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

Asked on by prontokla

### 2 Answers |Add Yours

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to determine the value of z=(2+3i)*(1+2i)^2

z=(2+3i)*(1+2i)^2

=> z=(2+3i)*(1 + 4i^2 + 4i)

=> z=(2 + 3i)*(1 - 4 + 4i)

=> z=(2 + 3i)*(-3 + 4i)

=> z = -6 - 9i + 8i + 12i^2

=> z = -6 - 9i + 8i - 12

=> z = -18 - i

The required value of z= -18 - i

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

We’ve answered 319,187 questions. We can answer yours, too.