Cálculos

Tem-se:

\[\begin{array}{c} a+bi\sqrt{5}=\left(c+di\sqrt{5}\right)\left(q_{1}+q_{2}i\sqrt{5}\right)+r_{1}+r_{2}i\sqrt{5}\Leftrightarrow\\ \Leftrightarrow\frac{r_{1}+r_{2}i\sqrt{5}}{c+di\sqrt{5}}=\frac{a+bi\sqrt{5}}{c+di\sqrt{5}}-\left(q_{1}+q_{2}i\sqrt{5}\right)\Leftrightarrow\\ \Leftrightarrow\frac{r_{1}+r_{2}i\sqrt{5}}{c+di\sqrt{5}}=\frac{\left(a+bi\sqrt{5}\right)\left(c-di\sqrt{5}\right)}{c^{2}+5d^{2}}-\left(q_{1}+q_{2}i\sqrt{5}\right)\Leftrightarrow\\ \Leftrightarrow\frac{r_{1}+r_{2}i\sqrt{5}}{c+di\sqrt{5}}=\frac{ac+5bd}{c^{2}+5d^{2}}-q_{1}+\left(\frac{bc-ad}{c^{2}+5d^{2}}-q_{2}\right)i\sqrt{5} \end{array}\]