Question

Sa se calculeze : a) (1-i) ( 1 +2i) - 3(2-i) ; b) (2+3i) ( 4+5i) (-7+22i)

Answer 1

a)(1-i)(1+2i)-3(2-i)=1+2i-i+2-6+3i=
-3+4i.
b)(2+3i)(4+5i)(-7+22i)=
(-7+22i)(-7+22i)=(-7+22i)^2
Se consideră i^2=-1.

Answer 2

a) (1-i) ( 1 +2i) - 3(2-i) =1+2i-i-2i^2-6+3i=-5+i+3i+2=-3+4i
b) (2+3i) ( 4+5i) (-7+22i)=(8+10i+12i+15i^2)(-7+22i)= (8+22i-15)(-7+22i)= (-7+22i)(-7+22i)= (-7+22i)^2= 49-308i+484i^2=49-308i-484=
-435-308i