Question

Să se calculeze:
a) (2 − 3i) + (−1+ i); b) (4 − 3i) − (2 − 5i); c) (1+ 3i) ⋅ (2 + 4i);
d) (2 − i) : (3 − 4i)

Answer 1

a) (2-3i)+ (-1+i)= 2-3i-1+i= 1-2i

b)(4-3i)-(2-5i)= 4-3i-2+5i= 2+2i

c)(1+3i) * (2+4i)=2+4i+6i+12i²= 2+10i+12* (-1)= 2-12+10i= -10+10i

d)(2-i):(3-4i)= (2-i)/(3-4i)= (2-i)(3+4i)/(3-4i)(3+4i)= (6+8i-3i-4i²)/(9-16i²)=

=(6+5i-4*(-1))/(9-16*(-1))= (6+5i+4)/(9+16)=(9+5i)/25