▶️ Ejercicio 263 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXXII

CANTIDADES IMAGINARIAS

Ejercicio 263

Dividir:

$\begin{matrix} (1 + \sqrt{- 1}) \div (1 - \sqrt{- 1}) \\ \frac{1 + \sqrt{- 1}}{1 - \sqrt{- 1}} & = \frac{1 + \sqrt{- 1}}{1 - \sqrt{- 1}} \times \frac{1 + \sqrt{- 1}}{1 + \sqrt{- 1}} \\ = \frac{{(1 + \sqrt{- 1})}^{2}}{1^{2} - {(\sqrt{- 1})}^{2}} \\ = \frac{1 + 2 \sqrt{- 1} + {(\sqrt{- 1})}^{2}}{1 - (- 1)} \\ = \frac{1 + 2 \sqrt{- 1} - 1}{1 + 1} \\ = \frac{2 \sqrt{- 1}}{2} \\ = i \end{matrix}$
$\begin{matrix} (3 + \sqrt{- 1}) \div (3 - \sqrt{- 1}) & = \frac{3 + \sqrt{- 1}}{3 - \sqrt{- 1}} \times \frac{3 + \sqrt{- 1}}{3 + \sqrt{- 1}} \\ = \frac{{(3 + \sqrt{- 1})}^{2}}{3^{2} - {(\sqrt{- 1})}^{2}} \\ = \frac{9 + 6 \sqrt{- 1} + {(\sqrt{- 1})}^{2}}{9 - (- 1)} \\ = \frac{9 + 6 \sqrt{- 1} - 1}{9 + 1} \\ = \frac{8 + 6 \sqrt{- 1}}{10} \\ = \frac{2 (4 + 3 i)}{} \\ = \frac{4 + 3 i}{5} \end{matrix}$
$\begin{matrix} (5 - 3 \sqrt{- 1}) \div (3 + 4 \sqrt{- 1}) & = \frac{5 - 3 \sqrt{- 1}}{3 + 4 \sqrt{- 1}} \times \frac{3 - 4 \sqrt{- 1}}{3 - 4 \sqrt{- 1}} \\ = \frac{(5 - 3 i) (3 - 4 i)}{3^{2} - {(4 i)}^{2}} \\ = \frac{15 - 20 i - 9 i + 12 i^{2}}{9 - 16 i^{2}} \\ = \frac{15 - 29 i + 12 (- 1)}{9 - 16 (- 1)} \\ = \frac{15 - 29 i - 12}{9 + 16} \\ = \frac{3 - 29 i}{25} \end{matrix}$
$\begin{matrix} (8 - 5 i) \div (7 + 6 i) & = \frac{8 - 5 i}{7 + 6 i} \times \frac{7 - 6 i}{7 - 6 i} \\ = \frac{(8 - 5 i) (7 - 6 i)}{7^{2} - {(6 i)}^{2}} \\ = \frac{56 - 48 i - 35 i + 30 i^{2}}{49 - 36 i^{2}} \\ = \frac{56 - 83 i + 30 (- 1)}{49 - 36 (- 1)} \\ = \frac{56 - 83 i - 30}{49 + 36} \\ = \frac{26 - 83 i}{85} \end{matrix}$
$\begin{matrix} (4 + \sqrt{- 3}) \div (5 - 4 \sqrt{- 3}) & = \frac{4 + \sqrt{- 3}}{5 - 4 \sqrt{- 3}} \times \frac{5 + 4 \sqrt{- 3}}{5 + 4 \sqrt{- 3}} \\ = \frac{(4 + \sqrt{3} i) (5 + 4 \sqrt{3} i)}{5^{2} - {(4 \sqrt{- 3})}^{2}} \\ = \frac{20 + 16 \sqrt{3} i + 5 \sqrt{3} i + 4 (3 i^{2})}{25 - 16 (- 3)} \\ = \frac{20 + 21 \sqrt{3} i + 12 i^{2}}{25 + 48} \\ = \frac{20 + 21 \sqrt{3} i + 12 (- 1)}{73} \\ = \frac{20 + 21 \sqrt{3} i - 12}{73} \\ = \frac{8 + 21 \sqrt{3} i}{73} \end{matrix}$
$\begin{matrix} (\sqrt{2} + 2 \sqrt{- 5}) \div (4 \sqrt{2} - \sqrt{- 5}) & = \frac{\sqrt{2} + 2 \sqrt{- 5}}{4 \sqrt{2} - \sqrt{- 5}} \times \frac{4 \sqrt{2} + \sqrt{- 5}}{4 \sqrt{2} + \sqrt{- 5}} \\ = \frac{\sqrt{2} + 2 \sqrt{5} i}{4 \sqrt{2} - \sqrt{5} i} \times \frac{4 \sqrt{2} + \sqrt{5} i}{4 \sqrt{2} + \sqrt{5} i} \\ = \frac{(\sqrt{2} + 2 \sqrt{5} i) (4 \sqrt{2} + \sqrt{5} i)}{{(4 \sqrt{2})}^{2} - {(\sqrt{5} i)}^{2}} \\ = \frac{4 (2) + \sqrt{10} i + 8 \sqrt{10} i + 2 (5) i^{2}}{16 (2) - 5 i^{2}} \\ = \frac{8 + 9 \sqrt{10} i + 10 (- 1)}{32 - 5 (- 1)} \\ = \frac{8 + 9 \sqrt{10} i - 10}{32 + 5} \\ = \frac{- 2 + 9 \sqrt{10} i}{37} \end{matrix}$