Complex and Irrational Roots

The plus-or-minus is not the reason

Start by watching the pairing show up where no single formula produced it. Take

The Rational Root Theorem lists $\pm 1$ and $\pm 13$ as the only candidates for a rational root, and $p(1) = 1 - 5 + 17 - 13 = 0$, so $x = 1$ is one. The Factor Theorem then says $x - 1$ divides $p$, and synthetic division returns the quotient:

The quadratic factor is the one from a moment ago, whose roots are $2 \pm 3i$. So the three roots of this cubic are

and the two nonreal ones are a conjugate pair again. No formula swept the cubic in one motion; the roots came out of three separate steps, and the pairing appeared anyway.

Now put the $\pm$ story on trial. If the $\pm$ were the cause, then any quadratic at all should produce conjugate answers, because every quadratic has that $\pm$. Test it on

Multiply out $(x - 1)(x - i) = x^2 - ix - x + i$ and you get exactly $q$, so the roots of $q$ are $1$ and $i$. Those are not conjugates of each other. The formula still cooperates, by the way: the discriminant is $(1+i)^2 - 4i = 2i - 4i = -2i$, and since $(1 - i)^2 = -2i$ as well, the formula’s $\pm$ delivers

The $\pm$ is right there doing its job, producing two answers, and they refuse to pair. So the $\pm$ was never the reason. The only thing that changed between $x^2 - 4x + 13$ and $q$ is that a coefficient stopped being real. That is the real suspect, and the rest of this lesson convicts it.

Conjugation slides through plus and times

Everything rests on three facts about the conjugate $\overline{z} = a - bi$ of $z = a + bi$, all of them proved when you learned complex arithmetic:

The first two say conjugation does not care whether you conjugate before or after adding, or before or after multiplying. Apply the sum rule repeatedly and the bar spreads over a sum of any length. Apply the product rule repeatedly and $\overline{z^k} = (\overline{z})^k$ for every whole number $k$, since $z^k$ is just $z$ multiplied by itself.

The third fact is a genuine two-way street, and both directions were established: a real number $r$ equals $r + 0i$, and flipping the sign of $0$ changes nothing, so $\overline{r} = r$. Conversely, for $z = a + bi$, the equation $\overline{z} = z$ forces $-b = b$, so $b = 0$ and $z$ is real. Conjugation leaves the real numbers alone, and it leaves nothing else alone. Hold on to that. It is about to become the hypothesis of a theorem.

The conjugate root theorem

Theorem. Let $p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$, where every coefficient $a_k$ is a real number. If $z$ is a complex number with $p(z) = 0$, then $p(\overline{z}) = 0$ too.

Look at what the proof never needed: the degree, a formula for the roots, which root you started from, or any special feature of $z$. It needed two rules about conjugation and one fact about the coefficients. Real coefficients are the only thing standing between $p(z) = 0$ and $p(\overline{z}) = 0$, which is why the theorem holds at degree $7$ exactly as firmly as at degree $2$.

Notice also what happens when $z$ is real: then $\overline{z} = z$ and the theorem says only that a root is a root. All the content is in the nonreal roots. They cannot travel alone.

A paired root is worth more than a second root. It hands you a real quadratic factor. Pair $z = a + bi$ (with $b \neq 0$) against $\overline{z} = a - bi$ and multiply the two linear factors:

Both new coefficients are real, because $z + \overline{z} = 2a$ and $z\,\overline{z} = a^2 + b^2$ are the two quantities that conjugate pairs are famous for making real. And that quadratic really is a factor of $p$, for a reason you can state in one line. Whenever $r$ and $s$ are distinct roots of $p$, the Factor Theorem gives $p(x) = (x - r)\,g(x)$; substituting $x = s$ turns $0 = p(s) = (s - r)\,g(s)$ into $g(s) = 0$, because $s - r \neq 0$. So the Factor Theorem applies once more, now to $g$, and the product $(x - r)(x - s)$ divides $p$. Take $r = z$ and $s = \overline{z}$, which are distinct precisely because $b \neq 0$.

The hypothesis is doing all the work

Read the theorem’s statement again and notice how much weight the words “every coefficient is real” are carrying. Take them away and the conclusion dies immediately. The smallest possible counterexample is degree $1$:

Its only root is $i$, since $p(i) = i - i = 0$, and $-i$ is not a root, since $p(-i) = -i - i = -2i \neq 0$. A root with no partner. The theorem is not being violated, because $p$ has the coefficient $-i$, which is not a real number, so the theorem never applied. The quadratic $q(x) = x^2 - (1 + i)x + i$ from the opening section fails for the same reason, with the same nonreal coefficient sitting in plain sight.

So “complex roots come in conjugate pairs” is false as a bare slogan. What is true is the conditional: if the coefficients are real, then the nonreal roots come in conjugate pairs. Whenever you invoke the pairing, check the coefficients first. It is a one-second check, and skipping it is how the mistake gets made.

Do not read the theorem backwards, either. Paired roots do not force real coefficients. Scaling a polynomial by a constant cannot move a single root, so

has the roots $i$ and $-i$, a perfectly matched conjugate pair, while its coefficients $i$ and $i$ are about as nonreal as coefficients get. The most you can conclude from a root list whose nonreal entries pair off with equal multiplicities is that your polynomial is a constant multiple of one with real coefficients. The reason is that pairing the nonreal roots into real quadratics and the real roots into real linear factors rebuilds a real polynomial. That leaves only the leading constant free to be anything it likes.

The multiplicities are not a detail you can wave through. Take $(x - i)^2(x + i) = x^3 - ix^2 + x - i$. Its nonreal roots are $i$ and $-i$, which look paired if you only read the list of values, but $i$ occurs twice and $-i$ once. That is something no constant multiple of a real polynomial can do. A pairing that ignores multiplicity is not a pairing.

The same proof over the rationals

Nothing in the proof cared that the swap was $i \mapsto -i$ in particular. It used only that the swap respects addition and multiplication, and that it leaves every coefficient alone. Any other swap with those two properties will prove its own theorem, with its own hypothesis on the coefficients. Here is the one that matters most.

Fix a rational number $d$ whose square root is irrational (for a positive integer $d$, that means $d$ is not a perfect square). With that $d$ fixed, look at the numbers of the form $a + b\sqrt{d}$ with $a$ and $b$ rational. The swap is $\sqrt{d} \mapsto -\sqrt{d}$, and the partner of $s = a + b\sqrt{d}$ is written

Give it a new symbol rather than reusing the bar, because the bar is already taken and it does something different here. The number $2 + \sqrt{5}$ is real, so its complex conjugate is $\overline{2 + \sqrt{5}} = 2 + \sqrt{5}$, unchanged. The surd partner $2 - \sqrt{5}$ comes from an entirely different swap.

Theorem. Let $p$ have rational coefficients, and let $a$, $b$, and $d$ all be rational, with $b \neq 0$ and $\sqrt{d}$ irrational. If $a + b\sqrt{d}$ is a root of $p$, then $a - b\sqrt{d}$ is a root of $p$.

That the number $d$ is itself rational is a hypothesis, not a formality, and it is exactly the kind of hypothesis this lesson keeps telling you to watch. Drop it, keep every other word of the theorem, and the conclusion dies. Take $d = 3 + 2\sqrt{2}$. Its square root is irrational, as required, and you can name it exactly:

Now take $a = 0$ and $b = 1$, both rational with $b \neq 0$. Then $a + b\sqrt{d} = 1 + \sqrt{2}$, and it is a root of the rational-coefficient polynomial $p(x) = x^2 - 2x - 1$. Every hypothesis you have left holds. Yet the alleged partner $a - b\sqrt{d} = -1 - \sqrt{2}$ is not a root:

The proof collapsed back at the product step, where $bed = d$ needed to be rational and was not. The theorem itself was never in danger: written honestly, $1 + \sqrt{2}$ is $1 + 1\cdot\sqrt{2}$ with the rational $d = 2$, and that reading pairs it with $1 - \sqrt{2}$, which really is a root of $p$. It was the illegal $d$ that failed, not the mathematics. A hypothesis you never noticed is still a hypothesis.

Two theorems, one skeleton. Setting them side by side is the point of this lesson, because it shows that conjugation is not really about $i$ at all. It is about which number system the coefficients are drawn from, and about a swap that leaves exactly that system alone.

The last row deserves a second look, because the two hypotheses are genuinely different and confusing them is the standard error. Real coefficients buy you the complex pairing and nothing more. The polynomial

has real coefficients, and the conjugate root theorem is perfectly happy with it (its roots $2$ and $\sqrt{2}$ are real, so each is its own conjugate). But $-\sqrt{2}$ is nowhere among its roots. The surd pairing needs the coefficients to be rational, and $2 + \sqrt{2}$ is not.

What the pairing forces

The pairing is not a curiosity. It constrains what a real polynomial can look like, and the constraints are strong enough to be useful. Everything below follows from the conjugate root theorem plus the Fundamental Theorem of Algebra. That last theorem guarantees that any polynomial of degree $1$ or more has at least one complex root to work with.

Two consequences fall straight out of that factorization, and both are worth stating on their own.

Nonreal roots are matched exactly, multiplicity and all. Every nonreal root of $p$ has to come from one of the quadratics $d_j$, and each $d_j$ contributes the pair $z_j$ and $\overline{z_j}$ once each. So $z$ and $\overline{z}$ appear the same number of times in the complete root list: they have equal multiplicity. Counting with multiplicity, a real polynomial has exactly $2m$ nonreal roots, an even number, always.

Every real polynomial of odd degree has at least one real root. Suppose $n$ is odd. Since $n = k + 2m$, the number $k$ of real linear factors is $n - 2m$, which is odd, so $k \geq 1$. At least one factor $x - r_i$ is there, and $r_i$ is a real root. Put the other way round: the nonreal roots come in twos, so they can never use up an odd count, and something real is always left over. This matches the picture you already have of an odd-degree graph, which runs off toward opposite ends and so has to cross the axis somewhere. But notice that we proved that result without drawing anything.

For a cubic this pins things down completely. Counting with multiplicity, it has $3$ roots, of which an even number are nonreal. So that cubic has either $0$ or $2$ nonreal roots among those three, which means either $3$ or $1$ real roots. A real cubic with no real root does not exist.

Common mistakes