Part 1 · Mathematical tools

1.1 Complex numbers

Source: Centini’s whiteboard of 2 Oct, p. 1–2, and 6 Oct, p. 3; the square against the square modulus, the conjugate of $e^{i θ}$ and the points on the circle from the 2024/25 board #3

Objectives

Read a complex number as a point of the plane, and as a length and a direction.
Recipe: go between $a + ib$ and $∣ c ∣ e^{i θ}$ , minding the quadrant.
Pictures: the conjugate as a mirror image; multiplication as a stretch and a turn.
Theorem: Euler’s identity; $∣ c ∣^{2} = c c^{*}$ is real.
Vocabulary: real and imaginary part, phase, polar form.
Essential vocabulary: complex conjugate, modulus.

A wavefunction is complex, and a probability is real. Everything in this section is about getting from one to the other.

The complex plane

Definition 1.1.1Complex number

$c = a + ib, i^{2} = - 1, a, b \in R .$ $a = Real (c)$ is the real part and $b = Imag (c)$ the imaginary part; both are real numbers. The set of all $c$ is $C$ .

So $c$ is the point $(a, b)$ of a plane, or the arrow from the origin to it: $a$ along the real axis, $b$ along the imaginary axis.

Definition 1.1.2Complex conjugate and modulusEssential

The complex conjugate of $c = a + ib$ is $c^{*} = a - ib$ : its mirror image in the real axis. The modulus of $c$ is the length of its arrow, $∣ c ∣ = a^{2} + b^{2} = Real (c)^{2} + Imag (c)^{2} .$

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Figure 1.1.1 Drag

c

. Its

real part

and

imaginary part

are its shadows on the axes, and

c^{*}

is its mirror image. The point snaps to the half-unit grid, to the dotted unit circle and to every

1 5^{\circ}

. Put it on the real axis and

c

is its own conjugate; take it into the left half and read the warning about the phase.

There is no “ $c > d$ ” for complex numbers: no ordering of the plane is compatible with arithmetic (beyond the boards). To say which of two complex numbers is bigger, compare their moduli.

Proposition 1.1.3The modulus from the conjugate

$∣ c ∣^{2} = c c^{*} .$

Proof

$(a + ib) (a - ib) = a^{2} - iab + iab - i^{2} b^{2} = a^{2} + b^{2}$ . The cross terms cancel and $- i^{2} = + 1$ .

The product of a number with its conjugate is always real and never negative. This is why quantum mechanics squares a wavefunction as $ψ^{*} ψ$ : it turns a complex amplitude into a real probability.

Warning

$∣ c ∣^{2}$ is not $c^{2}$ . $c^{2} = (a + ib)^{2} = a^{2} - b^{2} + 2 iab$ is complex in general, with imaginary part $2 ab$ ; $∣ c ∣^{2} = a^{2} + b^{2}$ is always real. (2024/25 board #3.)

Adding $c$ and $c^{*}$ cancels the imaginary parts; subtracting them cancels the real parts:

Proposition 1.1.4Real and imaginary part from

c

and

c^{*}

$Real (c) = \frac{c + c ^{*}}{2}, Imag (c) = \frac{c - c ^{*}}{2 i} .$

Polar form

Instead of the two coordinates, give the arrow’s length and its angle with the real axis.

Definition 1.1.5Phase

The phase (or argument) of $c \neq = 0$ is the angle $θ$ of its arrow with the real axis: $θ = arctg (\frac{Imag ( c )}{Real ( c )}) = arctg \frac{b}{a} (for a > 0) .$

Warning

$arctg$ only returns angles between $- π /2$ and $π /2$ , so it cannot tell $1 + i$ from $- 1 - i$ : both have $b / a = 1$ . When $a < 0$ , add $π$ . (Beyond the boards.)

Theorem 1.1.6Euler’s identity

$e^{i θ} = cos θ + i sin θ, e^{- i θ} = cos θ - i sin θ .$

$e^{i θ}$ is the point at angle $θ$ on the circle of radius $1$ : $∣ e^{i θ} ∣ = cos^{2} θ + sin^{2} θ = 1$ .
$e^{- i θ}$ is its conjugate. Conjugating flips the sign of the sine, and since $cos$ is even and $sin$ odd, $(e^{i θ})^{*} = cos (- θ) + i sin (- θ) = e^{- i θ}$ (2024/25 board #3).
So $e^{i θ} e^{- i θ} = e^{0} = 1$ , which is $∣ e^{i θ} ∣^{2} = 1$ again.

Five points of the circle worth knowing (2024/25 board #3):

$θ$	$0$	$π /2$	$π$	$3 π /2 = - π /2$	$2 π$
$e^{i θ}$	$1$	$i$	$- 1$	$- i$	$1$

A quarter turn multiplies by $i$ , half a turn by $- 1$ : $i \cdot i = - 1$ is two quarter turns.

Key observation 1.1.7Polar form

Every complex number is a length times a direction on the unit circle: $c = ∣ c ∣ e^{i θ} = a + ib .$ Same number, two forms. On the board: $x_{0} = 100 e^{i θ_{0}}$ has modulus $100$ and phase $θ_{0}$ .

Recipe 1.1.8Between the two forms

From $a + ib$ : $∣ c ∣ = a^{2} + b^{2}$ , $θ = arctg (b / a)$ , plus $π$ if $a < 0$ .
From $∣ c ∣ e^{i θ}$ : $a = ∣ c ∣ cos θ$ , $b = ∣ c ∣ sin θ$ (Euler).

Example 1.1.9

c = 3 + 4 i

beyond the boards

$∣ c ∣ = 9 + 16 = 5$ ; $θ = arctg (4/3) = 0.927$ rad $= 53. 1^{\circ}$ ; $c = 5 e^{0.927 i}$ .
$c^{*} = 3 - 4 i$ , and $c c^{*} = 9 + 16 = 25 = ∣ c ∣^{2}$ .
$\frac{c + c ^{*}}{2} = 3$ and $\frac{c - c ^{*}}{2 i} = \frac{8 i}{2 i} = 4$ .

Multiplying is stretching and turning

The polar form earns its keep in products.

Proposition 1.1.10Products in polar formbeyond the boards

$c_{1} c_{2} = ∣ c_{1} ∣ e^{i θ_{1}} ∣ c_{2} ∣ e^{i θ_{2}} = ∣ c_{1} ∣∣ c_{2} ∣ e^{i (θ_{1} + θ_{2})} :$ moduli multiply and phases add.

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Figure 1.1.2 Drag

c_{1}

and

c_{2}

and watch their

product

. Its length is the product of the lengths and its angle the sum of the angles. Put

c_{2}

on the unit circle: the product only turns.

So multiplying by a pure phase $e^{i α}$ rotates a number without changing its modulus. That is why a time factor like $e^{- iω t}$ never changes a probability $∣ \cdot ∣^{2}$ (Key observation 2.4.8).

Periodicity. Once more around the circle lands on the same point: $e^{i θ} = e^{i (θ + 2 πn)}, n \in Z .$ The 2024/25 board writes it as $e^{i θ} e^{i 2 kπ} = e^{i θ}$ , since $e^{i 2 kπ} = cos 2 kπ + i sin 2 kπ = 1$ . This is what makes momentum discrete in a box (§1.5 Momentum in a box, Fourier and the Dirac delta).

Exercises

Exercise 1.1.1beyond the boards

Write $c = - 1 - i$ in polar form.

Solution

$∣ c ∣ = 2$ . $arctg (b / a) = arctg 1 = π /4$ , but $a < 0$ , so $θ = π /4 + π = 5 π /4$ , or equivalently $- 3 π /4$ . $c = 2 e^{5 πi /4}$ .

Exercise 1.1.2beyond the boards

Compute $c^{2}$ and $∣ c ∣^{2}$ for $c = 1 + 2 i$ , and check that $∣ c^{2} ∣ = ∣ c ∣^{2}$ .

Solution

$c^{2} = 1 - 4 + 4 i = - 3 + 4 i$ , complex. $∣ c ∣^{2} = 1 + 4 = 5$ , real. $∣ c^{2} ∣ = 9 + 16 = 5$ : moduli multiply (Proposition 1.1.10).

Exercise 1.1.3beyond the boards

Show that $∣ c_{1} c_{2} ∣^{2} = ∣ c_{1} ∣^{2} ∣ c_{2} ∣^{2}$ using only Proposition 1.1.3 and $(c_{1} c_{2})^{*} = c_{1}^{*} c_{2}^{*}$ .

Solution

$∣ c_{1} c_{2} ∣^{2} = c_{1} c_{2} (c_{1} c_{2})^{*} = c_{1} c_{2} c_{1}^{*} c_{2}^{*} = (c_{1} c_{1}^{*}) (c_{2} c_{2}^{*}) = ∣ c_{1} ∣^{2} ∣ c_{2} ∣^{2}$ .

Exercise 1.1.4beyond the boards

Which complex numbers satisfy $c^{*} = 1/ c$ ?

Solution

$c^{*} = 1/ c$ means $c c^{*} = 1$ , so $∣ c ∣ = 1$ : exactly the points of the unit circle, $c = e^{i θ}$ .