complex numberswhat is the value of z=(2+3i)*(1+2i)^2?

First, we need to raise to square the second factor. For
this reason, we'll use the formula:

(a+b)^2 = a^2 + 2ab +
b^2

Let a = 1 and b  =
2i

(1+2i)^2 = 1 + 2*1*2i +
(2i)^2

(1+2i)^2 = 1 + 4i +
4i^2

But i^2 = -1

(1+2i)^2 = 1
+ 4i - 4

(1+2i)^2 = -3 +
4i

Now, we'll perform the
multiplication:

(2+3i)*(1+2i)^2 = (2+3i)*(-3 +
4i)

(2+3i)*(1+2i)^2 = -6 + 8i - 9i +
12i^2

(2+3i)*(1+2i)^2 = -6 - i -
12

(2+3i)*(1+2i)^2 = - 18 -
i

The result of multiplication is the complex
number z = - 18 - i.