Question

A)z1•z2. b) z1/z2 z1=(1-2i) z2=(-2+i)

Answer 1

$z1\cdot z2 = (1-2i)(-2+i) = 1\cdot (-2+i) - 2i\cdot (-2+i) =$
$= -2 + i +4i - 2\cdot i^2 = -2 + 5i -2\cdot (-1) = -2 + 5i + 2 = 5i$
$\frac{z1}{z2}=\frac{1-2i}{-2+i} = \frac{(1-2i)\cdot (-2-i)}{(-2+i)\cdot (-2-i)}=$
$= \frac{-2-i - 2i\cdot (-2-i)}{(-2)^2 - i^2}=\frac{-2-i+4i+2i^2}{4 - i^2} =$
$= \frac{-2+3i+2\cdot (-1)}{4 - (-1)} = \frac{-2+3i-2}{4+1} =$
$=\frac{-4+3i}{5} = -\frac{4}{5} + \frac{3}{5}i$