ok so
(3+5i)/(2-i)
try to make it into the form a+bi
make bottom number in real
remember that i^2=-1 and
(a-b)(a+b)=a^2-b^2
look at bottomm
(2-i)
we can make it into 2^2-i^2 by multiplying it by (2+i)
multiply top and bottom
(3+5i)/(2-i) times (2+i)/(2+i)=(6+10i+3i+5i^2)/(2^2-i^2)= (6+13i+5(-1))/(4-(-1))=(6+13i-5)/(5)= (1+13i)/5=(1/5)+(13i/5)
a+bi
1/5+13/5i
