3+6i / 6+i HELP HELP HELP? algebra?

Question

i need it in the a+bi form

Answer 1

you need to take 6+i off the bottom this is done by multiplying by 1, where 1 = (6-i)/(6-i)


so N = [(3+6i)/(6+i)] * [(6-i)/(6-i)]

= (3+6i)(6-i) / [(6+i)(6-i)] (note that (a-bi)(a+bi) = a^2 +b^2 since the i terms cancel out.)

= [18 +36i -3i +6 ]/ [36 +1]

= 24/37 + 33i/37

Answer 2

( 6 - i )( 3 + 6 i ) / ( 6 + i )( 6 - i )
(18 + 33 i + 6 ) / (36 + 1)
( 24 + 33 i ) / 37