Question

What is the value of the product (3 – 2i)(3 + 2i)?
5
9 + 4i
9 – 4i
13

Well... (3 - 2i)(3+2i) = (3 * 3) + (-2i * 2i) = 9 + -4i^2

i^2 = -1 so we have

9 +-4(-1)
= 9 + 4
= 13

Is my work correct here?

Answer 1

Hi friend

(3-2i)(3+2i)
as (a+b)(a-b) =a²-b²

3²-(2i)²
=9-(4*-1)
=9-(-4)
=9+4
=13
Hope this helps you!

Answer 2

we have

$(3-2i)(3+2i)$

we know that

A difference of square is equal to

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

remember that

$i^{2}=-1$

so

Applying difference of square

$(3-2i)(3+2i)=(3)^{2}-(2i)^{2}$

$=9-4i^{2}$

$=9-4(-1)$

$=9+4$

$=13$

therefore

the work is correct

the answer is $13$