Answer:
(7i-1)/5
Step-by-step explanation:
(4+2i)/(1-3i)
the opposite of 1-3i is 1+3i
multiply that
[(4+2i)(1+3i)]/[(1-3i)(1+3i)]
=[(4+2i+12i+6i^2)]/[(1-3i+3i-9i^2)]
=[(4+14i+6(-1))]/[(1-9(-1))]
=(4+14i-6)/(1+9)
=(-2+14i)/10
factor out the 2,
[-2(1-7i)]/10
-(1-7i)/5
(-1+7i)/5
(7i-1)/5
Remember that i^2=-1, because sqrt(-1)*sqrt(-1)=-1.
