Question

(Números complexos) @)Efetue as operações
$a)\frac{ - 2 + 5i}{7 - i }$
$b)\frac{3 + 4i}{ - 5 + 2i}$

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Answer 1

a)

$\frac{(-2+5i)(7+i)}{(7-i)(7+i)} =\frac{-14-2i+35i+5i^{2} }{49-i^{2} } =\frac{-14+33i+5.(-1)}{49-(-1)} =\frac{-14+33i-5}{49+1} =\frac{-19+33i}{50} =-\frac{19}{50} +\frac{33i}{50}$

b)

$\frac{(3+4i)(-5-2i)}{(-5+2i)(-5-2i)} =\frac{-15-6i-20i-8i^{2} }{25-4i^{2} } =\frac{-15-26i-8.(-1)}{25-4.(-1)} =\frac{-15-26i+8}{25+4} =\frac{-7-26i}{29} =-\frac{7}{29} -\frac{26i}{29}$

Answer 2

Resposta:

$\frac{( - 2 + 5i)(7 + i)}{(7 - i)(7 + i)}$

$\frac{ - 14 - 2i + 35i + 5 {i}^{2}}{49 - {i}^{2} }$

$\frac{ - 14 - 2i + 35i + 5 \times ( - 1)}{49 - ( - 1)}$

$\frac{ - 14 - 2i + 35i - 5}{50}$

$r = \frac{ - 19}{50} + \frac{33}{50}i$

b)

$\frac{(3 + 4i)( - 5 - 2i)}{( - 5 + 2i)( - 5 - 2i)}$

$\frac{ - 15 - 6i - 20i - 8 {i}^{2} }{25 - 4 {i}^{2} }$

$\frac{ - 7 - 26i}{29}$

$r = - \frac{7}{29} - \frac{26}{29} i$