Question

Me ajudem amigos divisão de números complexos
a) 1+2i/3-2i
b) 4-10i/2i
c) 5/6+6i
d)2+iV7/1+iV7

Answer 1

a)
$\frac{1+2i}{3-2i)} = \frac{(1+2i)(3+2i)}{(3-2i)(3+2i)} = \\ \\ \frac{3+2i+6i+4i^2}{9+6i-6i-4i^2} = \\ \\ \frac{3+8i+4(-1)}{9-4(-1)} = \\ \\ \frac{3-4+8i}{9+4} = \\ \\ \frac{-1+8i}{13} =- \frac{1}{13} + \frac{8i}{13}$


b)
$\frac{4-10i}{2i} = \frac{(4-10i)(-2i)}{(2i)(-2i)} = \\ \\ \frac{-8i+20i^2}{-4i^2} = \frac{-8i+20(-1)}{-4(-1)} = \\ \\ \frac{-20-8i}{4} = \frac{4(-2i-5)}{4} = \\ \\ -5-2i$


c)
$\frac{5}{6+6i} = \frac{5(6-6i)}{(6+6i)(6-6i)} = \\ \\ \frac{30-30i}{36-6i+6i-36i^2} \\ \\ \frac{30+30i}{36-36(-1)} = \frac{30+30i}{36+36} = \\ \\ \frac{30+30i}{72} = \frac{6(5-5i)}{72} = \\ \\ \frac{5-5i}{12} = \frac{5}{12} - \frac{5i}{12}$


d
$\frac{2+i \sqrt{7} }{1+i \sqrt{7} } = \frac{(2+i \sqrt{7})(1-i \sqrt{7}) }{(1+i \sqrt{7})(1-i \sqrt{7}) } = \\ \\ \frac{2-2i \sqrt{7}+i \sqrt{7}-i^2 \sqrt{7^2} }{1-i \sqrt{7}+i \sqrt{7} -i^2 \sqrt{7^2} } = \\ \\ \frac{2-i \sqrt{7}-(-1)(7) }{1-(-1)(7)} = \\ \\ \frac{2-i \sqrt{7}+7 }{1+7} = \\ \\ \frac{9-i \sqrt{7} }{8} = \frac{9}{8} - \frac{i \sqrt{7} }{8}$