Question about the Fundamental Theorem of Algebra

Question

I have been studying the Fundamental theorem of Algebra over the past few days, and I'm having a hard time finding the answers for my following question. I hope that someone will shine a light so I can move forward with my study.

From Wikipedia:

"The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number is a complex number with its imaginary part equal to zero.

The theorem is also stated as follows: every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots. The equivalence of the two statements can be proven through the use of successive polynomial division."

My question is:

Is there such a thing as a polynomial with complex (imaginary part included) coefficients? Can a polynomial be created such as $f(x) = (x-2)(x-i)(x+2i) = x^3 + (2+i)x^2 + (2+2i)x + 4$? If not, is the reason for it the fact that the coefficients of the polynomial contain an imaginary part, or because it simply defies the complex conjugate root theorem?

Thank you.

Answer 1

There is no barrier to considering polynomials with complex coefficients and complex roots. The fundamental theorem of algebra is an assertion about those polynomials - each one factors into a product of linear factors.

When the coefficients happen to be real the roots must occur in conjugate pairs.

Edit in response to comment.

If the roots occur in conjugate pairs then the coefficients are real, because $(x- r)(x-\bar r)$ has real coefficients. But the roots must pair up. The polynomial $$ (x-i)^2(x+i) $$ has $i$ and $-i$ as roots but nonreal coefficients.

Answer 2

For sure complex polynomials exist.

You can define a polynomial on any field $\mathbb F$. For example

$$ p(x) = x^2+x+1$$ is a polynomial of the field with two elements.

You can even define polynomials on rings like $\mathbb Z$ like the polynomial

$$q(x) = 3x^3-x+27.$$

Answer 3

A real polynomial $f(X)$ (formally, an element of the polynomial ring $\Bbb R[X]$) is of the form

$$f(X)=a_nX^n+\dots+a_0~~~a_j\in\Bbb R,\,a_n\ne0$$

A complex polynomial $f(X)$ (formally, an element of the polynomial ring $\Bbb C[X]$) is of the form

$$f(X)=a_nX^n+\dots+a_0~~~a_j\in\Bbb C,\,a_n\ne0$$

Alternatively, in the latter case every $a_j$ admits a representation of the form $a_j=x_j+iy_j$ where $x_j,y_j\in\Bbb R$.

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In particular, every real polynomial is a complex polynomial (since $\Bbb R\subset\Bbb C$) but not vice-versa. Your given polynomial is completely fine as polynomial over $\Bbb C$ but not over $\Bbb R$ as $i\notin\Bbb R$.