Verify if the product (3-2i)(3+2i) is complex?

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We know that multiplying a complex number by it's conjugate the result is a real number.

The special product returns a difference of two squares.

`(a - bi)(a + bi) = a^2 + abi - abi - b^2*i^2`

But `i^2` = -1

`(a - bi)(a + bi) = a^2 +b^2`

Let a = 3 and b = 2i

`(3 - 2i)(3 + 2i) = 3^2 +2^2`

(3 - 2i)(3 + 2i) = 9 + 4

(3 - 2i)(3 + 2i) = 13

**Therefore, the given product is a real number: (3 - 2i)(3 + 2i) = 13.**

The non-real component of the product is (-2i * 3) +(3 * 2i)

This is -6i + 6i =0

Therefore the product is real number. Real component being 9 +4 =13.

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