HL · De Moivre 1 · Conjugate roots of polynomial equations

Module 10 of 46Lesson 1 of 6 · HL · De Moivre 1 · Conjugate roots of polynomial equations

De Moivre's theorem, roots of complex numbers & polynomial roots · Essential understanding 1

Conjugate roots of polynomial equations

IB AA AHL 1.14with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: If a polynomial has real coefficients, its complex roots always arrive in conjugate pairs: if

a + bi

is a root then so is

a - bi

. Spot one, and you get the other for free. That makes it much easier to find every root.

By the end of this lesson you can:

state and use the conjugate root theorem for polynomials with real coefficients
build a quadratic factor $z^2 - 2az + (a^2 + b^2)$ from a pair of conjugate roots
find the remaining roots of a cubic or quartic when one complex root is given
find unknown real coefficients from given roots
explain why a polynomial of odd degree must have at least one real root
know when the theorem does not apply (non-real coefficients)

🧑🏾 Imagine this…

Think of a mirror standing along the real axis of an Argand diagram. Every point

a + bi

has a reflection

a - bi

on the other side. A polynomial with real coefficients is “fair” to this mirror: it cannot treat the point above the mirror differently from its reflection. So if the equation is satisfied by one, it is satisfied by the other. Real roots sit on the mirror, so they have no partner.

Conjugate

the conjugate of

a + bi

a - bi

(same real part, opposite imaginary part). It is written

z^*

Real coefficients

every number in front of a power of

z

is a real number, for example

z^3 - 5z^2 + 17z - 13

. No

i

anywhere.

Root (zero)

a number that makes the polynomial equal to zero. A polynomial of degree

n

has exactly

n

roots when we count complex ones (and repeated ones).

Quadratic factor

the pair of factors

(z - \alpha)(z - \alpha^*)

multiplies out to a quadratic with real coefficients.

★ The conjugate root theorem

If $p(z)$ has real coefficients and $p(a + bi) = 0$ , then $p(a - bi) = 0$ too.
The pair $a \pm bi$ gives the real quadratic factor $(z - a - bi)(z - a + bi) = z^2 - 2az + (a^2 + b^2)$ : sum of the pair $= 2a$ , product $= a^2 + b^2$ .
Why it works: conjugating a sum or a product gives the sum or product of the conjugates, and real numbers do not change. So $p(z^*) = \big(p(z)\big)^* = 0^* = 0$ .
A polynomial of odd degree has an odd number of roots, and the non-real ones come in pairs, so at least one root is real.
To find the remaining roots: write down the conjugate, build the quadratic factor, divide, and solve what is left.

⚠️ Watch out

The theorem needs real coefficients. The equation $z^2 - (2 + i)z + (1 + i) = 0$ has the roots $1$ and $1 + i$ . They are not conjugates, because the coefficient $-(2 + i)$ is not real.
A given root $2 + 3i$ tells you $2 - 3i$ too, but it tells you nothing about the other real roots. Always count: a cubic needs three roots in total.
Do not forget the sign when you build the factor: for roots $3 \pm 2i$ the middle term is $-6z$ (minus twice the real part).

💡 Command terms

Write down the other root: no working, just the conjugate. Find all the roots: show the quadratic factor and the division (or comparing coefficients). Show that: each line of the expansion must be visible. A complex root is a root of the equation, so check it by substitution if you have time.

🧪 Try it yourselfA cubic with one real root

This is the cubic

(x - r)(x^2 - 2ax + a^2 + b^2)

. The quadratic part has roots

a \pm bi

. Slide a and b (keep b above 0): the graph never gains a second crossing, because

a \pm bi

are complex. Only r moves the real root. The starting values are

r = 1

a = 2

b = 3

, the cubic of Example 1.2.

Worked examples

Example 1.1Building a quadratic from one complex root

A quadratic equation with real coefficients has the root

3 + 2i

. Write down the other root and find the quadratic in the form

z^2 + pz + q = 0

🧠 Think it through

The coefficients are real, so the second root is the conjugate.
Multiply out $(z - \alpha)(z - \alpha^*)$ , or use sum $= 2a$ and product $= a^2 + b^2$ .

✍️ By hand

Other root:

3 - 2i

. (A1)

Sum of roots

= (3 + 2i) + (3 - 2i) = 6

Product

= (3 + 2i)(3 - 2i) = 9 + 4 = 13

Quadratic:

z^2 - (\text{sum})z + \text{product} = 0

, so

z^2 - 6z + 13 = 0

. (M1)(A1)

🔢 On your TI-Nspire CX II

Check that the root satisfies the equation.

✅ Answer Other root

3 - 2i

; quadratic

z^2 - 6z + 13 = 0

The product of a conjugate pair is

|z|^2

, always a positive real number. Here

|3 + 2i|^2 = 13

Example 1.2Finding the remaining roots of a cubic

The equation

z^3 - 5z^2 + 17z - 13 = 0

has the root

2 + 3i

. Find the other two roots.

🧠 Think it through

The coefficients are real, so $2 - 3i$ is a root too.
Their quadratic factor is $z^2 - 4z + 13$ (sum 4, product $4 + 9 = 13$ ).
Divide the cubic by it, or compare coefficients. The quotient is a linear factor.

✍️ By hand
Second root: .   Quadratic factor: .   (A1)(M1)
Let .
Constant terms: , so .   (M1)
Check the  term:  ✓.   Factor: .
Roots: , , .   (A1)

🔢 On your TI-Nspire CX II

Check the factorisation by trying a test value, z = 2: both sides must agree. Then substitute the real root.

✅ Answer The other roots are

2 - 3i

and

1

Compare this cubic with the graph lab: with r = 1, a = 2, b = 3 the graph crosses the axis only at x = 1. The complex pair never shows on the graph.

Example 1.3Four coefficients from two roots

The equation

z^4 + az^3 + bz^2 + cz + d = 0

, where

a, b, c, d

are real, has the roots

1 + i

and

2 - i

. Find

a, b, c, d

🧠 Think it through

A quartic has four roots. The coefficients are real, so the other two are $1 - i$ and $2 + i$ .
Make two real quadratic factors, one from each pair.
Multiply them together and read off the coefficients.

✍️ By hand
Roots:  and .   (R1)
: sum 2, product , factor .
: sum 4, product , factor .   (A1)
   (M1)
So .   (A1)

🔢 On your TI-Nspire CX II

Check the expansion with a test value, z = 2: the product and the quartic must give the same number.

✅ Answer

a = -6,\; b = 15,\; c = -18,\; d = 10

Check:

d = 10

must equal the product of all four roots

2 \times 5

. ✓ And the sum of the roots is

2 + 4 = 6 = -a

✓.

Example 1.4A purely imaginary root

(a) Show that

i\sqrt{3}

is a root of

z^3 + 2z^2 + 3z + 6 = 0

. (b) Hence find all the roots.

🧠 Think it through

Substitute $z = i\sqrt{3}$ . Use $i^2 = -1$ , so $(i\sqrt{3})^2 = -3$ and $(i\sqrt{3})^3 = -3i\sqrt{3}$ .
Pure imaginary means $a = 0$ . The conjugate is $-i\sqrt{3}$ .
The pair has sum 0 and product 3, so the quadratic factor is $z^2 + 3$ .

✍️ By hand
(a) 
 ✓   (M1)(AG)
(b) Conjugate root: ; factor .   (R1)
   (M1)
Roots: .   (A1)

🔢 On your TI-Nspire CX II

Check with the calculator.

✅ Answer (b)

z = -2

z = i\sqrt{3}

z = -i\sqrt{3}

The graph of

y = z^3 + 2z^2 + 3z + 6

crosses the real axis only at

-2

: the imaginary roots do not show on it.

Example 1.5Finding unknown coefficients in a cubic

The cubic

z^3 + pz^2 + qz + 10 = 0

, with

p, q

real, has the root

2 + i

. Find

p

q

and the other two roots.

🧠 Think it through

The conjugate $2 - i$ is a root. The quadratic factor is $z^2 - 4z + 5$ .
The third factor is $(z + k)$ . The constant term of the cubic is $5k = 10$ .
Expand and read off $p$ and $q$ .

✍️ By hand
Quadratic factor from : sum 4, product 5, so .   (A1)
Constant term: , so .   (M1)
.   (A1)
So , ; other roots  and .   (A1)

🔢 On your TI-Nspire CX II

Check by substituting the given root.

✅ Answer

p = -2

q = -3

; the other roots are

2 - i

and

-2

A second method: substitute

z = 2 + i

into the cubic, then equate the real part and the imaginary part to zero. That gives two equations for

p

and

q

. It is longer but it also works when the theorem is not available.

Challenge questions

⭐ Challenge 1.6A cubic with a real root givenPaper 1 style

The polynomial

z^3 + pz^2 + qz + r

has real coefficients. Two of its roots are

1 + 2i

and

-3

. Find

p

q

and

r

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Name the third root using the conjugate pair.
Build the real quadratic factor for $1 \pm 2i$ , then multiply by $(z + 3)$ .

✍️ By hand

Third root:

1 - 2i

. Factor for the pair: sum 2, product 5, so

z^2 - 2z + 5

. (R1)(A1)

(z^2 - 2z + 5)(z + 3) = z^3 + 3z^2 - 2z^2 - 6z + 5z + 15

(M1)

= z^3 + z^2 - z + 15

p = 1,\; q = -1,\; r = 15

. (A1)

🔢 On your TI-Nspire CX II

Check with a test value, z = 1.

✅ Answer

p = 1

q = -1

r = 15

⭐ Challenge 1.7A quartic with an imaginary pairPaper 2 style

Let

f(z) = z^4 - 4z^3 + 14z^2 - 36z + 45

(a) Show that $3i$ is a root of $f(z) = 0$ .
(b) Hence write $f(z)$ as a product of two real quadratic factors.
(c) Solve $f(z) = 0$ and use your GDC to check.
(d) The four roots are plotted on an Argand diagram. Find the area of the quadrilateral they form.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

(a) Use $(3i)^2 = -9$ , $(3i)^3 = -27i$ , $(3i)^4 = 81$ .
(b) The pair $\pm 3i$ gives $z^2 + 9$ . Divide $f(z)$ by it.
(c) Solve the second quadratic with the formula.
(d) Plot the points, spot the shape (parallel sides) and use the trapezium area.

✍️ By hand
(a)    (M1)
 ✓   (AG)
(b) The conjugate  is also a root, so  is a factor.   (R1)
: the constant term gives  ✓; the  term gives .   (M1)
Check the  term:  ✓.   So .   (A1)
(c)  gives .
Roots: .   (A1)
(d) Points . Two vertical sides, lengths 6 and 2, a distance 2 apart.
Area  square units.   (M1)(A1)

🔢 On your TI-Nspire CX II

Substitute the root. Then use the polynomial tool menu💻 click menu on keypad▸3: Algebra▸3: Polynomial Tools▸1: Find Roots of Polynomial (command cPolyRoots) on the quadratic factor, or on the whole quartic.
The calculator lists the four roots: $-3i$ , $3i$ , $2 - i$ , $2 + i$ (in some order). Plot them as points to check the shape.

✅ Answer (a) shown. (b)

(z^2 + 9)(z^2 - 4z + 5)

. (c)

\pm 3i

2 \pm i

. (d) 8 square units.

Notice the sum of the four roots is

0 + 4 = 4

, the negative of the

z^3

coefficient, as for every polynomial.

Next: Lesson 2 · De Moivre’s theorem and powers

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · De Moivre 1 · Conjugate roots of polynomial equations

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

Conjugate roots of polynomial equations

Worked examples

Challenge questions