What's the value of the product (3-2i)(3+2i)
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:
\((a+b)(a-b) = a^2-b^2\)
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