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

Sagot :

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

Answer:

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

Step-by-step explanation:

Using the identity rule:

[tex](a+b)(a-b) = a^2-b^2[/tex]

Given that:

[tex](3-2i)(3+2i)[/tex]             ...[1]

here, i is the imaginary part.

[tex]i^2 = -1[/tex]

Apply the rule on [1] we have;

[tex](3)^2-(2i)^2[/tex]

⇒[tex]9-(4i^2)[/tex]

⇒[tex]9+4[/tex]

Simplify:

13

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