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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
(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
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