Algebra 2 · N.CN.C.7

Complex roots come in pairs

A quadratic with a negative discriminant hands back two complex roots that differ only in the sign of i. Is that an accident of the formula, or does every polynomial do it?

You need

The discriminantAlgebra 2

Where you are 5 in Complex numbers

Number9

Complex numbers5

Adding, subtracting and multiplying complex numbers Conjugates and dividing complex numbers Imaginary and complex numbers

Complex roots come in pairs

The complex plane, modulus and argument

Why this exists

When the discriminant of a quadratic is negative, the quadratic formula gives two complex roots, such as 1 + 2i and 1 - 2i. They are conjugates: the same real part, opposite imaginary parts.

That could be a quirk of the formula and its \pm. Cubics and quartics have roots too, and their formulas are far too long to read a pattern off. Yet the pairing survives, as long as every coefficient of the polynomial is a real number, and that has a practical payoff: one complex root hands you a second one free. For a cubic, two known roots out of three are enough to finish the job by hand.

Definition

A polynomial with real coefficients is one such as 2x^3 - x + 7, in which every number multiplying a power of x is real. A root of a polynomial p(x) is a number z, real or complex, with p(z) = 0.

The conjugate of z = a + bi, with a and b real, is \overline{z} = a - bi. When b \ne 0, z and \overline{z} are two different numbers, a non-real conjugate pair.

Conjugate pairs as roots. If a polynomial has real coefficients and z is a root, then \overline{z} is also a root. For a non-real root this is a second, different root. For a real root, \overline{z} = z, and it says nothing new.

Intuition

For a quadratic, the \pm makes the pair

The discriminant lesson showed where the pair comes from. When D = b^2 - 4ac is negative, the formula gives x = p \pm qi with \displaystyle p = -\frac{b}{2a} and \displaystyle q = \frac{\sqrt{-D}}{2a}, both real because a, b and c are. The \pm adds the pure imaginary part qi once and subtracts it once, so the two roots share the real part p and have opposite imaginary parts. On the complex plane they are mirror images in the real axis.

Going backwards, a pair builds a real quadratic

Multiply the two factors that belong to a pair. Treat x - a as one piece and use the difference of squares:

\begin{aligned} (x - a - bi)(x - a + bi) &= (x - a)^2 - (bi)^2 \\ &= x^2 - 2ax + a^2 + b^2 \end{aligned}

Every i is gone. A conjugate pair is exactly what a real quadratic factor needs, while one non-real root alone cannot give one: the factor x - i has the non-real coefficient -i.

The same picture holds at any degree. Here are the three roots of a cubic, found in Example 3.

Figure 1The three roots of x^3 - 6x^2 + 18x - 40 = 0. The real root sits on the real axis; the other two are mirror images across it.

Results

Conjugate root theorem. Let p(x) be a polynomial with real coefficients. Then

p(z) = 0 \implies p(\overline{z}) = 0

Three consequences follow.

Non-real roots come in pairs. If a + bi is a root, with b \ne 0, so is a - bi.
Each pair gives a real quadratic factor, x^2 - 2ax + (a^2 + b^2). Its middle coefficient, -2a, is minus the sum of the pair, and its constant, a^2 + b^2, is their product.
An odd degree forces a real root. A polynomial of degree n has n roots counting repeats, by the fundamental theorem of algebra. The non-real ones pair off, so there is an even number of them, and a cubic with real coefficients has one real root or three, never none.

Why it is true

The proof uses two facts about conjugates. For z = a + bi and w = c + di, with a, b, c, d real:

\begin{aligned} \overline{z + w} &= \overline{z} + \overline{w} \\ \overline{zw} &= \overline{z}\,\overline{w} \end{aligned}

For the sum, z + w = (a + c) + (b + d)i has conjugate (a + c) - (b + d)i, which is (a - bi) + (c - di). For the product,

\begin{aligned} zw &= (ac - bd) + (ad + bc)i \\ \overline{z}\,\overline{w} &= (a - bi)(c - di) \\ &= (ac - bd) - (ad + bc)i \end{aligned}

and the last line is the conjugate of the first. Using the product fact again and again gives \overline{z^k} = \overline{z}^{\,k} for every power. A real number r is its own conjugate, \overline{r} = r.

Now take p(x) = a_n x^n + \dots + a_1 x + a_0 with every a_k real, and a root z. Take the conjugate of both sides of p(z) = 0. The conjugate of 0 is 0, and the conjugate of a sum of products is the sum of the products of the conjugates:

\begin{aligned} 0 = \overline{p(z)} &= \overline{a_n}\,\overline{z}^{\,n} + \dots + \overline{a_0} \\ &= a_n\,\overline{z}^{\,n} + \dots + a_0 \\ &= p(\overline{z}) \end{aligned}

The middle step is the only place the real coefficients are used: \overline{a_k} = a_k. If a coefficient is not real, that step fails, and so can the theorem (Example 5).

A repeated root’s partner repeats as often: dividing out the real factor x^2 - 2ax + a^2 + b^2 by polynomial division leaves a quotient with real coefficients, and the argument runs again on it.

Worked examples

Example 1

The second root comes free

The equation x^2 - 10x + 41 = 0 has 5 + 4i as a root. Find the other root without solving, then confirm it.

Solution

Every coefficient is real, so the conjugate 5 - 4i is also a root, and a quadratic has room for no more. To confirm, use the sum and product of the roots: with a = 1 they must be \displaystyle -\frac{b}{a} = 10 and \displaystyle \frac{c}{a} = 41, and

\begin{aligned} (5 + 4i) + (5 - 4i) &= 10 \\ (5 + 4i)(5 - 4i) &= 25 + 16 = 41 \end{aligned}

Both match. The quadratic formula agrees: D = 100 - 164 = -64, so \displaystyle x = \frac{10 \pm 8i}{2} = 5 \pm 4i.

Example 2

One root is enough for a real quadratic

Find a quadratic x^2 + bx + c with real coefficients that has -4 + i as a root.

Solution

The coefficients are to be real, so the conjugate -4 - i must be a root too, and a quadratic has room for no more. The polynomial is the product of the two factors. Grouping x + 4, they are (x + 4) - i and (x + 4) + i, a difference of squares:

\begin{aligned} (x + 4 - i)(x + 4 + i) &= (x + 4)^2 - i^2 \\ &= x^2 + 8x + 16 + 1 \\ &= x^2 + 8x + 17 \end{aligned}

Check by sum and product: (-4 + i) + (-4 - i) = -8, which is -b, and (-4 + i)(-4 - i) = 16 + 1 = 17, which is c. Any non-zero real multiple, such as 3x^2 + 24x + 51, has the same roots; asking for the form x^2 + bx + c picks one.

Example 3

Using a known root to find the others

Given that 1 + 3i is a root of x^3 - 6x^2 + 18x - 40 = 0, find all three roots.

Solution

The coefficients are real, so 1 - 3i is a root as well. The pair has sum 2 and product 1 + 9 = 10, so it gives the real factor

(x - 1 - 3i)(x - 1 + 3i) = x^2 - 2x + 10

The factor theorem says the cubic is this times one more factor x - r. Match the part that is easiest to read: the constant term of the product is 10 \cdot (-r), and it must be -40, so r = 4. Now multiply out to confirm every coefficient, not just the one used:

\begin{aligned} &(x^2 - 2x + 10)(x - 4) \\ &\quad = x^3 - 4x^2 - 2x^2 + 8x + 10x - 40 \\ &\quad = x^3 - 6x^2 + 18x - 40 \end{aligned}

It matches, so this is the factorization. The roots are 4, 1 + 3i and 1 - 3i, as in Figure 1. Check the real one: 64 - 96 + 72 - 40 = 0.

Example 4

Building a polynomial with a given value

Find the polynomial p(x) of least degree, with real coefficients, that has roots 3 and -1 + 2i and satisfies p(1) = 32.

Solution

Real coefficients force -1 - 2i to be a root too, so there are at least three roots and the least degree is 3. The pair has sum -2 and product 1 + 4 = 5, giving the factor x^2 + 2x + 5. With an unknown leading number k:

\begin{aligned} p(x) &= k(x - 3)(x^2 + 2x + 5) \\ &= k(x^3 - x^2 - x - 15) \end{aligned}

Then p(1) = k(1 - 1 - 1 - 15) = -16k, and this must be 32, so k = -2:

p(x) = -2x^3 + 2x^2 + 2x + 30

Check: p(1) = -2 + 2 + 2 + 30 = 32, and p(3) = -54 + 18 + 6 + 30 = 0.

Example 5

When a coefficient is not real

Solve x^2 - 2ix - 2 = 0, and test whether the conjugate of a root is also a root.

Solution

Nothing in the quadratic formula needs real coefficients, so it still applies. Here a = 1, b = -2i and c = -2:

\begin{aligned} D &= (-2i)^2 - 4(1)(-2) \\ &= -4 + 8 = 4 \end{aligned}

so \displaystyle \sqrt{D} = 2 and

x = \frac{2i \pm 2}{2} = i \pm 1

The roots are 1 + i and -1 + i. This time the middle, \displaystyle -\frac{b}{2a} = i, is imaginary and the \pm adds a real number, so the roots share an imaginary part instead of a real one. Test 1 - i, the conjugate of 1 + i:

\begin{aligned} &(1 - i)^2 - 2i(1 - i) - 2 \\ &\quad = -2i - 2i + 2i^2 - 2 \\ &\quad = -4 - 4i \end{aligned}

That is not 0, so 1 - i is not a root. The theorem has not failed: its condition, real coefficients, was not met.

Figure 2The roots of x^2 - 2ix - 2 = 0. Both sit above the real axis, so neither is the mirror image of the other across it.

Example 6

Finding unknown coefficients

The equation x^3 + ax^2 + bx + 30 = 0, with a and b real, has -1 + 3i as a root. Find a, b and the other two roots.

Solution

Since a and b are real, every coefficient is real, so -1 - 3i is also a root. The pair has sum -2 and product 1 + 9 = 10, giving the factor x^2 + 2x + 10. The cubic starts with x^3, so it is that factor times some x + s:

\begin{aligned} &(x^2 + 2x + 10)(x + s) \\ &\quad = x^3 + (s + 2)x^2 + (2s + 10)x + 10s \end{aligned}

The constant must be 30, so 10s = 30 and s = 3. Matching the other coefficients, a = s + 2 = 5 and b = 2s + 10 = 16. The third root makes x + 3 = 0, so it is -3.

The roots are -1 + 3i, -1 - 3i and -3, and the equation is x^3 + 5x^2 + 16x + 30 = 0. Check: -27 + 45 - 48 + 30 = 0.

Boundaries

Every coefficient must be real. Example 5 has one non-real coefficient, and its roots 1 + i and -1 + i include no conjugate pair.

Real roots need no partner. A real root is its own conjugate. The cubic in Example 3 has the real root 4 on its own, and the two real roots 2 and 5 of (x - 2)(x - 5) are not related at all.

The pairing is about i, not about square roots in general. \displaystyle 1 + \sqrt{2} and \displaystyle 1 - \sqrt{2} are both roots of x^2 - 2x - 1, but that pairing needs rational coefficients and is a different theorem. The polynomial \displaystyle x - \sqrt{2} has real coefficients and the root \displaystyle \sqrt{2}, and \displaystyle -\sqrt{2} is not a root of it.

Forced partners count toward the degree. A quadratic with real coefficients cannot have both 2i and 3 + i as roots: their conjugates would make four roots, and a quadratic has at most two.

Where this goes next

The fundamental theorem of algebra says a polynomial of degree n has exactly n complex roots, counting repeats. Together with this lesson, it means every polynomial with real coefficients splits into real linear factors and real quadratic factors, one quadratic for each conjugate pair. Polynomial division and the factor theorem turn a known root into a factor that can be divided out. Writing a polynomial from its zeros uses the pairing every time a zero is complex.

Exercises

Which roots are forced

Each polynomial has real coefficients and the roots listed. Name every other root it must have, and give the least degree it can have.

7 - 2i
-5i and 1
3, -1 + 4i and 6i
\displaystyle 2 + i\sqrt{3} and \displaystyle \sqrt{5}

The second root of a quadratic

Each equation has the root shown. Name the other root without solving, and confirm the pair with its sum and product.

x^2 + 6x + 34 = 0, root -3 + 5i
4x^2 - 4x + 5 = 0, root \displaystyle \frac{1}{2} + i

Building real polynomials

Find x^2 + bx + c with real coefficients that has 6 + i as a root.
Find a quadratic with integer coefficients that has \displaystyle \frac{3}{2} - 2i as a root.
Find the polynomial of least degree, with real coefficients and leading coefficient 1, that has roots -2 and 3i.
Find the cubic p(x) with real coefficients that has roots 1 and 2 + 3i and satisfies p(0) = 26.

Finding the rest from one root

2 - i is a root of x^3 - 3x^2 + x + 5 = 0. Find the other roots.
3i is a root of x^3 - 4x^2 + 9x - 36 = 0. Find the other roots.
-1 + i is a root of 2x^3 + x^2 - 2x - 6 = 0. Find the other roots.
1 + 2i is a root of x^4 - 2x^3 + 6x^2 - 2x + 5 = 0. Find all four roots.
The equation x^3 + ax^2 + bx + 20 = 0, with a and b real, has 2 + 4i as a root. Find a, b and the other two roots.

Coefficients that are not real

Show that i is a root of x^2 + (1 - i)x - i = 0 and that -i is not. Find the other root.
Solve x^2 - 3ix - 2 = 0. Are the roots a conjugate pair?

Reasoning

Explain why every cubic with real coefficients has at least one real root.
Yelena wants a quadratic with real coefficients whose roots are 2 + i and 3 + i. Explain why none exists, and find the quadratic x^2 + bx + c, with non-real b and c, that has these roots.
Zygmunt says that since real coefficients force conjugate roots, any polynomial with real coefficients that has \displaystyle \sqrt{3} as a root must also have \displaystyle -\sqrt{3}. Give a polynomial that shows he is wrong, and say what the conjugate rule is really about.
Viveka knows that a quartic with real coefficients has the roots 1 + i, 1 - i and i. Wladimir says its fourth root could be real. Is he right?
Show that for real a and b, the product (x - (a + bi))(x - (a - bi)) has real coefficients, and that a single factor x - (a + bi) with b \ne 0 does not.

Answers

7 + 2i; degree 2.
5i; degree 3. The real root 1 needs no partner.
-1 - 4i and -6i; degree 5.
\displaystyle 2 - i\sqrt{3}; degree 3. \displaystyle \sqrt{5} is real, so it is its own conjugate.
-3 - 5i, the conjugate, since the coefficients are real. Sum \displaystyle -6 = -\frac{b}{a}, product \displaystyle 9 + 25 = 34 = \frac{c}{a}.
\displaystyle \frac{1}{2} - i, the conjugate. Sum \displaystyle 1 = \frac{4}{4}, product \displaystyle \frac{1}{4} + 1 = \frac{5}{4}.
x^2 - 12x + 37. The partner is 6 - i; sum 12, product 36 + 1 = 37.
4x^2 - 12x + 25. The partner is \displaystyle \frac{3}{2} + 2i; sum 3, product \displaystyle \frac{9}{4} + 4 = \frac{25}{4}. So \displaystyle x^2 - 3x + \frac{25}{4}, times 4.
x^3 + 2x^2 + 9x + 18. The partner -3i is forced, and (x + 2)(x^2 + 9) multiplies out to it.
p(x) = -2x^3 + 10x^2 - 34x + 26. The pair 2 \pm 3i gives x^2 - 4x + 13, and (x - 1)(x^2 - 4x + 13) is x^3 - 5x^2 + 17x - 13. Then p(0) = -13k = 26, so k = -2.
2 + i and -1. The pair gives x^2 - 4x + 5, and (x^2 - 4x + 5)(x + 1) multiplies out to the cubic.
-3i and 4. The pair gives x^2 + 9, and (x^2 + 9)(x - 4) multiplies out to the cubic.
-1 - i and \displaystyle \frac{3}{2}. The pair gives x^2 + 2x + 2. The other factor is 2x + s with 2s = -6, and (x^2 + 2x + 2)(2x - 3) multiplies out to the cubic.
1 + 2i, 1 - 2i, i and -i. The pair gives x^2 - 2x + 5. Write the other factor as x^2 + px + q: the constant gives 5q = 5 and the x^3 term gives p - 2 = -2, so it is x^2 + 1. Check: (x^2 - 2x + 5)(x^2 + 1) multiplies out to the quartic.
a = -3 and b = 16; the other roots are 2 - 4i and -1. The pair gives x^2 - 4x + 20, and (x^2 - 4x + 20)(x + s) has constant 20s = 20, so s = 1. Then a = s - 4 and b = 20 - 4s.
i^2 + (1 - i)i - i is -1 + i + 1 - i = 0, but (-i)^2 + (1 - i)(-i) - i gives -1 - i - 1 - i = -2 - 2i. The other root is -1: the product of the roots is \displaystyle \frac{c}{a} = -i, and \displaystyle \frac{-i}{i} = -1. The coefficient 1 - i is not real, so the theorem does not apply.
x = 2i or x = i. D = (-3i)^2 - 4(1)(-2), which is -9 + 8 = -1, so \displaystyle x = \frac{3i \pm i}{2}. Not a pair: the conjugate of i is -i, not 2i. The coefficient -3i is not real.
A cubic has three roots, counting repeats, by the fundamental theorem of algebra. The non-real ones come in conjugate pairs, so there are 0 or 2 of them, and the rest, at least one, are real.
Real coefficients would force 2 - i and 3 - i to be roots as well, four roots for a quadratic. The quadratic is (x - 2 - i)(x - 3 - i), which is x^2 - (5 + 2i)x + (5 + 5i).
\displaystyle x - \sqrt{3} has real coefficients and the root \displaystyle \sqrt{3}, but at \displaystyle -\sqrt{3} it is \displaystyle -2\sqrt{3}, not 0. The rule pairs a + bi with a - bi, and \displaystyle \sqrt{3} is real, its own conjugate. Pairing \displaystyle p + \sqrt{q} with \displaystyle p - \sqrt{q} needs rational coefficients.
No. i is a root and the coefficients are real, so -i is a root too. That makes four roots, 1 \pm i and \pm i, with no room left for a real one.
(x - a - bi)(x - a + bi) is (x - a)^2 - (bi)^2, which is x^2 - 2ax + a^2 + b^2. Its coefficients 1, -2a and a^2 + b^2 are real. A single factor x - a - bi has the constant term -a - bi, which is not real when b \ne 0.