Question

Calcule:

$\frac{(2 + 3i).(2-4i)}{1+i}$

Answer 1

.
$\frac{(2+3i)(2-4i)}{1+i} = \\ \\ \frac{4-8i+6i-12i^2}{1+i} = \\ \\ \frac{4-2i-12(-1)}{1+i} = \\ \\ \frac{4-2i+12}{1+i} = \\ \\ \frac{16-2i}{1+i} = \\ \\ \frac{(16-2i)(1-i)}{(1+i)(1-i)} = \\ \\ \frac{16-16i-2i+2i^2}{1^2-i^2} =$

$\frac{16-18i+2(-1)}{1-(-1)} = \\ \\ \frac{16-18i-2}{1+1} = \\ \\ \frac{14-18i}{2} = \\ \\ \frac{\not2(7-9i)}{\not2} =\fbox{ 7-9i }$