Question

A)z1•z2. b) z1/z2 z1=(1-2i) z2=(-2+i)

Answer 1

$z1cdot z2 = (1-2i)(-2+i) = 1cdot (-2+i) - 2icdot (-2+i) =$
$= -2 + i +4i - 2cdot i^2 = -2 + 5i -2cdot (-1) = -2 + 5i + 2 = 5i$
$frac{z1}{z2}=frac{1-2i}{-2+i} = frac{(1-2i)cdot (-2-i)}{(-2+i)cdot (-2-i)}=$
$= frac{-2-i - 2icdot (-2-i)}{(-2)^2 - i^2}=frac{-2-i+4i+2i^2}{4 - i^2} =$
$= frac{-2+3i+2cdot (-1)}{4 - (-1)} = frac{-2+3i-2}{4+1} =$
$=frac{-4+3i}{5} = -frac{4}{5} + frac{3}{5}i$