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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 $z-i$ . . .

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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 8 hours ago
    
every monic polynomial f(z) is the product of $z-\alpha_k$, where $\alpha_k$ are its roots. – Dima Pasechnik 3 hours ago
    
@DimaPasechnik Yes - so? – Noah Schweber 3 hours ago
    
so you can get polynomials of degree n with n prescribed roots... – Dima Pasechnik 3 hours ago
    
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 3 hours ago

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