Complex Roots of Quadratics

Square roots of negative numbers, pinned down

The Imaginary Unit and Complex Numbers lesson already answered the question the real numbers could not: for any positive real $k$, the negative number $-k$ has exactly two square roots, $i\sqrt{k}$ and $-i\sqrt{k}$, and the radical symbol $\sqrt{-k}$ names the choice $i\sqrt{k}$, the same convention that makes $\sqrt{9}$ mean $3$ rather than $-3$. So $\sqrt{-16} = 4i$, and the other root, $-4i$, is never lost: the $\pm$ in the quadratic formula supplies it, exactly as it supplied $-3$ alongside $3$ in the real case.

One habit from that lesson is worth repeating here, because everything below leans on it: convert every $\sqrt{-k}$ to $i$-form before doing any radical arithmetic. The familiar rule $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ holds only for non-negative $a$ and $b$, and multiplying negative radicals directly, without converting first, gives the wrong sign.

The formula runs to completion

Return to the derivation of the quadratic formula. Completing the square on $ax^2 + bx + c = 0$ reached the line

and in Chapter 4 the case $\Delta < 0$ stopped there: no real square could equal a negative number. Now nothing stops. Try it on a specific equation first.

Nothing about $x^2 - 6x + 13$ was special. The same two moves, read $\sqrt{\Delta}$ as $i\sqrt{|\Delta|}$ and let the $\pm$ do the rest, finish every quadratic with $\Delta < 0$. Here is why, stated in general.

Two features of that answer deserve a hard look. First, the real part $p = -\tfrac{b}{2a}$ is the axis of symmetry, the same axis the real roots straddled in Chapter 4, and it is exactly the $3$ that Example 1 found. Second, the two roots have the same real part and opposite imaginary parts, so each is the conjugate of the other, as $3 + 2i$ and $3 - 2i$ were. That is not a small remark, and the next section proves it is no accident of the formula.

The left graph below and the right graph are two different pictures of the same equation, not one picture: the left one plots $y$ against $x$, and no point on that curve has an imaginary part. The right one plots the complex plane, where the axes are the real and imaginary parts of a root, not $x$ and $y$. The only thing the two pictures share is the number $3$, the parabola’s axis of symmetry and the roots’ common real part.

For real coefficients, non-real roots always arrive in conjugate pairs

The formula handed us conjugate pairs, but that could in principle be a habit of the formula rather than a law of the equations. It is a law. Real coefficients alone force it, by an argument that never mentions the quadratic formula and will survive, word for word, for polynomials of every degree.

Read the theorem’s fine print, because both halves of it matter. When $z$ is real, $\overline{z} = z$ and the statement says nothing new; its full force is for non-real roots, which it forbids from ever appearing alone. A quadratic with real coefficients therefore has exactly three possible root inventories: two distinct real roots, one repeated real root, or a conjugate pair off the real line. “One real root and one non-real root” is not on the list, for any real coefficients whatsoever. And the hypothesis is not decoration: the quadratic $x^2 - 2ix - 1 = (x - i)^2$ has the double root $i$, and its conjugate $-i$ is not a root at all. No contradiction, because the coefficient $-2i$ is not real. State the theorem with its hypothesis attached, every time.

Sum and product still tell the truth

Chapter 4 closed with the relations $r_1 + r_2 = -\tfrac{b}{a}$ and $r_1 r_2 = \tfrac{c}{a}$, but the proof there leaned on a factorization over the reals, which assumed the roots were real. Do the relations survive the move off the real line? They do, and the cleanest way to see it is to compute both quantities straight from the pair $p \pm qi$. The sum is immediate, because the imaginary parts cancel:

The product is a conjugate product, the difference of squares in disguise:

and substituting the formulas for $p$ and $q$ collapses it to a single fraction:

Both relations hold verbatim, so everything you learned to do with sums and products of roots still works when the roots are complex. The previous lesson measured a complex number’s distance from the origin by its modulus, with $|z|^2 = p^2 + q^2$, so the product you just computed is exactly the squared modulus of either root:

The practical payoff is reconstruction. One non-real root of a real quadratic determines everything, because the conjugate theorem hands you the other root, and then the sum and product hand you the coefficients. A monic quadratic with roots $p \pm qi$ must be

Every quadratic factors completely now

The factoring lesson in Chapter 4 insisted that “factors” is meaningless until you name the number system. Over the reals a quadratic with $\Delta < 0$ is genuinely irreducible: no real roots, so no real linear factors. Over the complex numbers the last holdouts fall.

For any two roots $r_1$ and $r_2$ of $ax^2 + bx + c$, real or not, the sum and product relations turn $a(x - r_1)(x - r_2) = a\left(x^2 - (r_1 + r_2)x + r_1 r_2\right)$ straight back into $ax^2 + bx + c$, so $a(x - r_1)(x - r_2)$ is always the factored form. Chapter 4 already showed this when the roots are real. When they are the non-real conjugate pair $z$ and $\overline{z}$, it is the identity just proved:

The simplest cases are the most striking: $x^2 + 25 = x^2 - (5i)^2 = (x - 5i)(x + 5i)$ is a difference of squares in disguise, and $x^2 - 4x + 5$, irreducible over the reals, factors as $(x - 2 - i)(x - 2 + i)$.

Step back and look at what happened to the discriminant’s trichotomy for a quadratic with real coefficients. Over the reals it counted roots: two, one, or none. Over the complex numbers the count is always the same, exactly two roots counted with multiplicity, and the discriminant’s job shrinks to saying where they live. If $\Delta > 0$ they sit at two points of the real line; if $\Delta = 0$ they coincide at one real point, giving the repeated factor $a(x - r)^2$. If $\Delta < 0$ they leave the line and stand as a conjugate pair, one above it and one below. Nothing is ever missing: this is the degree-two case of the Fundamental Theorem of Algebra, which a later chapter states for every degree.

Common mistakes