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 it were the cause, every quadratic should produce conjugate roots, because every quadratic has that $\pm$. Test it on

Multiplying out $(x - 1)(x - i) = x^2 - ix - x + i$ gives exactly $q$, so the roots of $q$ are $1$ and $i$, and those are not conjugates of each other. The quadratic formula still applies to $q$ and still hands back two answers; nothing about the formula stops that. What it cannot do is make $1$ and $i$ conjugates, because they simply are not. The only thing that changed between $x^2 - 4x + 13$ and $q$ is that a coefficient stopped being real. That is the real cause, and the rest of this lesson tracks exactly how.

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. That quadratic is genuinely a factor of $p$. 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 $g(s) = 0$, since $s - r \neq 0$. So the Factor Theorem applies again, now to $g$, and the product $(x - r)(x - s)$ divides $p$. Take $r = z$ and $s = \overline{z}$, 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. Without real coefficients, the pairing is simply not guaranteed, one way or the other. Whenever you invoke the pairing, check the coefficients first. It is a one-second check, and skipping it is how the mistake gets made.

That conditional does not run backwards, either. Paired roots do not, by themselves, force real coefficients. Scaling a polynomial by a constant cannot move a root, so

has the perfectly matched pair $i$ and $-i$, even though its coefficients $i$ and $i$ are as nonreal as coefficients get.

The same proof over the rationals

Nothing in the proof above 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.

Watch what happens when a square root, not $i$, sits inside a root. Take the two numbers $2 + \sqrt{5}$ and $2 - \sqrt{5}$ and multiply the matching factors:

The $\sqrt5$ is gone from both new coefficients, even though it sat in both roots a moment ago. A number like $2 + \sqrt5$, built from a square root that will not simplify to a whole number or a fraction, is called a surd. What just happened is the surd conjugate theorem at work, and it happens for the same underlying reason: a rational coefficient cannot tell $\sqrt5$ apart from $-\sqrt5$.

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 $p$ as well. Write the partner of $s = a + b\sqrt{d}$ as $s^{*} = a - b\sqrt{d}$: flipping the sign in front of $\sqrt{d}$ and nothing else.

Here is why it works, in the same shape as the complex proof. Multiply two numbers of the form $a + b\sqrt{d}$ together and use $\sqrt d \cdot \sqrt d = d$:

Because $d$ is rational, both the new rational part and the new $\sqrt{d}$ part stay rational. So every sum and product of numbers built from $\sqrt{d}$ produces another number of that same two-part shape. That means $p$ evaluated at $a + b\sqrt{d}$ comes out as (some rational number) plus (some rational number) times $\sqrt{d}$. Flipping the sign on every $\sqrt{d}$ term is the partner swap. It turns the rational part into itself and the $\sqrt{d}$ part into its negative, and it passes through addition and multiplication exactly the way complex conjugation does. So applying the swap to $p(a + b\sqrt{d}) = 0$ turns it into $p(a - b\sqrt{d}) = 0$. Every rational coefficient of $p$ survives the swap unchanged, because a rational number is already its own partner, and only the copies of $\sqrt{d}$ flip sign.

That $\sqrt{d}$ is irrational is what keeps the two-part representation $a + b\sqrt{d}$ unique, which is what makes the swap well defined. Rationality of $d$ is a separate hypothesis, spent somewhere else. Drop it and the sign-flip can stop respecting multiplication, so it may no longer turn $p\big(a + b\sqrt{d}\big) = 0$ into $p\big(a - b\sqrt{d}\big) = 0$, and the theorem can fail.

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 conjugate theorem 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 theorem guarantees that any polynomial of degree $1$ or more has at least one complex root to work with.

Every real polynomial breaks apart into real linear factors and real quadratic factors with negative discriminant, one piece per root or per conjugate pair. Here is why. You already know $p$ factors completely into $n$ linear factors over the complex numbers, counted with multiplicity: $p(x) = a_n(x - r_1)(x - r_2)\cdots(x - r_n)$. Conjugating that whole equation the same way turns the left side back into $p(x)$, since every coefficient is real, and turns the right side into $a_n(x - \overline{r_1})(x - \overline{r_2})\cdots(x - \overline{r_n})$. A degree-$n$ polynomial factors into linear pieces in only one way, except for reordering. So that second list is just $r_1, \ldots, r_n$ rearranged: conjugation shuffles the root list instead of changing which roots appear or how often each one does. In particular, a nonreal root and its conjugate occur exactly as many times as each other. Group that list by pairing every nonreal root with its conjugate, using the equal counts just shown so no root is left without a partner. Each real root stays a lone factor $(x - r)$. Each conjugate pair $z = u + vi$ and $\overline{z}$ multiplies into one real quadratic factor $x^2 - 2ux + (u^2+v^2)$. That quadratic has no real roots of its own, since its discriminant is $-4v^2$, negative. Multiplying every grouped piece back together reconstructs $p$ exactly, because that product is just the same $n$ linear factors, regrouped.

That breakdown pins down how many nonreal roots a real polynomial can have. Every conjugate pair uses up exactly two roots at once, counted with multiplicity, so the nonreal roots always come in a whole number of pairs. Counted with multiplicity, a real polynomial has an even number of nonreal roots, always.

Odd degree pins it down further. If $p$ has odd degree $n$, its $n$ roots (counted with multiplicity) split into an even number of nonreal roots and the rest real. An even number subtracted from an odd number is odd, so at least one real root is left over.

Every real polynomial of odd degree has at least one real root. Put another way, 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. Notice that this argument reached the same conclusion 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 either $0$ or $2$ of them, leaving either $3$ or $1$ real roots. A real cubic with no real root does not exist.

Common mistakes