
g)
\( \dfrac{4-3i}{-1+i}-\dfrac{1-i}{\sqrt{2}-i} \)
Multiplicando o numerador e o denominador do primeiro termo pelo complexo conjugado do denominador
(
\( \overline{z}=-1-i \)),
e multiplicando o numerador e o denominador do segundo termo pelo complexo conjugado do denominador
(
\( \overline{z}=\sqrt{2}+i \))
\[
\begin{gather}
\frac{4-3i}{(-1+i)}.\frac{(-1-i)}{-1-i}-\frac{1-i}{\sqrt{2}-i}.\frac{\left(\sqrt{2}+i\right)}{\left(\sqrt{2}+i\right)}\\[5pt]
\frac{[4.(-1)+4.(-i)-3 i .(-1)-3 i .(-i)]}
{[-1.(-1)-1.(-i)+ i .(-1)+ i. (-i)]}
-\frac{\left[1.\sqrt{2}+1.i-i.\sqrt{2}- i . i\right]}
{\left[\sqrt{2}.\sqrt{2}+\sqrt{2}.i-i.\sqrt{2}- i . i\right]}\\[5pt]
\frac{[-4-4i+3i+3i^{2}]}{[1+i-i-i^{2}]}-\frac{\left[\sqrt{2}+i-\sqrt{2}\;i-i^{2}\right]}{\left[\left(\sqrt{2}\right)^{2}+\sqrt{2}\;i-\sqrt{2}`i.-i^{2}\right]}
\end{gather}
\]
sendo
\( i^{2}=-1 \)
\[
\begin{gather}
\frac{[-4-i+3.(-1)]}{[1-(-1)]}-\frac{\left[\sqrt{2}+\left(1-\sqrt{2}\right)\;i-(-1)\right]}{[2-(-1)]}\\[5pt]
\frac{[-4-i-3]}{[1+1]}-\frac{\left[\sqrt{2}+\left(1-\sqrt{2}\right)\;i+1\right]}{[2+1]}\\[5pt]
\frac{-7-i}{2}-\frac{1+\sqrt{2}+\left(1-\sqrt{2}\right)\;i}{3}\\[5pt]
-{\frac{7}{2}}-\frac{i}{2}-\frac{1+\sqrt{2}}{3}+\frac{\left(1-\sqrt{2}\right)\;i}{3}\\[5pt]
\left[-{\frac{7}{2}}-\frac{1+\sqrt{2}}{3}\right]+\left[-{\frac{1}{2}}+\frac{\left(1-\sqrt{2}\right)}{3}\right]i\\[5pt]
\left[\frac{-3.7-2.\left(1+\sqrt{2}\right)}{6}\right]+\left[\frac{-3.1+2.\left(1-\sqrt{2}\right)}{6}\right]i\\[5pt]
\left[\frac{-21-2-2\sqrt{2}}{6}\right]+\left[\frac{-3-2+2\sqrt{2}}{6}\right]i
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{\frac{-23-2\sqrt{2}}{6}+\frac{-5+2\sqrt{2}}{6}i}
\]