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Find the product of the complex number and its conjugate. 1 + 3i
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Do they mean (1 + 3i)^2?
that equals to 4
The conjugate of 1 + 3i is 1 - 3i (1 + 3i)(1 - 3i) = (1)^2 + 3i - 3i - (3i)^2 The two middle terms drop out, and you are left with (1)^2 - (3i)^2 1^2 = 1 3^2 = 9 i^2 = (√-1)^2 = -1 You end up with 1 - {(-1)(9)} So it becomes 1 - (-9) Which is 1 + 9 which equals 10
If you are given complex number \(a+bi\), then its conjugate would be \(a-bi\) So you have \((1 + 3i)(1 - 3i)\)
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Oh, ok. Just my lack of vocabulary.
ya it is 10
@jazminjones1252 you are wrong
wait 1 plus 3 equals to 4i and that is the answer
\((1 + 3i)(1 - 3i)\\~\\1^2 - (3i)^2\\~\\1 - (-9)\\~\\1+9\\~\\\boxed{10}\)
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@jazminjones1252
ok
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