Question

Como resolvo
a) (3+3i)+(5-3i)=
b) (2+3i)-(5-3i)=
c) (3+3i)(5-3i)=
d) 2+3i/5-3i=

Answer 1

a) (3+3i)+(5-3i)= 8+0i = 8
b) (2+3i)-(5-3i)=   -3+6i
c) (3+3i)(5-3i)=   15-9i+15i-9i²  =  15+6i-9(-1) = 15+9+6i  = 24+6i

d) 2+3i/5-3i=
$\frac{2+3i}{5-3i}. \frac{5+3i}{5+3i}= \frac{10+6i+15i+9i^2}{25+15i-15i-9i^2}= \frac{10+21i+9(-1)}{25-9(-1)}= \frac{10-9+21i}{25+9}= \frac{1+21i}{34}$