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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.
Posted by giorgiana1976 on September 8, 2011 at 11:56 PM (Answer #1)
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