The Complex Conjugate Root Theorem requires a polynomial function with real coefficients. This seems to imply the possibility that a complex polynomial can exist with an odd number of complex roots. True, or am I making an assumption based on facts not in evidence?
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Hint: think about $z-i$ . . .
Noah Schweber
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@Deusovi I assumed "complex roots" referred to complex non-real roots. (Note that $f(z)=z$ has only real coefficients.) – Noah Schweber Apr 03 '16 at 00:45
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every monic polynomial f(z) is the product of $z-\alpha_k$, where $\alpha_k$ are its roots. – Dima Pasechnik Apr 03 '16 at 05:48
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@DimaPasechnik Yes - so? – Noah Schweber Apr 03 '16 at 05:50
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so you can get polynomials of degree n with n prescribed roots... – Dima Pasechnik Apr 03 '16 at 05:52
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Well, you don't need that fact though - just look at $\prod (z-\alpha_k)$ for distinct $\alpha_k$s. We don't need to know that every monic has this form. – Noah Schweber Apr 03 '16 at 05:56
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I agree, I was just thinking aloud – Dima Pasechnik Apr 03 '16 at 05:59