Question

calcule os quocientes. a=1+7i/1-3i b= 6i/2+4i c=25/3-4i d=27+18i/9i.

Answer 1

$a = \frac{1 + 7i}{1 - 3i} = \frac{(1 + 7i)(1 + 3i)}{(1 - 3i)(1 + 3i)} \\ a = \frac{1 + 3i + 7i + 21i^{2}}{1 - 9i^{2}} = \frac{1 + 10i - 21}{1 + 9} \\ a = \frac{10i - 20}{10} = i - 2$

b)
$\frac{6i}{2 + 4i} = \frac{3i}{1 + 2i} = \frac{3i(1 - 2i)}{(1 + 2i)(1 - 2i)} = \frac{3i - 6i^{2}}{1 - 4i^{2}} \\ a = \frac{3i + 6}{1 + 4} = \frac{3i + 6}{5}$

c)
$\frac{25}{3 - 4i} = \frac{25(3 + 4i)}{(3 - 4i)(3 + 4i)} = \frac{75 + 100i}{9 - 16i^{2}} \\ a = \frac{75 + 100i}{9 + 16} = \frac{75 + 100i}{25} = 3 + 4i$

d)
$d = \frac{27 + 18i}{9i} = \frac{3 + 2i}{i} = \frac{- i(3 + 2i)}{- i^{2}} = \frac{- 3i - 2i^{2}}{ - 1} \\ d = \frac{- 3i + 2}{- 1} = 3i - 2$