Question

Efetue:
a) (2+3i)•(1+5i)
b) (1+2i)•(2+i)
c) (4-3i)•(5-i)•(1+i)
d) (7+2i)•(7-2i)

Answer 1

OBS.: lembrando que nos números complexos i^2=-1

a) (2+3i).(1+5i) = 2+10i+3i+15i^2

= 2+13i+15.(-1) = 2+13i-15

= 13i-13

b) (1+2i).(2+i) = 2+i+4i+2i^2

+ 2+5i+2.(-1) = 2+5i-2

=5i

c) (4-3i).(5-i).(1+i) = (4-3i).(5+5i-i-i^2)

= (4-3i).(5+4i-(-1)) = (4-3i).(5+4i+1)

= (4-3i).(6+4i) = 24+16i-18i-12i^2

= 24-2i-12.(-1) = 24-2i+12

= 36-2i

d) (7+2i).(7-2i) = 49-14i+14i-4i^2

= 49-4.(-1) = 49+4 = 53