0.6 Complex numbers and polynomials, done properly

A complex number is a pair of reals with a chosen product, which turns i squared equals minus one into a theorem and multiplication into "turn and stretch". Polynomials over R and C then divide with remainder, have at most n roots, and over C always have one.

In school you met $i$ as a symbol with $i^2 = -1$ and then used a list of rules. This chapter answers two questions the rules never asked. Does such a number exist, or are we pretending? And why does the same trick (multiply the moduli, add the angles) work for every product? The second half moves to polynomials, where complex numbers earn their keep: over $\mathbb C$ every polynomial of degree $n$ has exactly $n$ roots, and that is the reason eigenvalues, differential equations and Fourier transforms all work.

Before you start

Chapters 0.1 and 0.2: reading a theorem, direct proof, contradiction, and induction. From school: the quadratic formula, the remainder theorem, and $\sin(\alpha + \beta)$ , $\cos(\alpha + \beta)$ .

Quick checkSolve

x^2 + 2x + 5 = 0

using the quadratic formula.

Quick checkWhat is the remainder when

x^3 - 2x + 1

is divided by

x - 3

Quick checkTo prove a statement about every

n \ge 1

by induction, what do you prove besides the base case?

Where this came from· Square roots of negatives were forced on algebra by the cubic

In 1545 Girolamo Cardano published the solution of the cubic equation in his Ars Magna (the method came from Scipione del Ferro and Niccolò Tartaglia). It had an embarrassing flaw. For $x^3 = 15x + 4$ , which has the root $x = 4$ (check: $64 = 60 + 4$ ), the formula returns $\sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}}$ , a sum involving square roots of a negative number. Rafael Bombelli’s L’Algebra (1572) took the next step: he set the cube root of $2 + \sqrt{-121}$ equal to $2 + \sqrt{-1}$ , and checked that the cubes work out. In modern notation, $(2 + i)^3 = 2 + 11i$ and $(2 - i)^3 = 2 - 11i$ , so the two cube roots add up to $(2 + i) + (2 - i) = 4$ . Bombelli himself called such quantities “sophistic”, yet he accepted them as a calculating device that produced correct real answers, and that was why they could not be thrown away.

The picture

A complex number is a point in the plane. Adding two of them is the vector addition you know: the parallelogram. Multiplying is new. The visualizer below shows a quarter turn of the plane. Look where the two basic arrows go: $(1, 0)$ goes to $(0, 1)$ and $(0, 1)$ goes to $(-1, 0)$ .

This is multiplication by $i$ . Call the point $(1, 0)$ the number $1$ and the point $(0, 1)$ the number $i$ . Multiplying by $i$ sends $1$ to $i$ and $i$ to $-1$ . So $i \cdot i = -1$ says “two quarter turns make a half turn”, and a half turn sends $1$ to $-1$ . There is nothing mysterious left in $i^2 = -1$ .

Here are the sixth roots of unity (the solutions of $z^6 = 1$ ), placed on the unit circle, one every $60^\circ$ . Section “Roots of unity” proves that these are all of them.

Corner	Angle	Coordinates
$\omega_0$	$0^\circ$	$1$
$\omega_1$	$60^\circ$	$\tfrac12 + \tfrac{\sqrt3}{2} i$
$\omega_2$	$120^\circ$	$-\tfrac12 + \tfrac{\sqrt3}{2} i$
$\omega_3$	$180^\circ$	$-1$
$\omega_4$	$240^\circ$	$-\tfrac12 - \tfrac{\sqrt3}{2} i$
$\omega_5$	$300^\circ$	$\tfrac12 - \tfrac{\sqrt3}{2} i$

The picture starts at $n = 6$ , the table above. Slide $n$ to see the solutions of $z^n = 1$ for other $n$ : they always sit at the corners of a regular $n$ -sided polygon on the circle of radius $1$ .

The idea, in words

Why· Complex numbers are the plane with a product

The plane already has addition (vectors). It has no natural product of two points, so we choose one. We want three things from it: it must agree with ordinary multiplication on the horizontal axis, it must obey the usual laws (order does not matter, brackets can be moved, it distributes over addition), and every non-zero point must have an inverse. Essentially one choice does all this (up to renaming which point we call $i$ ), and the surprise is that it also gives a point whose square is $-1$ for free. So $i^2 = -1$ is not an assumption we make. It is something we find, the way you find that a rotation by $90^\circ$ done twice is a rotation by $180^\circ$ .

In JEE you went the other way round. You assumed $i^2 = -1$ and then expanded $(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i$ . That expansion is a rule that defines the product, with the rule $i^2 = -1$ hidden inside it. Here we write the answer, $(ac - bd, ad + bc)$ , down as the definition, and then every rule you used becomes something we can prove.

For polynomials the idea is the same as for whole numbers in chapter 0.5: you can divide with a remainder, and the remainder is smaller than the divisor. That one fact gives the factor theorem, the bound “at most $n$ roots”, and (with the conjugate trick) the shape of every real polynomial.

Definitions and theorems

Complex numbers as pairs

Definition B

Complex number

A complex number is an ordered pair $z = (a, b)$ of real numbers. We call $a$ the real part and $b$ the imaginary part. The set of all of them is $\mathbb C$ . Addition and multiplication are

(a, b) + (c, d) = (a + c,\ b + d), \qquad (a, b)\,(c, d) = (ac - bd,\ ad + bc).

We identify the real number $a$ with the pair $(a, 0)$ . This is safe because $(a, 0) + (c, 0) = (a + c, 0)$ and $(a, 0)(c, 0) = (ac, 0)$ : the pairs on the horizontal axis add and multiply like the real numbers they stand for. We write $i = (0, 1)$ . Then

(a, b) = (a, 0) + (b, 0)(0, 1) = a + bi,

because $(b, 0)(0, 1) = (b \cdot 0 - 0 \cdot 1,\ b \cdot 1 + 0 \cdot 0) = (0, b)$ . So the school notation $a + bi$ is just another way to write the pair.

In one lineAn ordered pair $(a, b)$ of real numbers, added by $(a,b) + (c,d) = (a+c, b+d)$ and multiplied by $(a,b)(c,d) = (ac - bd,\ ad + bc)$ .

Why it existsIt gives the plane a product that extends real multiplication, obeys all the usual laws, and has a square root of $-1$ .

Theorem B

The square of i

In $\mathbb C$ , $i^2 = -1$ .

In one lineWith the product above, $i \cdot i = -1$ .

Why it is trueIt turns the school assumption $i^2 = -1$ into a one-line computation.

Proofdirect from the definition

The idea: Multiply (0,1) by (0,1) with the definition and read off the answer.

i = (0, 1)

Definition of

i

i \cdot i = (0 \cdot 0 - 1 \cdot 1,\ 0 \cdot 1 + 1 \cdot 0) = (-1, 0)

The product rule with

a = 0

b = 1

c = 0

d = 1

(-1, 0) = -1

, so

i^2 = -1

\blacksquare

We identified the real number

-1

with the pair

(-1, 0)

Quick checkIn the pair definition of

\mathbb C

, why is

i^2 = -1

The next theorem checks that the pair product really behaves like the multiplication you have always used, so that every manipulation from school is legal.

Theorem B

Laws of complex arithmetic

For all complex numbers $z, w, u$ :

$zw = wz$ (commutative).
$(zw)u = z(wu)$ (associative).
$z(w + u) = zw + zu$ (distributive).
$1 \cdot z = z$ .
If $z = (a, b) \ne (0, 0)$ , then $z \cdot \left(\dfrac{a}{a^2 + b^2},\ \dfrac{-b}{a^2 + b^2}\right) = 1$ , so $z$ has an inverse $z^{-1}$ .

(Addition obeys the same laws because it works coordinate by coordinate on real numbers.)

In one lineAddition and multiplication of complex numbers are commutative and associative, multiplication distributes over addition, $1 = (1, 0)$ is the identity, and every $z \ne 0$ has an inverse.

Why it is trueEvery algebraic manipulation from school (expanding brackets, collecting terms, dividing) is justified by these laws.

Proofdirect computation

The idea: Every law is a statement about two pairs of real numbers. Write both sides in coordinates and compare; the real numbers already obey the laws.

Fix coordinates.

Write

z = (a, b)

w = (c, d)

u = (e, f)

with real entries.

Commutative:

zw = (ac - bd,\ ad + bc)

and

wz = (ca - db,\ cb + da)

. These are equal.

Definition of the product, then the commutativity of real multiplication and addition.

Associative:

(zw)u = \big((ac - bd)e - (ad + bc)f,\ (ac - bd)f + (ad + bc)e\big) = (ace - bde - adf - bcf,\ acf - bdf + ade + bce)

. Also

z(wu) = \big(a(ce - df) - b(cf + de),\ a(cf + de) + b(ce - df)\big) = (ace - adf - bcf - bde,\ acf + ade + bce - bdf)

. The two lists have the same four terms in each coordinate.

Expand

(zw)u

using the product rule twice, then

z(wu)

the same way, and compare the real numbers that appear.

Distributive:

z(w + u) = \big(a(c + e) - b(d + f),\ a(d + f) + b(c + e)\big) = (ac - bd + ae - bf,\ ad + bc + af + be) = zw + zu

Distributive law for real numbers, applied in each coordinate.

Identity:

(1, 0)(c, d) = (1 \cdot c - 0 \cdot d,\ 1 \cdot d + 0 \cdot c) = (c, d)

Product rule with

(1, 0)

Inverse:

(a, b)\left(\tfrac{a}{a^2 + b^2}, \tfrac{-b}{a^2 + b^2}\right) = \left(\tfrac{a^2 + b^2}{a^2 + b^2},\ \tfrac{-ab + ab}{a^2 + b^2}\right) = (1, 0)

\blacksquare

Product rule. The condition

z \ne 0

means

a^2 + b^2 > 0

, so we may divide by it.This is the only place we divide, and it is why

\mathbb C

has inverses: a sum of squares is zero only when both squares are zero.

Because of these laws, you may now expand $(a + bi)(c + di)$ by the school method. The step $i^2 = -1$ is the theorem above, and everything else is the laws.

The plane, modulus and conjugate

Definition B

Modulus

The modulus (or absolute value) of $z = a + bi$ is $|z| = \sqrt{a^2 + b^2}$ . In the plane it is the length of the arrow from $(0, 0)$ to $(a, b)$ , by Pythagoras. For real $z$ it is the usual absolute value.

In one lineThe modulus of $z = a + bi$ is $|z| = \sqrt{a^2 + b^2}$ , its distance from the origin.

Why it existsIt measures size, so that 'small', 'close' and 'unit length' make sense in $\mathbb C$ .

Definition B

Conjugate

The conjugate of $z = a + bi$ is $\bar z = a - bi$ : reflect the point in the horizontal axis.

In one lineThe conjugate of $z = a + bi$ is $\bar z = a - bi$ , its mirror image in the real axis.

Why it existsMultiplying by the conjugate turns a complex number into a real one, which is how division and moduli are computed.

Proposition B

Rules for conjugates

For all complex numbers $z$ and $w$ (with $z \ne 0$ in the last formula):

$z\bar z = |z|^2$ , a non-negative real number.
$\overline{z + w} = \bar z + \bar w$ .
$\overline{zw} = \bar z\,\bar w$ .
$z^{-1} = \bar z / |z|^2$ .

In one line $z\bar z = |z|^2$ , $\overline{z + w} = \bar z + \bar w$ , $\overline{zw} = \bar z\,\bar w$ , and $z^{-1} = \bar z / |z|^2$ .

Why it is trueThey let you compute with moduli and quotients without ever taking square roots, and the last two are the engine of the conjugate-pair theorem for polynomials.

Proofdirect computation

The idea: All four are coordinate computations; the product rule does the work.

(1) For

z = (a, b)

z\bar z = (a \cdot a - b(-b),\ a(-b) + b \cdot a) = (a^2 + b^2, 0) = |z|^2

Product rule with

\bar z = (a, -b)

(2) With

w = (c, d)

\overline{z + w} = (a + c,\ -(b + d)) = (a, -b) + (c, -d) = \bar z + \bar w

Both sides add coordinate by coordinate, and negation respects addition.

(3)

zw = (ac - bd,\ ad + bc)

, so

\overline{zw} = (ac - bd,\ -ad - bc)

. And

\bar z\,\bar w = (a, -b)(c, -d) = (ac - bd,\ -ad - bc)

. Equal.

Product rule on both sides.

(4)

z \cdot \dfrac{\bar z}{|z|^2} = \dfrac{z\bar z}{|z|^2} = 1

, so

\bar z / |z|^2

is the inverse of

z

\blacksquare

Part (1) gives

z \bar z = |z|^2

, and

|z|^2 \ne 0

because

z \ne 0

. Divide by it.This is the school trick of multiplying top and bottom by the conjugate, now with its reason.

Theorem B

The modulus is multiplicative

For all complex numbers $z, w$ : $|zw| = |z|\,|w|$ .

In one line $|zw| = |z|\,|w|$ : the length of a product is the product of the lengths.

Why it is trueIt says multiplication scales lengths consistently, the first half of 'multiplication is a turn and a stretch'.

Proofstandard

The idea: Square both sides to remove the square roots. Then |zw| squared is zw times its conjugate, and the conjugate of a product is the product of the conjugates, so the factors can be regrouped.

|zw|^2 = (zw)\,\overline{zw}

Rules for conjugates (1), applied to the number

zw

= (zw)(\bar z\,\bar w)

Rules for conjugates (3): the conjugate of a product is the product of the conjugates.

= (z\bar z)(w\bar w)

Laws of arithmetic: multiplication is commutative and associative, so we may reorder the four factors.This is where we use the laws we proved, not just the formula.

= |z|^2\,|w|^2 = (|z|\,|w|)^2

Rules for conjugates (1) twice.

|zw| = |z|\,|w|

\blacksquare

Both

|zw|

and

|z||w|

are non-negative, and a non-negative number is determined by its square.

Proof workshopGive the reason for each line: the modulus of a product

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

$|zw|^2 = (zw)\,\overline{zw}$
$= (zw)(\bar z\,\bar w)$
$= (z\bar z)(w\bar w)$
$= |z|^2\,|w|^2$
$|zw| = |z|\,|w|$

Beyond the course: for understanding· A famous identity hiding in this proof

Write $z = a + bi$ and $w = c + di$ in the theorem and square: $(ac - bd)^2 + (ad + bc)^2 = (a^2 + b^2)(c^2 + d^2)$ . A product of two sums of two squares is again a sum of two squares. This is the Brahmagupta–Fibonacci identity, and a number theorist will use it to study which integers are sums of two squares. It drops out of the theorem with no extra work: it is $|zw|^2 = |z|^2|w|^2$ written in coordinates.

Quick checkWhat is

|(1 + i)(2 - 2i)|

Worked example NCERT Class 11 style

Divide

(2 + 3i)

(1 - i)

and check the modulus.

Solution.

Multiply top and bottom by the conjugate of the bottom. $\dfrac{2 + 3i}{1 - i} = \dfrac{(2 + 3i)(1 + i)}{(1 - i)(1 + i)}$ . The conjugate of $1 - i$ is $1 + i$ . This is the trick $z^{-1} = \bar z / |z|^2$ in disguise.
Bottom. $(1 - i)(1 + i) = 1 + 1 = 2$ . $z\bar z = |z|^2$ , here $1^2 + (-1)^2$ . The bottom is now a real number.
Top. $(2 + 3i)(1 + i) = 2 + 2i + 3i + 3i^2 = (2 - 3) + 5i = -1 + 5i$ . Expand with the laws, and use $i^2 = -1$ .
Divide. $\dfrac{-1 + 5i}{2} = -\dfrac12 + \dfrac52 i$ .

Check. $|2 + 3i| = \sqrt{13}$ and $|1 - i| = \sqrt 2$ , so the quotient should have modulus $\sqrt{13}/\sqrt 2 = \sqrt{6.5}$ . And $\sqrt{\tfrac14 + \tfrac{25}{4}} = \sqrt{6.5}$ . By the multiplicativity theorem, the modulus of a quotient is the quotient of the moduli.

Answer. $-\tfrac12 + \tfrac52 i$ .

Your answerWhat is

\left|\dfrac{1}{3 + 4i}\right|

Polar form: why multiplication turns and stretches

Every non-zero point has a distance $r$ from the origin and a direction. The direction is an angle $\theta$ measured anticlockwise from the positive horizontal axis, in radians.

Definition B

Polar form

The polar form of a non-zero complex number $z = a + bi$ is

z = r(\cos\theta + i\sin\theta), \qquad r = |z| = \sqrt{a^2 + b^2},

where $\theta$ is an angle with $\cos\theta = a/r$ and $\sin\theta = b/r$ . Such a $\theta$ exists because $(a/r, b/r)$ is a point on the unit circle, and every point on the unit circle is $(\cos\theta, \sin\theta)$ for some angle. The angle $\theta$ is an argument of $z$ . It is only determined up to adding a multiple of $2\pi$ , so $\theta$ and $\theta + 2\pi$ describe the same $z$ . The zero number has no argument.

In one lineWriting a non-zero $z$ as $z = r(\cos\theta + i\sin\theta)$ , where $r = |z|$ and $\theta$ is an angle, called an argument of $z$ .

Why it existsIt separates a complex number into size and direction, which is exactly what multiplication treats separately.

We abbreviate $\cos\theta + i\sin\theta$ as $\operatorname{cis}\theta$ for the next two pages. Notice that $|\operatorname{cis}\theta| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$ : it is a point on the unit circle.

Theorem B

Multiplying in polar form

For $r, s > 0$ and angles $\alpha, \beta$ :

r(\cos\alpha + i\sin\alpha) \cdot s(\cos\beta + i\sin\beta) = rs\big(\cos(\alpha + \beta) + i\sin(\alpha + \beta)\big).

In one line $r\operatorname{cis}\alpha \cdot s\operatorname{cis}\beta = rs\operatorname{cis}(\alpha + \beta)$ : multiply the lengths and add the angles.

Why it is trueThis is why multiplication by a complex number is a stretch combined with a turn, and why De Moivre's formula and the roots of unity exist.

Proofstandard; Needham, ch. 1

The idea: Multiply out with the laws of arithmetic, group the real and imaginary parts, and recognise the addition formulas for cosine and sine.

r\operatorname{cis}\alpha \cdot s\operatorname{cis}\beta = rs\big(\cos\alpha\cos\beta + i\cos\alpha\sin\beta + i\sin\alpha\cos\beta + i^2\sin\alpha\sin\beta\big)

Laws of arithmetic: pull out the real factor

rs

and expand the brackets.

= rs\big[(\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta)\big]

i^2 = -1

(the square of

i

), then collect the real terms and the terms with

i

= rs\big[\cos(\alpha + \beta) + i\sin(\alpha + \beta)\big]

\blacksquare

The addition formulas

\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta

and

\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta

.We are not proving the addition formulas here. They are the school facts that say two rotations in a row make one rotation by the sum of the angles. In fact the theorem and the formulas are the same statement, one written with matrices and the other with complex numbers.

Drag $z$ and $w$ below and watch the rule happen: the purple arrow $zw$ always has length $|z|\,|w|$ and angle equal to the sum of the two angles.

In the quarter-turn visualizer earlier, multiplying by $i = \operatorname{cis}(\pi/2)$ adds $\pi/2$ to every angle and leaves lengths alone. That is the quarter turn. Multiplying by a general $z = a + bi$ is the map with matrix $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ : it sends $1$ to $(a, b)$ and $i$ to $(-b, a)$ , which is a rotation by $\arg z$ followed by a stretch by $|z|$ .

Quick checkPoint

z

has modulus 3 and angle

70^\circ

. Point

w

has modulus 2 and angle

50^\circ

. What are the modulus and angle of

zw

Corollary B

De Moivre's formula

For every integer $n$ and angle $\theta$ : $(\operatorname{cis}\theta)^n = \operatorname{cis}(n\theta)$ .

In one line $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for every integer $n$ .

Why it is trueIt is the JEE rule for powers of a unit complex number, now proved, and it is the tool for finding roots.

Proofstandard

The idea: Induction on n for n at least 1 (each extra factor adds one more theta), then n = 0, then negative n by inverting.

For

n = 1

the formula is trivial.

Mathematical induction, base case: the statement for

n = 1

says

\operatorname{cis}\theta = \operatorname{cis}\theta

(\operatorname{cis}\theta)^{k+1} = \operatorname{cis}(k\theta + \theta) = \operatorname{cis}((k+1)\theta)

Inductive step: assume

(\operatorname{cis}\theta)^k = \operatorname{cis}(k\theta)

. Write

(\operatorname{cis}\theta)^{k+1} = (\operatorname{cis}\theta)^k \cdot \operatorname{cis}\theta = \operatorname{cis}(k\theta)\operatorname{cis}\theta

, then apply multiplying in polar form with

r = s = 1

The case

n = 0

holds.

Any non-zero number to the power

0

1

, and

\operatorname{cis}(0) = \cos 0 + i\sin 0 = 1

For

n = -m

with

m \ge 1

(\operatorname{cis}\theta)^{-m} = (\operatorname{cis}(-\theta))^m = \operatorname{cis}(-m\theta)

, using the case already proved.

\blacksquare

By multiplying in polar form,

\operatorname{cis}\theta \cdot \operatorname{cis}(-\theta) = \operatorname{cis}(0) = 1

, so

\operatorname{cis}(-\theta)

is the inverse of

\operatorname{cis}\theta

Worked example JEE style

Write

-\sqrt3 + i

in polar form and compute its sixth power.

Solution.

Modulus. $r = \sqrt{(-\sqrt3)^2 + 1^2} = \sqrt{3 + 1} = 2$ . Distance from the origin.
Find the quadrant first. $a = -\sqrt3 < 0$ and $b = 1 > 0$ , so the point is in the second quadrant. $\tan^{-1}(b/a)$ alone cannot tell the second quadrant from the fourth. Decide the quadrant, then the angle.
Angle. $\cos\theta = a/r = -\sqrt3/2$ and $\sin\theta = b/r = 1/2$ , so $\theta = 5\pi/6$ ( $150^\circ$ ). Both equations must hold, not just one.
Apply De Moivre. $z^6 = 2^6\operatorname{cis}(6 \cdot 5\pi/6) = 64\operatorname{cis}(5\pi) = 64\operatorname{cis}(\pi) = -64$ . Lengths to the sixth power, angles times six. $5\pi$ is $\pi$ plus two full turns.

Check by a second method. $(-\sqrt3 + i)^2 = 3 - 2\sqrt3\,i + i^2 = 2 - 2\sqrt3\,i = 4\operatorname{cis}(-\pi/3)$ . Cubing that gives $64\operatorname{cis}(-\pi) = -64$ .

Answer. $z = 2\operatorname{cis}(5\pi/6)$ and $z^6 = -64$ .

Your answerWhat is

(1 + i)^8

Your answerGive the argument of

-1 - i

in radians, in the range from

-\pi

\pi

Worked example

Use De Moivre's formula to express

\cos 2\theta

and

\sin 2\theta

in terms of

\cos\theta

and

\sin\theta

Solution.

Write the power. $(\cos\theta + i\sin\theta)^2 = \cos 2\theta + i\sin 2\theta$ . De Moivre’s formula with $n = 2$ .
Expand the left side. Write $c = \cos\theta$ and $s = \sin\theta$ . Then $(c + is)^2 = c^2 + 2ics + i^2s^2 = (c^2 - s^2) + i(2cs)$ . Ordinary expansion, with $i^2 = -1$ . Collect the real part and the part multiplying $i$ .
Compare real and imaginary parts. $\cos 2\theta = c^2 - s^2$ and $\sin 2\theta = 2cs$ . Two complex numbers are equal exactly when both coordinates agree, because a complex number is a pair.
Check with a number. At $\theta = \frac\pi3$ : $c = \frac12$ , $s = \frac{\sqrt3}2$ , so $c^2 - s^2 = \frac14 - \frac34 = -\frac12 = \cos\frac{2\pi}3$ and $2cs = \frac{\sqrt3}{2} = \sin\frac{2\pi}3$ . Both formulas agree with values you know.

Answer. $\cos 2\theta = \cos^2\theta - \sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$ . The same method gives the triple-angle formulas in the problems.

Euler’s formula

Beyond the course: for understanding· Euler's formula (label B; the full proof is in chapter 1.6)

The polar form $\cos\theta + i\sin\theta$ obeys the same rule as an exponential: it multiplies by adding exponents, $\operatorname{cis}\alpha \cdot \operatorname{cis}\beta = \operatorname{cis}(\alpha + \beta)$ . That is not a coincidence. The exponential function has a power series, $e^x = \sum_{n \ge 0} x^n / n!$ , and series make sense for complex $x$ too. Putting $x = i\theta$ gives the next theorem. Everything depends on knowing that a power series may be rearranged term by term and that $e^{z+w} = e^z e^w$ holds for complex $z, w$ . Both are proved in chapter 1.6 (power series). Here we take them as given, so label this section B, beyond what you can prove yet.

Theorem B

Euler's formula

For every real $\theta$ , with $e^z$ defined by $\sum_{n \ge 0} z^n / n!$ :

e^{i\theta} = \cos\theta + i\sin\theta.

In particular $e^{i\pi} = -1$ .

In one line $e^{i\theta} = \cos\theta + i\sin\theta$ for every real $\theta$ .

Why it is trueIt explains the pattern 'multiply lengths, add angles' as the exponent rule $e^{i\alpha}e^{i\beta} = e^{i(\alpha + \beta)}$ , and it turns trigonometry into algebra.

ProofEuler 1748; assumes rearrangement from chapter 1.6

The idea: Put i theta into the series. The powers of i cycle through 1, i, -1, -i, which sorts the terms into the series of cosine (even powers) and sine (odd powers).

e^{i\theta} = \sum_{n \ge 0} \dfrac{(i\theta)^n}{n!} = \sum_{n \ge 0} \dfrac{i^n\theta^n}{n!}

Definition of the exponential by its power series, with

z = i\theta

Split the sum into even powers

n = 2k

and odd powers

n = 2k + 1

i^2 = -1

, so

i^3 = -i

i^4 = 1

, and the cycle repeats every four steps.The pattern is

i^{2k} = (-1)^k

and

i^{2k+1} = (-1)^k i

e^{i\theta} = \sum_{k \ge 0} \dfrac{(-1)^k\theta^{2k}}{(2k)!} + i\sum_{k \ge 0} \dfrac{(-1)^k\theta^{2k+1}}{(2k+1)!}

Rearranging a convergent series (chapter 1.6) lets us add the even and odd parts separately.

= \cos\theta + i\sin\theta

. Putting

\theta = \pi

gives

e^{i\pi} = \cos\pi + i\sin\pi = -1

\blacksquare

These are the Taylor series of

\cos\theta

and

\sin\theta

that you know from school calculus.

From now on $\operatorname{cis}\theta$ will be written $e^{i\theta}$ . Multiplying in polar form now reads $re^{i\alpha} \cdot se^{i\beta} = rs\,e^{i(\alpha + \beta)}$ , which is the exponent rule you already know. The same rule gives De Moivre: $(e^{i\theta})^n = e^{in\theta}$ .

A preview· Preview: why this matters for differential equations (Unit 7)

If $a$ and $b$ are real, then $e^{(a + ib)t} = e^{at}(\cos bt + i\sin bt)$ . The real part $a$ says the solution grows or dies, and the imaginary part $b$ says it oscillates. A second-order equation like $y'' + y = 0$ has the solutions $e^{it}$ and $e^{-it}$ , which are the cosine and sine in disguise.

Roots of unity

A root of unity is a solution of $z^n = 1$ for some whole number $n \ge 1$ . School taught you that there are $n$ of them. Here is the proof, and the picture.

Theorem B

The n-th roots of unity

For every integer $n \ge 1$ :

The solutions of $z^n = 1$ are exactly the $n$ numbers $\omega_k = \cos\dfrac{2\pi k}{n} + i\sin\dfrac{2\pi k}{n}$ for $k = 0, 1, \dots, n - 1$ .
For $n \ge 2$ , $\omega_0 + \omega_1 + \dots + \omega_{n-1} = 0$ .

In one line $z^n = 1$ has exactly $n$ complex solutions, $\omega_k = e^{2\pi i k / n}$ for $k = 0, 1, \dots, n - 1$ , the corners of a regular $n$ -gon on the unit circle; for $n \ge 2$ they add up to $0$ .

Why it is trueThese numbers are the building blocks of the Fourier transform, and they show how the geometry of multiplication produces algebra.

Proofstandard

The idea: Take the modulus to see that a solution lies on the unit circle. Then De Moivre turns z^n = 1 into a statement about the angle: n times the angle must be a whole number of full turns.

z^n = 1

then

|z|^n = 1

. Since

|z|

is a positive real number,

|z| = 1

Take the modulus of both sides of

z^n = 1

; the modulus is multiplicative, so

|z^n| = |z|^n

. A positive real number

x

with

x^n = 1

1

: if

x > 1

then

x^n > 1

, and if

x < 1

then

x^n < 1

z = e^{i\theta}

for some angle

\theta

Polar form with

r = 1

z^n = e^{in\theta} = 1

forces

n\theta

to be a whole number of full turns:

n\theta = 2\pi k

for some integer

k

De Moivre:

(e^{i\theta})^n = e^{in\theta}

. This equals

1 = e^{i \cdot 0}

exactly when

\cos n\theta = 1

and

\sin n\theta = 0

\theta = 2\pi k / n

, so

z = \omega_k

Solve for the angle.This is the 'where did the idea come from': we found what angle the answer must have, then checked it works.

So every

\omega_k

is a solution, and every solution is some

\omega_k

The converse: each

\omega_k

does solve the equation, by De Moivre,

\omega_k^n = e^{2\pi i k} = 1

Exactly

n

different numbers.

For

k

from

0

n - 1

the angles

2\pi k/n

are different and all lie in

[0, 2\pi)

, and two points on the unit circle with different angles in that range are different. Adding

n

k

adds a full turn and gives the same point.

Then

(\omega - 1)(\omega_0 + \dots + \omega_{n-1}) = \omega^n - 1 = 0

For the sum:

\omega_k = \omega_1^k

by De Moivre. Let

\omega = \omega_1

. The geometric-series identity

(\omega - 1)(1 + \omega + \dots + \omega^{n-1}) = \omega^n - 1

holds by expanding.

\omega - 1 \ne 0

, hence

\omega_0 + \dots + \omega_{n-1} = 0

\blacksquare

For

n \ge 2

\omega = \omega_1

has angle

2\pi/n

strictly between

0

and

2\pi

, so

\omega \ne 1

. If a product

(\omega - 1)S

0

and

\omega - 1 \ne 0

, multiply both sides by the inverse of

\omega - 1

(it exists by the laws theorem) to get

S = 0

. This is the no-zero-divisors fact, proved in full in the lemma below.Picture: the corners of a regular polygon centred at the origin balance, like equal weights on a wheel.

Worked example JEE style

Find all cube roots of

8i

Solution.

Write the target in polar form. $8i = 8\,e^{i\pi/2}$ . Modulus $8$ , angle $90^\circ$ . A root $z = se^{i\varphi}$ must satisfy $z^3 = 8i$ .
Match the moduli. $s^3 = 8$ , so $s = 2$ . The modulus is multiplicative, so $|z^3| = |z|^3 = s^3$ . A positive real number has exactly one positive cube root.
Match the angles, remembering full turns. $3\varphi = \pi/2 + 2\pi k$ , so $\varphi = \pi/6 + 2\pi k/3$ . Two angles describe the same point when they differ by a multiple of $2\pi$ . The $k$ is how we make sure we find every root.
Take $k = 0, 1, 2$ . $\varphi = \pi/6,\ 5\pi/6,\ 3\pi/2$ . Taking $k = 3$ would add a full turn and repeat $k = 0$ .
Convert back. $z_0 = 2(\cos 30^\circ + i\sin 30^\circ) = \sqrt3 + i$ ; $z_1 = 2(\cos 150^\circ + i\sin 150^\circ) = -\sqrt3 + i$ ; $z_2 = 2(\cos 270^\circ + i\sin 270^\circ) = -2i$ .

Check. $(-2i)^3 = -8i^3 = -8(-i) = 8i$ . The other two are the same point rotated by $120^\circ$ each time, as the theorem predicts.

Answer. $\sqrt3 + i,\ -\sqrt3 + i,\ -2i$ .

Your answerThe 8th roots of unity are

\omega_k = e^{2\pi i k/8}

. What is the real part of

\omega_1

Polynomials

From here until the fundamental theorem we work with coefficients in $\mathbb R$ or in $\mathbb C$ . Everything holds for either, because the only properties we use are the laws of arithmetic and the existence of inverses.

Definition B

Polynomial

A polynomial is an expression $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$ , where $n$ is a whole number and each coefficient $a_k$ is a real (or complex) number. Two polynomials are equal when all their coefficients are equal. We can add and multiply polynomials by the usual rules.

In one lineAn expression $f(x) = a_nx^n + \dots + a_1x + a_0$ with coefficients in $\mathbb R$ or $\mathbb C$ .

Why it existsPolynomials are the simplest functions, and equations like eigenvalue equations, and the characteristic equation of a differential equation, are polynomial equations.

Definition B

Degree

The degree $\deg f$ of a non-zero polynomial is the largest $n$ for which the coefficient $a_n$ is not zero. That coefficient is the leading coefficient. The zero polynomial has no degree; we write ” $r = 0$ or $\deg r < \deg g$ ” when we mean ” $r$ is smaller than $g$ ”.

In one lineThe degree of a non-zero polynomial is the largest $n$ with $a_n \ne 0$ .

Why it existsDegree measures the size of a polynomial, and division with remainder says the remainder is smaller than the divisor in degree.

Definition B

Root

A root of a polynomial $f$ is a number $c$ (real or complex) with $f(c) = 0$ .

In one lineA root of $f$ is a number $c$ with $f(c) = 0$ .

Why it existsSolving an equation $f(x) = 0$ is finding the roots, and the factor theorem links roots to factors.

Lemma B

No zero divisors, and the degree of a product

If $z, w$ are complex numbers and $zw = 0$ , then $z = 0$ or $w = 0$ .
If $f, g$ are non-zero polynomials, then $fg$ is non-zero and $\deg(fg) = \deg f + \deg g$ .

In one lineIf $zw = 0$ then $z = 0$ or $w = 0$ ; and $\deg(fg) = \deg f + \deg g$ for non-zero polynomials.

Why it is trueThe second statement is the fact behind division with remainder and 'at most n roots'. It fails when the coefficients are not in a field, such as integers mod 8.

Proofstandard

The idea: Part 1: if z is not zero, multiply by its inverse. Part 2: look only at the top-degree terms, whose product is a product of two non-zero numbers.

(1) Suppose

zw = 0

. If

z = 0

we are done, so let

z \ne 0

Proof by cases: either

z = 0

(done) or

z \ne 0

Then

w = (z^{-1}z)w = z^{-1}(zw) = z^{-1} \cdot 0 = 0

Laws of arithmetic: every non-zero

z

has an inverse

z^{-1}

, and multiplication is associative.

(2) In the product

fg

, no term has a power above

x^{m+n}

Definition of degree: let

f

have leading term

ax^m

and

g

have leading term

bx^n

, with

a, b \ne 0

The coefficient of

x^{m+n}

fg

ab

, and

ab \ne 0

by part (1).

The only way to get

x^{m+n}

ax^m \cdot bx^n

.This is the step that fails if the coefficients are, say, integers mod 8: then

2x \cdot 4x = 8x^2 = 0

fg \ne 0

and

\deg(fg) = m + n

\blacksquare

A non-zero coefficient at

x^{m+n}

and none above it means exactly this.

Division with remainder

You divided one polynomial by another in school, subtracting multiples until the degree dropped. The next theorem says that this process always works, and has one answer.

Theorem B

Polynomial division algorithm

Let $f$ and $g$ be polynomials with coefficients in $\mathbb R$ (or $\mathbb C$ ), and $g \ne 0$ . Then there exist unique polynomials $q$ (the quotient) and $r$ (the remainder) such that

f = gq + r, \qquad r = 0 \ \text{ or } \ \deg r < \deg g.

If $f$ and $g$ have real coefficients, so do $q$ and $r$ .

In one lineFor polynomials $f$ and $g \ne 0$ there are unique polynomials $q$ and $r$ with $f = gq + r$ and either $r = 0$ or $\deg r < \deg g$ .

Why it is trueIt makes 'remainder' meaningful for polynomials, and every later result in this chapter (remainder theorem, factor theorem, at most n roots) is a consequence.

Proofstandard; Axler 4.8

The idea: Existence by strong induction on the degree of f: subtract the multiple of g that cancels the top term of f, and the result has smaller degree. Uniqueness: two answers would give a multiple of g whose degree is too small to be a non-zero multiple.

Existence. If

f = 0

\deg f < \deg g

, take

q = 0

and

r = f

Strong induction on

m = \deg f

(chapter 0.2). The statement

S(m)

: every

f

of degree

m

has a

q, r

as claimed. We also treat

f = 0

directly.

Otherwise

\deg f = m \ge n = \deg g

This is the only other case. Write the leading terms

f = ax^m + \dots

and

g = bx^n + \dots

with

a, b \ne 0

and

m \ge n

Define

h = f - \dfrac{a}{b}x^{m-n}g

This is the school move: multiply

g

by the term that matches the top term of

f

. It needs

b \ne 0

to be invertible, which holds because we work with numbers.

h = 0

\deg h < m

The top term of

\tfrac{a}{b}x^{m-n}g

\tfrac{a}{b}x^{m-n} \cdot bx^n = ax^m

, which cancels the top term of

f

There are

q_1, r_1

with

h = gq_1 + r_1

and

r_1 = 0

\deg r_1 < n

The induction hypothesis:

S

holds for every degree smaller than

m

, so

h

has a quotient and remainder.Strong induction is needed because the degree of

h

can drop by more than one.

f = \tfrac{a}{b}x^{m-n}g + h = g\left(\tfrac{a}{b}x^{m-n} + q_1\right) + r_1

, which has the required form.

Substitute

h

back into its definition and regroup.

Uniqueness. Subtracting the two expressions for

f

g(q - q') = r' - r

Proof by contradiction for uniqueness: suppose

f = gq + r = gq' + r'

with both remainders small, and suppose

q \ne q'

The left side is non-zero (since

g \ne 0

and

q - q' \ne 0

) and has degree at least

n

Lemma: degrees add when we multiply non-zero polynomials, so the left side has degree

\deg g + \deg(q - q') \ge \deg g

The right side is zero or has degree below

n

. This is a contradiction.

Each of

r

and

r'

is zero or has degree below

n

, and so does their difference.

q = q'

and

r = r'

. The real-coefficient statement holds because the construction only uses

+

-

\times

\div

on coefficients.

\blacksquare

Since

q = q'

, the equation

g(q - q') = r' - r

says

r' - r = 0

Worked example NCERT Class 11 style

Divide

f(x) = x^3 - 2x^2 + 3x + 5

g(x) = x^2 + x - 1

Solution. The steps below follow the existence proof: cancel the top term, repeat.

Answer. Quotient $x - 3$ , remainder $7x + 2$ .

Remainder and factor theorems

You used both of these as rules in JEE. Now they are consequences of the division algorithm.

Corollary B

Remainder theorem for polynomials

For a polynomial $f$ and a number $c$ , the remainder when $f$ is divided by $x - c$ is $f(c)$ . That is, $f(x) = (x - c)q(x) + f(c)$ for some polynomial $q$ .

In one lineDividing $f$ by $x - c$ leaves the constant remainder $f(c)$ .

Why it is trueYou can find a remainder by evaluating, with no division.

Proofstandard

The idea: The remainder must have degree below 1, so it is a constant. Put x = c to find which constant.

f(x) = (x - c)q(x) + r

, where

r = 0

\deg r < 1

Division algorithm with

g = x - c

, which has degree 1.

r

is a constant number.

A polynomial of degree less than 1 (or zero) is a constant.

f(c) = (c - c)q(c) + r = r

\blacksquare

Evaluate both sides at

x = c

. The product term has a factor

c - c = 0

Corollary B

Factor theorem for polynomials

A number $c$ is a root of $f$ if and only if $f(x) = (x - c)q(x)$ for some polynomial $q$ . In that case $\deg q = \deg f - 1$ .

In one line $c$ is a root of $f$ exactly when $x - c$ divides $f$ .

Why it is trueIt turns finding a root into finding a factor, and a factor lowers the degree, which is what makes induction on the degree work.

Proofstandard

The idea: Both directions come from the remainder theorem.

(

\Rightarrow

) If

f(c) = 0

then

f(x) = (x - c)q(x)

Remainder theorem:

f(x) = (x - c)q(x) + f(c)

. If

c

is a root,

f(c) = 0

(

\Leftarrow

) If

f(x) = (x - c)q(x)

then

f(c) = 0 \cdot q(c) = 0

Evaluate at

x = c

; the factor

c - c = 0

For the degree,

\deg f = 1 + \deg q

\blacksquare

Lemma: degrees add, and

\deg(x - c) = 1

Worked example NCERT Class 11 style

Find a remainder without dividing, and factor a cubic given one root.

Solution.

Remainder of $f(x) = x^3 - 2x^2 + 3x + 5$ on division by $x - 2$ . $f(2) = 8 - 8 + 6 + 5 = 11$ . Remainder theorem with $c = 2$ .
Check it a second way. Synthetic division with coefficients $1, -2, 3, 5$ and $c = 2$ : bring down $1$ ; $-2 + 1 \cdot 2 = 0$ ; $3 + 0 \cdot 2 = 3$ ; $5 + 3 \cdot 2 = 11$ . The last number is the remainder; the others are the coefficients of the quotient, here $x^2 + 0x + 3$ .
Factor $p(x) = x^3 - 6x^2 + 11x - 6$ . Try $x = 1$ : $p(1) = 1 - 6 + 11 - 6 = 0$ . By the factor theorem, a root means a factor. Candidates are factors of the constant term $6$ .
Divide out the factor. $p(x) = (x - 1)(x^2 - 5x + 6)$ . The factor theorem promises a quotient of degree 2. Check by expanding: $x^3 - 5x^2 + 6x - x^2 + 5x - 6 = x^3 - 6x^2 + 11x - 6$ .
Factor the quadratic. $x^2 - 5x + 6 = (x - 2)(x - 3)$ .

Answer. The remainder is $11$ , and $p(x) = (x - 1)(x - 2)(x - 3)$ , with roots $1, 2, 3$ .

Your answerWhat is the remainder when

x^4 - 3x + 5

is divided by

x + 2

At most n roots

Theorem B

A polynomial of degree n has at most n roots

A non-zero polynomial $f$ of degree $n$ has at most $n$ distinct roots.

In one lineA non-zero polynomial of degree $n$ has at most $n$ roots (real or complex).

Why it is trueIt says a polynomial is pinned down by few values, and it is the reason a polynomial equation never has more solutions than its degree.

Proofstandard; Axler 4.12

The idea: Induction on n. Pull out a factor (x - c) at a root c; what is left has degree one less, so it has at most n - 1 roots, and every other root of f is a root of it.

A non-zero constant has no roots, and

0 \le 0

Mathematical induction on

n

(chapter 0.2). Base case

n = 0

f

is a non-zero constant.

f

has no root, we are done, since

0 \le k + 1

Inductive step: assume every non-zero polynomial of degree

k

has at most

k

roots. Let

f

have degree

k + 1

Otherwise let

c

be a root of

f

Proof by cases: this is the other case. Pick one root and call it

c

f(x) = (x - c)q(x)

with

\deg q = k

, so

q \ne 0

Factor theorem.

Let

r \ne c

be any other root of

f

. Then

q(r) = 0

Evaluate at

r

0 = f(r) = (r - c)q(r)

. Since

r - c \ne 0

, the lemma (no zero divisors) gives

q(r) = 0

.This is the step that needs numbers with no zero divisors. It is false over the integers mod 8.

q

has at most

k

roots.

The induction hypothesis applied to

q

, which has degree

k

f

has at most

1 + k

roots.

\blacksquare

The roots of

f

are

c

together with some of the roots of

q

Proof workshopRebuild the proof that a polynomial of degree n has at most n roots

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Quick checkHow many different complex roots can a polynomial of degree 3 have?

Worked example

A polynomial

f

of degree at most 3 has

f(0) = f(1) = f(2) = f(3) = 0

. Show

f

is the zero polynomial.

Solution.

Count the roots. The numbers $0, 1, 2, 3$ are four different roots of $f$ . A root is a number where $f$ is $0$ .
Suppose $f$ is not zero. Then $f$ is a non-zero polynomial of degree at most 3. Proof by contradiction. We compare the number of roots with the degree.
Apply the root bound. A non-zero polynomial of degree $m \le 3$ has at most $m \le 3$ roots. But $f$ has at least 4. This is the theorem ‘a polynomial of degree $n$ has at most $n$ roots’.
Contradiction. So $f$ cannot be non-zero. The assumption in step 2 made the root count exceed the degree.

Answer. $f = 0$ . The same argument shows two polynomials of degree at most $n$ that agree at $n + 1$ points are equal (take $f = g - h$ ), which is why three points fix a parabola.

Conjugate pairs

Theorem B

Complex roots of real polynomials come in conjugate pairs

Let $f(x) = a_nx^n + \dots + a_0$ have real coefficients. If $f(z) = 0$ for a complex number $z$ , then $f(\bar z) = 0$ .

In one lineIf a polynomial with real coefficients has root $z$ , then $\bar z$ is also a root.

Why it is trueIt explains why the quadratic formula gives a pair $p \pm qi$ , and it is the step that forces real polynomials to factor into real linear and quadratic pieces.

Proofstandard

The idea: Conjugation respects sums and products and does not change real numbers, so conjugating f(z) = 0 gives f of the conjugate equal to 0.

\overline{z^k} = \bar z^{\,k}

for every whole number

k \ge 1

Rules for conjugates (3), used

k - 1

times (induction on

k

Conjugate

f(z) = \sum a_kz^k

\overline{f(z)} = \sum \bar a_k\,\overline{z^k} = \sum a_k\bar z^{\,k} = f(\bar z)

Rules for conjugates (2) and (3), plus

\bar a_k = a_k

because each

a_k

is real.This is the only place we use that the coefficients are real. For

x - i

the root is

i

and not

-i

, because the coefficient

-i

is not real.

f(\bar z) = \overline{f(z)} = \bar 0 = 0

\blacksquare

f(z) = 0

and

\bar 0 = 0

Quick checkA cubic polynomial with real coefficients has the root

2 + 3i

. Which number must also be a root?

Worked example JEE style

A real cubic has the root

1 + i

and the form

x^3 - 3x^2 + 4x - 2

. Find all roots.

Solution.

Use the conjugate. The coefficients are real, so $1 - i$ is also a root. Conjugate pairs theorem.
Build the real quadratic with these roots. $(x - (1 + i))(x - (1 - i)) = x^2 - 2x + 2$ . The sum of the pair is $2$ and the product is $|1 + i|^2 = 2$ . The imaginary parts cancel, which is why the factor is real.
Divide. Long division of $x^3 - 3x^2 + 4x - 2$ by $x^2 - 2x + 2$ : the first quotient term is $x$ , and $x(x^2 - 2x + 2) = x^3 - 2x^2 + 2x$ leaves $-x^2 + 2x - 2$ ; the next term is $-1$ , and $-1\cdot(x^2 - 2x + 2)$ leaves remainder $0$ . So $x^3 - 3x^2 + 4x - 2 = (x^2 - 2x + 2)(x - 1)$ . The division algorithm gives a quotient and a remainder. The remainder is $0$ , so the quadratic divides the cubic exactly. Check by expanding: $x^3 - x^2 - 2x^2 + 2x + 2x - 2$ matches.
Read the third root. $x - 1 = 0$ , so $x = 1$ .

Answer. The roots are $1,\ 1 + i,\ 1 - i$ .

The fundamental theorem of algebra

Counting: a polynomial of degree $n$ has at most $n$ roots. Can it have fewer, for example none? Over $\mathbb R$ it can ( $x^2 + 1$ ). Over $\mathbb C$ it cannot.

Theorem B

Fundamental theorem of algebra

Every polynomial $f$ with complex coefficients and $\deg f \ge 1$ has at least one root in $\mathbb C$ .

In one lineEvery non-constant polynomial with complex coefficients has at least one complex root.

Why it is trueIt says $\mathbb C$ is complete for algebra: you never need to invent numbers beyond it to solve a polynomial equation.

We do not prove this. The standard proofs come from complex analysis (Liouville’s theorem), from topology, and from algebra (Galois theory). They belong to MATH 521, 551 and 541. The picture behind the proof is worth having.

Why· Why it must be true: winding around zero

Take $f(z) = z^2 + 1$ and walk once around the circle $|z| = R$ . Watch the point $f(z)$ .

At $R = 2$ : $z = 2e^{i\theta}$ gives $f(z) = 1 + 4e^{2i\theta}$ . That is a circle of radius 4 centred at $1$ , which surrounds $0$ and goes round it twice (the $2\theta$ ).
At $R = \tfrac12$ : $f(z) = 1 + \tfrac14e^{2i\theta}$ , a tiny circle of radius $\tfrac14$ about $1$ . It never surrounds $0$ . It winds zero times.

As $R$ shrinks from $2$ to $\tfrac12$ the curve moves continuously. The number of times it winds around $0$ can change from $2$ to $0$ only if, at some moment, the curve goes through $0$ itself, meaning $f(z) = 0$ for a $z$ on that circle. Here it happens at $R = 1$ , where $z = \pm i$ . For any polynomial of degree $n \ge 1$ the same story holds: for huge $R$ the term $z^n$ dominates, so $f(z) \approx z^n$ winds $n \ge 1$ times round $0$ , and for tiny $R$ the curve stays near $f(0)$ and winds $0$ times (if $f(0) = 0$ we already have a root). So some circle’s image passes through $0$ . Making “winds” and “continuously” exact is topology, and that is the gap this sketch leaves.

Where this came from· Pictures of the plane, and the hunt for a proof of the fundamental theorem

Around 1800 several people drew complex numbers as points in the plane. Caspar Wessel, a Norwegian working in Denmark, presented the idea in 1797 and it was published in 1799 in Danish, where almost nobody read it. Jean-Robert Argand, an amateur mathematician in Paris, published his version in 1806. The algebraic question was whether every polynomial equation has a root. Euler and d’Alembert had given arguments in the 1700s. In his doctoral thesis of 1799 Gauss criticised them and gave his own proof, for polynomials with real coefficients (that they factor into real linear and quadratic pieces, the last corollary of this chapter). A 2017 paper by Basu and Velleman shows that his first argument has a gap in how it treats algebraic curves, and fills it. Gauss gave further proofs in 1816 and 1849. Euler had published the formula $e^{ix} = \cos x + i\sin x$ in his Introductio in analysin infinitorum in 1748.

Corollary B

Every complex polynomial splits into linear factors

Every complex polynomial $f$ of degree $n \ge 1$ can be written $f(x) = a(x - r_1)\cdots(x - r_n)$ with $a \ne 0$ and $r_1, \dots, r_n \in \mathbb C$ (not necessarily distinct).

In one lineA complex polynomial of degree $n \ge 1$ equals $a(x - r_1)(x - r_2)\cdots(x - r_n)$ , where $a$ is its leading coefficient.

Why it is trueIt says that over $\mathbb C$ a polynomial has exactly $n$ roots when you count repeats, which is what eigenvalue counting in chapter 2.8 relies on.

Proofstandard

The idea: Induction on n: the fundamental theorem gives one root, the factor theorem pulls it out, and what is left has degree one lower.

f(x) = ax + b = a\left(x - (-b/a)\right)

Base case

n = 1

: mathematical induction.

f

has a root

r

\mathbb C

Inductive step: assume the claim for degree

k

and let

\deg f = k + 1 \ge 2

. The fundamental theorem gives a root

r

f(x) = (x - r)q(x)

Factor theorem:

\deg q = k \ge 1

q(x) = a(x - r_1)\cdots(x - r_k)

, so

f(x) = a(x - r)(x - r_1)\cdots(x - r_k)

\blacksquare

The induction hypothesis for

q

. The leading coefficient of

q

equals that of

f

Corollary B

Real polynomials factor into real linear and quadratic factors

Every polynomial $f$ with real coefficients and $\deg f \ge 1$ can be written as a real constant times a product of real polynomials, each of degree 1 or degree 2.

In one lineEvery polynomial with real coefficients and degree at least 1 is a constant times a product of real polynomials of degree 1 or 2.

Why it is trueIt is why a real polynomial never needs anything beyond quadratics, and why every rational function can be integrated by partial fractions.

Proofstandard; Axler 4.16

The idea: Strong induction on the degree. Find a root; if it is real, pull out a linear factor. If it is not, its conjugate is also a root, and the two together make a real quadratic factor.

Let

f

be real of degree

n \ge 1

. It has a root

z \in \mathbb C

Strong induction on

n = \deg f

(chapter 0.2). Fundamental theorem of algebra:

f

has a complex root

z

Case 1:

z

is real.

Proof by cases on whether

z

is real.

Then

f = (x - z)q

with

q

real of degree

n - 1

Factor theorem. The division algorithm uses only the real coefficients, so

q

is real.

f

is a constant times real factors of degree 1 and 2.

n - 1 = 0

q

is a constant and we are done. Otherwise the induction hypothesis applies to

q

Case 2:

z = a + bi

with

b \ne 0

Case 2:

z

is not real.

Then

\bar z

is also a root of

f

Conjugate pairs theorem. Also

\bar z \ne z

because

b \ne 0

Define

m(x) = (x - z)(x - \bar z)

. Its coefficients are real.

Expand:

m(x) = x^2 - (z + \bar z)x + z\bar z = x^2 - 2ax + (a^2 + b^2)

.This is the real quadratic of the worked example above.

Divide:

f = mq + ux + v

with

u, v

real.

Division algorithm for real polynomials:

f = mq + r

with

q, r

real and

r = ux + v

of degree at most 1.

0 = 0 + uz + v

, so

uz + v = 0

Evaluate at

z

. Both

f(z) = 0

and

m(z) = 0

Hence

r = 0

and

f = mq

with

q

real.

u \ne 0

then

z = -v/u

would be real, contradicting

b \ne 0

. So

u = 0

, then

v = 0

f

is a constant times real linear and quadratic factors.

\blacksquare

\deg q = n - 2 \ge 0

. If it is

0

we are done; otherwise the induction hypothesis applies to

q

Worked example

Explain why every real polynomial of degree 7 has at least one real root, and why

x^6 + 1

shows the claim fails for degree 6.

Solution.

Factor over the reals. By the last corollary, $f = c \cdot p_1 p_2 \cdots p_k$ with $c$ real and each $p_j$ a real polynomial of degree 1 or 2. This is the factorisation theorem for real polynomials.
Add the degrees. The degree of a product is the sum of the degrees, so the degrees of the $p_j$ add up to $7$ . The lemma on the degree of a product.
Use that 7 is odd. Degree-2 factors contribute even numbers, and a sum of even numbers is even. So at least one factor has degree $1$ . If every factor had degree 2, the total would be even, not 7.
Read off the root. A degree-1 real factor is $ux + v$ with $u \ne 0$ , and it vanishes at the real number $-v/u$ , which is therefore a root of $f$ . If a factor is $0$ at a point, so is the whole product.
The even case. $x^6 + 1 \ge 1 > 0$ for every real $x$ , so it has no real root. Its six roots are non-real, in conjugate pairs. One counterexample shows the claim needs odd degree.

Answer. Odd degree forces a real linear factor, hence a real root. Even degree does not.

Worked example

Factor

x^3 - 1

over

\mathbb R

and over

\mathbb C

Solution.

Over $\mathbb R$ . $x^3 - 1 = (x - 1)(x^2 + x + 1)$ . Difference of cubes. The quadratic has discriminant $1 - 4 = -3 < 0$ , so it has no real roots and cannot be split further over $\mathbb R$ .
Over $\mathbb C$ . The roots of $x^2 + x + 1$ are $\omega = \tfrac{-1 + i\sqrt3}{2}$ and $\bar\omega = \tfrac{-1 - i\sqrt3}{2}$ . From the cube-roots-of-unity computation above. They are a conjugate pair.
Combine. $x^3 - 1 = (x - 1)(x - \omega)(x - \bar\omega)$ .
Check. $(x - \omega)(x - \bar\omega) = x^2 - (\omega + \bar\omega)x + \omega\bar\omega = x^2 + x + 1$ , because $\omega + \bar\omega = -1$ and $\omega\bar\omega = |\omega|^2 = 1$ . This reproduces the real quadratic from step 1.

Answer. Over $\mathbb R$ : $(x - 1)(x^2 + x + 1)$ . Over $\mathbb C$ : $(x - 1)(x - \omega)(x - \bar\omega)$ .

A preview· Preview: eigenvalues are roots (chapter 2.8)

A square matrix $A$ has an eigenvalue $\lambda$ when $\det(A - \lambda I) = 0$ . The left side is a polynomial in $\lambda$ , called the characteristic polynomial. The fundamental theorem says that a complex $n \times n$ matrix has at least one eigenvalue, and the conjugate-pair theorem says that a real matrix has complex eigenvalues in pairs. The rotation by $90^\circ$ in the picture has matrix $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ and characteristic polynomial $\lambda^2 + 1$ , so its eigenvalues are $i$ and $-i$ : a conjugate pair, and the reason it has no real eigenvector.

Where it is used

Used inside maths· Inside maths: eigenvalues, partial fractions, differential equations

Eigenvalues (2.8). Eigenvalues are the roots of the characteristic polynomial. Over $\mathbb C$ there are always $n$ of them counted with repeats, which is why every theory of eigenvalues is cleanest over the complex numbers, and why a real matrix of odd size always has a real eigenvalue (odd degree forces a linear real factor).
Partial fractions in integration. In JEE you were told that the denominator of a rational function splits into real linear and irreducible quadratic factors. That is exactly the last corollary. For example $x^4 + 1 = (x^2 + \sqrt2x + 1)(x^2 - \sqrt2x + 1)$ , which is what makes $\int \frac{dx}{x^4 + 1}$ computable.
Differential equations (Unit 7). The equation $y'' + y = 0$ has characteristic polynomial $\lambda^2 + 1$ , with roots $\pm i$ . The solutions $e^{\pm it}$ are the sine and cosine through Euler’s formula.

Used in the real world· Real world: AC circuits, the FFT, and quantum amplitudes

AC circuits and phasors. In an alternating-current circuit the voltage is $V\cos(\omega t)$ . Engineers write it as the real part of $Ve^{i\omega t}$ , called a phasor (they use $j$ for $i$ , because $i$ already means current). Differentiating $e^{i\omega t}$ just multiplies it by $i\omega$ . So an inductor ( $v = L\,di/dt$ ) becomes $V = i\omega L\,I$ , and a capacitor ( $i = C\,dv/dt$ ) becomes $V = I/(i\omega C)$ . Calculus turns into complex multiplication:

Component	Impedance $Z$ (voltage divided by current)
Resistor $R$	$R$
Inductor $L$	$i\omega L$
Capacitor $C$	$\dfrac{1}{i\omega C} = \dfrac{-i}{\omega C}$

Real numbers: an Indian mains supply of $230$ V at $50$ Hz across a series circuit with $R = 100\ \Omega$ , $L = 0.5$ H, $C = 50\ \mu$ F.

Angular frequency. $\omega = 2\pi \cdot 50 = 314.16$ rad/s.
Reactances. $\omega L = 157.08\ \Omega$ and $1/(\omega C) = 1/(314.16 \cdot 50 \times 10^{-6}) = 63.66\ \Omega$ .
Impedance. In series the impedances add: $Z = 100 + i\,157.08 - i\,63.66 = 100 + 93.42i\ \Omega$ . Adding complex numbers is vector addition, so the resistive part and the reactive part are at right angles.
Size and phase of $Z$ . $|Z| = \sqrt{100^2 + 93.42^2} = 136.85\ \Omega$ and $\arg Z = \tan^{-1}(93.42/100) = 43.05^\circ$ .
Current. $I = V/Z$ . By the polar rule, $|I| = 230/136.85 = 1.681$ A, and the angle of $I$ is the angle of $V$ minus $43.05^\circ$ : the current lags the voltage by $43.05^\circ$ . Dividing complex numbers divides moduli and subtracts angles (this chapter’s rule).
Power. The resistor is the only part that uses energy: $P = |I|^2R = 1.681^2 \cdot 100 = 282.5$ W. Check: $VI\cos\varphi = 230 \cdot 1.681 \cdot \cos 43.05^\circ = 282.5$ W, the same answer.
Voltages across parts. $|V_R| = 168.1$ V, $|V_L| = 264.0$ V, $|V_C| = 107.0$ V. The inductor voltage exceeds the supply, and that is correct: $V_L$ and $V_C$ point in opposite directions (angles $+90^\circ$ and $-90^\circ$ from $V_R$ ), so $\sqrt{168.1^2 + (264.0 - 107.0)^2} = 230$ V.

Signal processing and the FFT. The discrete Fourier transform of $N$ samples uses the $N$ -th roots of unity $\omega = e^{-2\pi i/N}$ . Squaring the $N$ -th roots of unity gives each of the $(N/2)$ -th roots of unity twice, so a transform of size $N$ splits into two of size $N/2$ , glued by one multiplication per output (James Cooley and John Tukey published this in 1965 as the fast Fourier transform). For $N = 1024$ the direct method needs about $N^2 = 1{,}048{,}576$ multiplications and the FFT about $N\log_2N = 10{,}240$ , roughly a hundred times fewer. That saving is why audio, JPEG-style compression, radar and Wi-Fi all use it.

Quantum amplitudes (label B). The state of a qubit is $a|0\rangle + b|1\rangle$ with complex $a, b$ and $|a|^2 + |b|^2 = 1$ . The probability of measuring $0$ is $|a|^2$ , the modulus squared, which is the quantity our theorem controls. The complex phase matters through interference: the state $\tfrac1{\sqrt2}\left(|0\rangle + e^{i\varphi}|1\rangle\right)$ gives a 50% chance of each outcome for every $\varphi$ , but after one Hadamard gate the amplitude of $|0\rangle$ is $(1 + e^{i\varphi})/2$ and the probability of $0$ is $|1 + e^{i\varphi}|^2/4 = \cos^2(\varphi/2)$ . That is $1$ for $\varphi = 0$ , $\tfrac12$ for $\varphi = \pi/2$ , and $0$ for $\varphi = \pi$ . Quantum computing is built on this cancellation.

Traps

Watch out· Do not use $\sqrt{ab} = \sqrt a\sqrt b$ for negative numbers

$\sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = -6$ , but $\sqrt{(-4)(-9)} = \sqrt{36} = 6$ . The rule $\sqrt{ab} = \sqrt a\sqrt b$ is only for non-negative $a, b$ . The cure is to write $i$ before doing anything: $\sqrt{-4} = 2i$ . Taking a square root of a negative number is a choice of one of two roots, not a function you may combine freely.

Watch out· The argument depends on the quadrant

For $z = -1 - i$ , $\tan^{-1}(b/a) = \tan^{-1}(1) = \pi/4$ , but the point is in the third quadrant, so $\arg z = -3\pi/4$ (equivalently $5\pi/4$ ). Decide the quadrant first, or use both $\cos\theta = a/r$ and $\sin\theta = b/r$ . Also, an argument is only defined up to adding $2\pi$ , so “the” argument means a chosen one.

Watch out· $|z|^2$ is not $z^2$

$|z|^2 = z\bar z$ is a non-negative real number. $z^2$ is another complex number. For $z = i$ : $|i|^2 = 1$ but $i^2 = -1$ . The same slip appears as " $|z + w|^2 = |z|^2 + |w|^2$ ", which ignores the cross terms $z\bar w + \bar zw$ .

Watch out· Conjugate pairs need real coefficients

$x - i$ has the root $i$ and not $-i$ . $x^2 - 2ix - 1 = (x - i)^2$ also has only the root $i$ . The conjugate-pair theorem is about polynomials whose coefficients are all real. When a complex number sits in a coefficient, the roots need not pair up.

Watch out· 'At most n roots' needs a field, and a non-zero polynomial

The zero polynomial has every number as a root, which is why the theorem says non-zero. And the proof used “a product is zero only if a factor is”: over the integers mod 8, $x^2 - 1$ has four roots ( $1, 3, 5, 7$ ) because $(x - 1)(x + 1)$ can be $0$ with neither factor $0$ . Over $\mathbb R$ and $\mathbb C$ this cannot happen.

Problems

check

Compute $(3 + 4i)(1 - 2i)$ . Give its modulus and its conjugate, and check the modulus against $|3 + 4i|\,|1 - 2i|$ .

Solution

$(3 + 4i)(1 - 2i) = 3 - 6i + 4i - 8i^2 = 3 + 8 - 2i = 11 - 2i$ . Its conjugate is $11 + 2i$ (check: $\overline{3 + 4i}\cdot\overline{1 - 2i} = (3 - 4i)(1 + 2i) = 3 + 6i - 4i - 8i^2 = 11 + 2i$ ). Its modulus is $\sqrt{121 + 4} = \sqrt{125} = 5\sqrt5$ , and $|3 + 4i|\,|1 - 2i| = 5\cdot\sqrt5 = 5\sqrt5$ . They agree.

check

Write $-1 - i$ in polar form. Which quadrant is it in, and what goes wrong if you use $\tan^{-1}(b/a)$ alone?

Solution

$r = \sqrt{1 + 1} = \sqrt2$ . Both coordinates are negative, so it is in the third quadrant, with $\cos\theta = -1/\sqrt2$ and $\sin\theta = -1/\sqrt2$ , so $\theta = -3\pi/4$ (or $5\pi/4$ ). So $-1 - i = \sqrt2\,e^{-3\pi i/4}$ . The formula $\tan^{-1}(b/a) = \tan^{-1}(1) = \pi/4$ would give the point $1 + i$ in the first quadrant, a half-turn away.

check

Find the remainder when $x^5 - 3x^2 + 2$ is divided by $x + 1$ , without dividing.

Solution

Write $x + 1 = x - (-1)$ , so $c = -1$ . The remainder theorem gives $f(-1) = (-1)^5 - 3(-1)^2 + 2 = -1 - 3 + 2 = -2$ .

check

A real polynomial has degree 5. What is the largest number of different complex roots it can have, and why must at least one of them be real?

Solution

At most 5, by the theorem on at most $n$ roots. By the last corollary it is a constant times a product of real factors of degree 1 or 2 whose degrees add up to 5. Quadratic factors contribute 2 each, an even number, so at least one factor has degree 1, and a real linear factor $x - r$ gives a real root $r$ .

use

Divide $f(x) = x^4 + 2x^3 - x + 3$ by $g(x) = x^2 + 1$ . Check your remainder by evaluating $f$ at $i$ .

Solution

Divide: $x^4 \div x^2 = x^2$ , and $f - x^2g = 2x^3 - x^2 - x + 3$ . Next $2x^3 \div x^2 = 2x$ , and subtracting $2xg = 2x^3 + 2x$ leaves $-x^2 - 3x + 3$ . Next $-x^2 \div x^2 = -1$ , and subtracting $-g = -x^2 - 1$ leaves $-3x + 4$ , of degree 1, less than 2. So $q = x^2 + 2x - 1$ and $r = -3x + 4$ . Check by expanding: $(x^2 + 1)(x^2 + 2x - 1) + (-3x + 4) = x^4 + 2x^3 + 2x - 1 - 3x + 4 = x^4 + 2x^3 - x + 3$ . Second check: $i$ is a root of $g$ , so $f(i) = g(i)q(i) + r(i) = r(i)$ . Now $f(i) = 1 - 2i - i + 3 = 4 - 3i$ and $r(i) = -3i + 4$ . They agree.

useJEE style

Find all four complex solutions of $z^4 = -16$ and mark where they sit in the plane.

Solution

$-16 = 16e^{i\pi}$ . A solution $se^{i\varphi}$ needs $s^4 = 16$ , so $s = 2$ , and $4\varphi = \pi + 2\pi k$ , so $\varphi = \pi/4 + k\pi/2$ for $k = 0, 1, 2, 3$ . The solutions are $2e^{i\pi/4} = \sqrt2(1 + i)$ , $2e^{3i\pi/4} = \sqrt2(-1 + i)$ , $2e^{5i\pi/4} = \sqrt2(-1 - i)$ , $2e^{7i\pi/4} = \sqrt2(1 - i)$ . They are the corners of a square of radius 2 rotated $45^\circ$ . Check: $(\sqrt2(1 + i))^4 = 4(1 + i)^4 = 4(2i)^2 = -16$ .

use

The real cubic $x^3 - 5x^2 + 9x - 5$ has the root $2 + i$ . Find all its roots and its factorisation into real factors.

Solution

The coefficients are real, so $2 - i$ is also a root. The real quadratic with these roots is $(x - 2 - i)(x - 2 + i) = (x - 2)^2 + 1 = x^2 - 4x + 5$ . Dividing gives $x^3 - 5x^2 + 9x - 5 = (x^2 - 4x + 5)(x - 1)$ . Check by expanding: $x^3 - 4x^2 + 5x - x^2 + 4x - 5$ . The roots are $1,\ 2 + i,\ 2 - i$ , and the real factorisation is $(x - 1)(x^2 - 4x + 5)$ . Second check: for a monic cubic the product of the three roots is minus the constant term, here $-(-5) = 5$ , and indeed $1\cdot(2 + i)(2 - i) = 5$ .

use

A $230$ V, $50$ Hz supply feeds a series circuit with resistance $3\ \Omega$ and inductive reactance $4\ \Omega$ . Find the impedance, the current, the phase between voltage and current, and the power used.

Solution

$Z = 3 + 4i\ \Omega$ , so $|Z| = 5\ \Omega$ . The current has size $230/5 = 46$ A. The phase angle is $\tan^{-1}(4/3) = 53.13^\circ$ , with the current lagging the voltage (the reactance is inductive, $+i$ ). The only part that uses power is the resistor: $P = |I|^2R = 46^2 \cdot 3 = 6348$ W. Second check: $P = VI\cos\varphi = 230\cdot 46\cdot (3/5) = 6348$ W.

prove

Prove the triangle inequality $|z + w| \le |z| + |w|$ for complex numbers $z, w$ .

Solution

Plan. Square both sides to avoid square roots and expand with conjugates; the cross terms are twice a real part, which is at most a modulus.

$|z + w|^2 = (z + w)(\bar z + \bar w) = z\bar z + w\bar w + z\bar w + \bar zw$ , by the rules for conjugates.
Now $\bar zw = \overline{z\bar w}$ , and $u + \bar u = 2\operatorname{Re}u$ for any $u$ . So $z\bar w + \bar zw = 2\operatorname{Re}(z\bar w)$ .
For $u = a + bi$ , $\operatorname{Re}u = a \le \sqrt{a^2} \le \sqrt{a^2 + b^2} = |u|$ . So $2\operatorname{Re}(z\bar w) \le 2|z\bar w| = 2|z|\,|\bar w| = 2|z|\,|w|$ , using the multiplicative property and $|\bar w| = |w|$ .
So $|z + w|^2 \le |z|^2 + |w|^2 + 2|z||w| = (|z| + |w|)^2$ .
Both sides are non-negative, so take square roots: $|z + w| \le |z| + |w|$ . $\blacksquare$

prove

Prove that for $z, w \in \mathbb C$ with $w \ne 0$ : $|z/w| = |z|/|w|$ , and the argument of $z/w$ is the argument of $z$ minus the argument of $w$ (up to $2\pi$ ) when $z \ne 0$ .

Solution

Let $q = z/w$ , so $z = qw$ . By the multiplicative property, $|z| = |q|\,|w|$ , so $|q| = |z|/|w|$ since $|w| \ne 0$ . For the angle, write $q = se^{i\gamma}$ and $w = te^{i\beta}$ with $s, t > 0$ . Multiplying in polar form, $z = qw = st\,e^{i(\gamma + \beta)}$ . If $z = ue^{i\alpha}$ , then $e^{i\alpha} = e^{i(\gamma + \beta)}$ (dividing by the positive number $u = st$ ), so $\alpha$ and $\gamma + \beta$ differ by a multiple of $2\pi$ . Hence $\gamma \equiv \alpha - \beta \pmod{2\pi}$ . $\blacksquare$

prove

Prove De Moivre’s formula for all integers, then use it to prove $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ and $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ .

Solution

De Moivre was proved above for all integers (induction for $n \ge 1$ , then $n = 0$ and $n < 0$ ). For the triple angle, take $n = 3$ and write $c = \cos\theta$ , $s = \sin\theta$ . Then $\cos 3\theta + i\sin 3\theta = (c + is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 = (c^3 - 3cs^2) + i(3c^2s - s^3)$ , using $i^2 = -1$ and $i^3 = -i$ . Comparing real and imaginary parts (two complex numbers are equal when both coordinates agree): $\cos 3\theta = c^3 - 3cs^2 = c^3 - 3c(1 - c^2) = 4c^3 - 3c$ , and $\sin 3\theta = 3c^2s - s^3 = 3(1 - s^2)s - s^3 = 3s - 4s^3$ . Check at $\theta = 0.7$ : both sides of the first formula are $-0.50485$ . $\blacksquare$

proveAxler 4A style

Prove that if $f$ and $g$ are polynomials of degree at most $n$ and $f(c_j) = g(c_j)$ at $n + 1$ different numbers $c_1, \dots, c_{n+1}$ , then $f = g$ .

Solution

Plan. Subtract, and compare the number of roots with the degree.

Let $h = f - g$ . Then $h = 0$ or $\deg h \le n$ , and $h(c_j) = f(c_j) - g(c_j) = 0$ for each of the $n + 1$ different numbers $c_j$ . If $h$ were non-zero, it would be a non-zero polynomial of degree at most $n$ with at least $n + 1$ different roots, contradicting the theorem that a non-zero polynomial of degree $m$ has at most $m \le n$ roots. So $h = 0$ and $f = g$ . (For example, a polynomial of degree at most 2 is determined by its values at three points.) $\blacksquare$

prove

Use the factorisation theorem for real polynomials to prove that every real polynomial of odd degree has a real root. Then show by an example that the statement is false for even degree.

Solution

Let $f$ be real of odd degree $n$ . By the corollary, $f$ is a constant times a product of real factors of degree 1 or 2 whose degrees add to $n$ (the degree of a product is the sum of degrees). If every factor had degree 2, then $n$ would be even. So some factor has degree 1: it is $ux + v$ with $u \ne 0$ real, and $-v/u$ is a real root of $f$ . $\blacksquare$ For even degree, $x^2 + 1$ has roots $\pm i$ only.

stretchclassic

Let $n \ge 2$ and let $\omega_0 = 1, \omega_1, \dots, \omega_{n-1}$ be the $n$ -th roots of unity. Prove that $x^n - 1 = \prod_{k=0}^{n-1}(x - \omega_k)$ , and deduce that the product of the distances from the corner $1$ to the other corners of the regular $n$ -gon is $n$ : $\prod_{k=1}^{n-1}|1 - \omega_k| = n$ .

Solution

Plan. Show the two sides are equal polynomials by counting roots, then cancel a common factor and set $x = 1$ .

Let $P(x) = \prod_{k=0}^{n-1}(x - \omega_k)$ . Both $P$ and $x^n - 1$ are monic of degree $n$ , so $x^n - 1 - P$ is zero or has degree less than $n$ .
Both vanish at each of the $n$ different numbers $\omega_k$ (the roots of unity theorem), so the difference has $n$ roots. A non-zero polynomial of degree less than $n$ has fewer than $n$ roots, so the difference is zero: $x^n - 1 = P(x)$ .
Also $x^n - 1 = (x - 1)(1 + x + \dots + x^{n-1})$ . Since $\omega_0 = 1$ , comparing with step 2: $(x - 1)(1 + x + \dots + x^{n-1}) = (x - 1)\prod_{k=1}^{n-1}(x - \omega_k)$ .
Cancel $x - 1$ . This is allowed: if $(x - 1)A = (x - 1)B$ then $(x - 1)(A - B) = 0$ , and a product of non-zero polynomials is non-zero by the lemma, so $A = B$ . So $1 + x + \dots + x^{n-1} = \prod_{k=1}^{n-1}(x - \omega_k)$ .
Put $x = 1$ : $n = \prod_{k=1}^{n-1}(1 - \omega_k)$ . Take moduli, using multiplicativity: $\prod_{k=1}^{n-1}|1 - \omega_k| = n$ . $\blacksquare$

Check with $n = 4$ : $|1 - i|\cdot|1 - (-1)|\cdot|1 + i| = \sqrt2\cdot 2\cdot\sqrt2 = 4$ . With $n = 3$ : $\sqrt3\cdot\sqrt3 = 3$ .

stretchclassic (a JEE integration favourite)

Factor $x^4 + 1$ into two real quadratics, using its complex roots. Why is there no factorisation into real linear factors?

Solution

The roots of $x^4 = -1$ are $e^{i(\pi + 2\pi k)/4}$ for $k = 0, 1, 2, 3$ : that is $e^{i\pi/4}$ , $e^{3i\pi/4}$ , $e^{5i\pi/4}$ , $e^{7i\pi/4}$ . They form two conjugate pairs: $e^{i\pi/4}$ with $e^{-i\pi/4} = e^{7i\pi/4}$ , and $e^{3i\pi/4}$ with $e^{5i\pi/4}$ . A pair $e^{\pm i\varphi}$ gives the real quadratic $(x - e^{i\varphi})(x - e^{-i\varphi}) = x^2 - 2\cos\varphi\,x + 1$ . With $\varphi = \pi/4$ : $x^2 - \sqrt2x + 1$ . With $\varphi = 3\pi/4$ : $x^2 + \sqrt2x + 1$ . Check: $(x^2 + 1 - \sqrt2x)(x^2 + 1 + \sqrt2x) = (x^2 + 1)^2 - 2x^2 = x^4 + 1$ . There are no real linear factors because $x^4 + 1 \ge 1 > 0$ for every real $x$ , so it has no real roots; its four roots are all non-real. $\blacksquare$

Summary

A complex number is a pair of reals $(a, b)$ with $(a,b)(c,d) = (ac - bd, ad + bc)$ . Then $i = (0, 1)$ and $i^2 = -1$ is a one-line theorem. The pair product obeys the usual laws and every non-zero number has an inverse, so $\mathbb C$ is a field.
$z\bar z = |z|^2$ , and $\overline{zw} = \bar z\bar w$ . Together they prove $|zw| = |z||w|$ without coordinates.
Polar form: multiplying multiplies lengths and adds angles. The proof is the addition formulas for sine and cosine. De Moivre and roots of unity follow. $z^n = 1$ has exactly $n$ solutions, the corners of a regular $n$ -gon on the unit circle.
Euler’s formula (label B): $e^{i\theta} = \cos\theta + i\sin\theta$ from the power series, so “multiply lengths, add angles” is the exponent rule $e^{i\alpha}e^{i\beta} = e^{i(\alpha + \beta)}$ .
Polynomials divide with remainder. The remainder theorem and factor theorem follow, and so does at most $n$ roots (by induction, factoring one root out at a time). This needs numbers with no zero divisors.
Real polynomials have conjugate pairs of non-real roots. The fundamental theorem of algebra (label B, stated and sketched with winding, proved in complex analysis) says every complex polynomial of degree $n$ has $n$ roots counted with repeats. So every real polynomial is a product of real linear and quadratic factors.
Used in: eigenvalues of the characteristic polynomial (2.8), partial fractions, differential equations $e^{(a+ib)t}$ (Unit 7), AC impedance, the FFT, quantum amplitudes.

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0.6 Complex numbers and polynomials, done properly

The picture

The idea, in words

Definitions and theorems

Complex numbers as pairs

Complex number

The square of i

Laws of complex arithmetic

The plane, modulus and conjugate

Modulus

Conjugate

Rules for conjugates

The modulus is multiplicative

Polar form: why multiplication turns and stretches

Polar form

Multiplying in polar form

De Moivre's formula

Euler’s formula

Euler's formula

Roots of unity

The n-th roots of unity

Polynomials

Polynomial

Degree

Root

No zero divisors, and the degree of a product

Division with remainder

Polynomial division algorithm

Remainder and factor theorems

Remainder theorem for polynomials

Factor theorem for polynomials

At most n roots

A polynomial of degree n has at most n roots

Conjugate pairs

Complex roots of real polynomials come in conjugate pairs

The fundamental theorem of algebra

Fundamental theorem of algebra

Every complex polynomial splits into linear factors

Real polynomials factor into real linear and quadratic factors

Where it is used

Traps

Problems

Summary

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Where this sits

Comes from

Leads to

Same idea elsewhere