Question

(7+8i):(2+6i) qual é o resultado podem me ajudar?.

Answer 1

Sabendo que i² = -1

E que

$\large\fbox{ z=x+yi~\Longleftrightarrow~\overline{z}=x-yi }$

E que

$\large\fbox{ \dfrac{z_2}{z_1}=\dfrac{z_2}{z_1}\cdot\dfrac{\overline{z_1}}{\overline{z_1}} }$

E ainda

$\fbox{ z\cdot \overline{z}=(a+bi)\cdot(a-bi)=a^2-b^2i^2=a^2+b^2 }$

O que temos é

$~~~~\dfrac{7+8i}{2+6i}~\Leftrightarrow~\dfrac{(7+8i)}{(2+6i)}\cdot\dfrac{(2-6i)}{(2-6i)}\\\\\\\\=\dfrac{\begin{Bmatrix}(7\cdot2)-[8\cdot(-6)]\end{Bmatrix}}{2^2+6^2}+\dfrac{\begin{Bmatrix}[7\cdot(-6)]+(8\cdot2)\end{Bmatrix}}{2^2+6^2}i\\\\\\=\dfrac{\begin{Bmatrix}14-[-48]\end{Bmatrix}}{4+36}+\dfrac{\begin{Bmatrix}[-42]+16\end{Bmatrix}}{4+36}i\\\\\\=\dfrac{\begin{Bmatrix}14+48\end{Bmatrix}}{40}+\dfrac{\begin{Bmatrix}-42+16\end{Bmatrix}}{40}i$

$=\dfrac{62}{40}+\begin{pmatrix}\dfrac{-26}{40}i\end{pmatrix}\\\\\\=\dfrac{62}{40}+\begin{pmatrix}-\dfrac{26}{40}i\end{pmatrix}\\\\\\=\dfrac{62}{40}-\dfrac{26}{40}i~~~~(simplificando...)\\\\\\=\dfrac{62\div 2}{40\div 2}-\dfrac{26\div2}{40\div2}i\\\\\\\fbox{ =\dfrac{31}{20}-\dfrac{13}{20}i }$