Answer:
0 + 5i
Step-by-step explanation:
Multiply these the way you would any binomials, then replace i² with -1.
(1 +2i)(2 +i) = 1·2 +1·i +(2i)·2 +(2i)·i
= 2 + i + 4i + 2i² . . . . . the partial products
= 2 +5i -2 . . . . . . . . . . after replacement of i² = -1
= 0 +5i
