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.

  • $\begingroup$ That's a polynomial, with complex (but not real) coefficients. $\endgroup$ Sep 7, 2020 at 15:04
  • $\begingroup$ There are polynomials with complex coefficients, though they don't necessarily follow the complex conjugate root theorem $\endgroup$ Sep 7, 2020 at 15:07
  • $\begingroup$ @J.W.Tanner so how do we generalize that statement? "For the polynomial to follow the complex root theorem, the coefficients cannot have an imaginary part"? Is that valid? $\endgroup$
    – bru1987
    Sep 7, 2020 at 15:09
  • 2
    $\begingroup$ the complex conjugate root theorem is about polynomials with real coefficients $\endgroup$ Sep 7, 2020 at 15:13
  • 1
    $\begingroup$ @bru1987 - that is essentially the complex root theorem $\endgroup$
    – Henry
    Sep 7, 2020 at 15:23

3 Answers 3


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.

  • $\begingroup$ thank you for your answer. Therefore, from your statement, can I generalize "For the polynomial to follow the complex conjugate root theorem, the coefficients cannot have an imaginary part"? $\endgroup$
    – bru1987
    Sep 7, 2020 at 15:10
  • $\begingroup$ @bru1987: No, because the complex conjugate root theorem is about real polynomials to begin with. If the source you are reading does not state that clearly, then it is a bad source. It is just like many high-school texts that state "$a^{b·c} = (a^b)^c$" without stating what $a,b,c$ are. That is bad, because it is false for $(a,b,c)=(-1,6,1/2)$. The theorem is: Take any real polynomial $f$ and $z∈ℂ$ such that $f(z) = 0$. Then $f(z^*) = 0$. Proof sketch: $f(z^*) = \cdots = (f(z))^* = 0$ (using the fact that $c·(z^*)^k = c^*·(z^*)^k = (c·z^k)^*$ for every $c∈ℝ$ and $z∈ℂ$ and $k∈ℕ$). $\endgroup$
    – user21820
    Sep 28, 2020 at 4:46

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

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


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 $x_j,y_j\in\Bbb R$.

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 $i\notin\Bbb R$.


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