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Stupid question about this topic: I don't understand the rule that says "A poly with degree N will have N complex roots". eg: $P(X)=x^5 + x^3 - 1$ is a 5th degree polynomial function, so P(x) has exactly 5 complex zeros. Sorry, but how do you know it will have 5 complex zeros? Maybe it factors into only real/integer roots? Or, is every real root also a complex root? ie: $x = x + 0i$

What is a complex polynomial function (which has at least one complex zero)? Just a poly with an "i" in the coefficient? eg: $4ix^2 + ...$

Disclaimer: I am not a student trying to get free internet homework help. I am an adult who is learning Calculus from a textbook. I am deeply grateful to the members of this community for their time.

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    $\begingroup$ Real roots are special complex roots, those with imaginary part $0$. The mention of "complex" only indicates that one doesn't have to leave $\mathbb{C}$ to find the zeros. $\endgroup$ – Daniel Fischer Nov 4 '13 at 22:23
  • $\begingroup$ Your textbook should also say something about the roots being "counted according to their multiplicity". $(x - 1)^2(x - i)^2$ is a polynomial of degree 4 with just 2 distinct roots, $1$ and $i$. $\endgroup$ – Rob Arthan Nov 4 '13 at 22:35
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Yes, real numbers are complex numbers with imaginary part $0$.

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Every real number is a complex number. You aren't guaranteed that every n-degree polynomial will have n roots, i.e. x^2+1= 0 has no REAL roots but two complex roots, namely -i and i. x^2-1=0 has two complex roots as well, 1 and -1, they just happen to have no imaginary part and are thus entirely real.

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The real numbers are a subset of the complex numbers ($\mathbb{R}\subset\mathbb{C}$), so any element of the reals is also complex.

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