Question

What's the value of the product (3-2i)(3+2i)

Answer 1

Answer:

the product (3-2i)(3+2i) is, 13

Step-by-step explanation:

Using the identity rule:

$(a+b)(a-b) = a^2-b^2$

Given that:

$(3-2i)(3+2i)$             ...[1]

here, i is the imaginary part.

$i^2 = -1$

Apply the rule on [1] we have;

$(3)^2-(2i)^2$

⇒$9-(4i^2)$

⇒$9+4$

Simplify:

13

Therefore, the product (3-2i)(3+2i) is, 13

Answer 2

We have to multiply given (3-2i)(3+2i)
(3-2i)(3+2i)=9-6i+6i+4i^2
9+4i^2
We know that i^2=-1, so lets substitute it
9+4*(-1)=9-4=5