LElizabeth02
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What can you say about what the roots of cubic with complex coefficients will be ?
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RDKGames
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(Original post by LElizabeth02)
What can you say about what the roots of cubic with complex coefficients will be ?
They will not necessarily occur in conjugate pairs.
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DFranklin
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(Original post by RDKGames)
They will not occur in conjugate pairs.
Some of this is going to be a bit pedantic, but I think this question is actually a bit of a minefield.

Firstly, \mathbb{R} \subset \mathbb{C}, so arguably x^2 + 1 is still a poly with complex coefficients.

But even if we exclude this bit of pedantry, obviously ix^2 + i is a poly with complex coefficients.

To get around that, let's insist that our poly is monic. Then what about x^3 - ix^2 + x - i = (x-i)^2(x+i). At a minimum, this has (a pair of) roots that come in conjugate pairs; arguably, *all* of its roots come in conjugate pairs (if you say that the only *roots* are x = i, x = -i).

It's probably one of those questions where you really need to ask the OP why they're asking.
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ghostwalker
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(Original post by DFranklin)
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Always a pleasure to read your expositions - PRSOM.
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