Interactive guide · complex numbers

Complex Numbers

Drag z (and w) around the plane or type their coordinates. You'll see their rectangular, polar and exponential forms at once, and what happens when you operate with them. Tap any card to see its step-by-step proof.

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wz · wabρz
zwresultRe(z) = aIm(z) = b
Complex number
z = 2 + i
|z| = √5 ≈ 2.236 · Arg z ≈ 0.464 (26.6°)
z
w
Type a and b of z = a + bi. Integers, decimals, fractions or roots all work: 3, -1.5, 1/2, √3/2.
Operation
Examples from the notes
Result of the operation
(a + bi)(c + di) = (ac − bd) + (ad + bc)i
= (2·(−1) − 1·1) + (2·1 + 1·(−1))i
= −3 + i
(ρ1)θ1 · (ρ2)θ2 = (ρ1ρ2)θ1 + θ2
= (√5 · √2)0.464 + 3π/4 = √102.82
= √102.82
Every form of z
a + bi Rectangular
2 + i
Re(z) = a Real part
2
Im(z) = b Imaginary part
1
|z| = √a² + b² Modulus
√2² + 1² = √5 ≈ 2.236
θ = Arg z Argument
arctan(0.5) ≈ 26.6°
ρθ Polar
√50.464
ρ(cos θ + i sin θ) Trigonometric
√5(cos 0.464 + i sin 0.464)
ρeiθ Exponential
√5e0.464i
z = a − bi Conjugate
2 − i
01 · Definition

What is a complex number?

The equation x² + 1 = 0 has no real solution, because no real number squared gives −1. We add a new number, i, with i² = −1, and build all complex numbers from it.

Imaginary uniti = √−1, that is, i² = −1.
Rectangular (binomial) formz = a + bi, with a and b real numbers. It has two terms, hence “binomial”.
Real partRe(z) = a: the term without i.
Imaginary partIm(z) = b: the number multiplying i (without the i).
The set ℂℂ = {a + bi : a, b ∈ ℝ}. Real numbers are the complex numbers with b = 0.

Axiomatic definition: ℂ = (ℝ², +, ·)

(a,b)⋅(c,d)=(ac−bd, ad+bc)(a, b) \cdot (c, d) = (ac - bd,\ ad + bc)

A complex number is a pair of reals with a special multiplication rule. i² = −1 follows from it.

See proof

Powers of i

i0=1,i1=i,i2=−1,i3=−i,i4=1i^0 = 1,\quad i^1 = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1

Powers of i repeat every 4: 1, i, −1, −i, 1, …

See proof
02 · Structure

ℂ is a field… but it can't be ordered

A field is a set where you can add, subtract, multiply and divide (except by 0) with the usual rules. ℂ is one, but it loses something ℝ has: order.

ℂ is a field

z≠0  ⇒  z−1=a−bia2+b2z \neq 0 \;\Rightarrow\; z^{-1} = \dfrac{a - bi}{a^2 + b^2}

Addition and multiplication follow the usual rules and every z ≠ 0 has an inverse.

See proof

Is 2 < i? No: ℂ can't be ordered

2<i ?No2 < i\,?\quad \text{No}

There is no order on ℂ compatible with its operations. Comparing complex numbers with < makes no sense.

See proof
03 · Graphical representation

The complex plane

Since a complex number is a pair (a, b), it's drawn as a point of the plane or as an arrow (vector) from the origin.

The X axis is the real axis: points (x, 0) are the real numbers x. The Y axis is the imaginary axis: points (0, y) are the pure imaginaries yi. The drawing shows the examples from your notes: z = 2 − i and w = 1 + 2i.

Phasor: a complex number of modulus 1

∣z∣=1  ⇔  a2+b2=1  ⇔  z=cos⁡θ+isin⁡θ|z| = 1 \;\Leftrightarrow\; a^2 + b^2 = 1 \;\Leftrightarrow\; z = \operatorname{cos}\theta + i\operatorname{sin}\theta

Complex numbers with |z| = 1 form the unit circle: they are the trigonometric circle.

See proof
04 · Conjugate

The conjugate z̄

The conjugate of z = a + bi is z̄ = a − bi: flip the sign of the imaginary part. On the plane it's the reflection of z across the real axis (like a mirror).

The conjugate of the conjugate

z‾‾=z\overline{\overline{z}} = z

Conjugating twice leaves the number as it was.

See proof

Conjugate of a sum

z+w‾=z‾+w‾\overline{z + w} = \overline{z} + \overline{w}

Adding then conjugating is the same as conjugating then adding.

See proof

Conjugate of a product

z⋅w‾=z‾⋅w‾⇒zn‾=z‾ n\overline{z \cdot w} = \overline{z} \cdot \overline{w} \quad\Rightarrow\quad \overline{z^n} = \overline{z}^{\,n}

The conjugate of a product is the product of the conjugates. In particular, conj(zⁿ) = z̄ⁿ.

See proof
05 · Modulus

The modulus |z|

The modulus |z| = √(a² + b²) is the length of the arrow, that is, the distance from z to the origin. It comes from Pythagoras' theorem with legs a and b.

The modulus is ≥ 0 and only 0 at z = 0

∣z∣=a2+b2≥0|z| = \sqrt{a^2 + b^2} \ge 0

|z| is a distance: never negative, and 0 only at the origin.

See proof

Triangle inequality

∣z+w∣≤∣z∣+∣w∣|z + w| \le |z| + |w|

The size of a sum never exceeds the sum of the sizes.

z = 2 + i, w = −1 + i: |z + w| = 2.236 ≤ 2.236 + 1.414 = 3.65
See proof

Modulus of a product

∣zw∣=∣z∣⋅∣w∣⇒∣zn∣=∣z∣n|zw| = |z| \cdot |w| \quad\Rightarrow\quad |z^n| = |z|^n

When multiplying complex numbers, the moduli multiply: |zw| = |z||w|.

|zw| = 3.162 = 2.236 · 1.414
See proof

A number and its conjugate have the same size

∣z∣=∣z‾∣|z| = |\overline{z}|

|z| = |z̄|: reflecting doesn't change the distance to the origin.

See proof

z · z̄ = |z|²

z⋅z‾=∣z∣2z \cdot \overline{z} = |z|^2

A complex number times its conjugate is always a real number ≥ 0.

(2 + i)(2 − i) = 5 = |z|²
See proof

Division: z/w = z·w̄ / |w|²

zw=z w‾∣w∣2,1w=w‾∣w∣2(w≠0)\dfrac{z}{w} = \dfrac{z\,\overline{w}}{|w|^2},\qquad \dfrac{1}{w} = \dfrac{\overline{w}}{|w|^2}\quad (w \neq 0)

To divide, multiply top and bottom by the conjugate of the denominator.

z / w = −0.5 − 1.5i
See proof
06 · Polar form

Polar form ρθ

A point is also fixed by two pieces of data: how far it is from the origin and the angle it makes with the positive real axis. Like giving directions as “5 km, heading north-east”.

ρ (rho): modulusρ := |z|, the distance to the origin.
θ (theta): principal argumentθ := Arg(z) ∈ (−π, π], the angle measured from the positive real axis; counterclockwise is positive.
arg(z): every argumentarg(z) = {θ + 2πn : n ∈ ℤ}. Full turns (2π = 360°) lead to the same point.
Polar notationz = ρθ, where ρ = |z| and θ = Arg(z).
FormNotationz = 1 + iw = 1 − i
Rectangulara + bi1 + i1 − i
Polarρθ√2π/4√2−π/4
Trigonometricρ(cos θ + i sin θ)√2(cos π/4 + i sin π/4)√2(cos (−π/4) + i sin (−π/4))
Exponentialρeiθ√2eiπ/4√2e−iπ/4

From rectangular to polar (and the arctan trap)

ρ=a2+b2,θ=Arg⁡z∈(−π,π]\rho = \sqrt{a^2 + b^2},\qquad \theta = \operatorname{Arg} z \in (-\pi, \pi]

ρ = √(a² + b²) and θ = arctan(b/a), corrected by quadrant.

Arg(2 + i) ≈ 0.464 ≈ 26.6°
See proof

From polar to rectangular

a=ρcos⁡θ,b=ρsin⁡θ⇒z=ρ(cos⁡θ+isin⁡θ)a = \rho\operatorname{cos}\theta,\quad b = \rho\operatorname{sin}\theta \quad\Rightarrow\quad z = \rho(\operatorname{cos}\theta + i\operatorname{sin}\theta)

a = ρ cos θ and b = ρ sin θ. This gives the trigonometric form.

See proof
07 · Operations in polar form

Multiplying is rotating

In polar form, operations are very simple: when multiplying, the moduli multiply and the angles add. That's why multiplying by a complex number rotates and stretches.

Product in polar form: multiply moduli, add angles

(ρ1)θ1⋅(ρ2)θ2=(ρ1ρ2)θ1+θ2(\rho_1)_{\theta_1} \cdot (\rho_2)_{\theta_2} = (\rho_1\rho_2)_{\theta_1 + \theta_2}

(ρ₁)θ₁ · (ρ₂)θ₂ = (ρ₁ρ₂)θ₁+θ₂

√50.464 · √23π/4 = √102.82
See proof

Quotient in polar form: divide moduli, subtract angles

(ρ1)θ1(ρ2)θ2=(ρ1ρ2)θ1−θ2\dfrac{(\rho_1)_{\theta_1}}{(\rho_2)_{\theta_2}} = \left(\dfrac{\rho_1}{\rho_2}\right)_{\theta_1 - \theta_2}

(ρ₁)θ₁ / (ρ₂)θ₂ = (ρ₁/ρ₂)θ₁−θ₂

See proof

Conjugate and inverse in polar form

ρθ‾=ρ−θ,(ρθ)−1=(1ρ)−θ\overline{\rho_\theta} = \rho_{-\theta},\qquad (\rho_\theta)^{-1} = \left(\dfrac{1}{\rho}\right)_{-\theta}

z̄ = ρ−θ and z⁻¹ = (1/ρ)−θ

See proof

Multiplying by 1β rotates by β

(x+yi)⋅1β  =  rotate β(x + yi) \cdot 1_\beta \;=\; \text{rotate}\ \beta

Multiplying by cos β + i sin β rotates the whole plane by β around the origin.

See proof
08 · De Moivre

De Moivre's theorem

Raising to a power means multiplying the same number several times, so the angle gets added to itself n times.

De Moivre's theorem

(cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ(\operatorname{cos}\theta + i\operatorname{sin}\theta)^n = \operatorname{cos} n\theta + i\operatorname{sin} n\theta

(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. Raising to n multiplies the angle by n.

z³ = (2 + i)³ = 2 + 11i
See proof
09 · Exponential form

Euler's formula: eiθ

With the exponential form, the rules of complex numbers become the rules of powers you already know.

What does a power mean? 2³ = 2 · 2 · 2 makes sense by counting. But π² is no longer “π times”, and 2π even less: they are definitions that extend the idea while keeping its rules. eiθ is the same: it’s defined so that eaeb = ea+b still holds.

Euler's formula

eiθ≔cos⁡θ+isin⁡θ=1θe^{i\theta} \coloneqq \operatorname{cos}\theta + i\operatorname{sin}\theta = 1_\theta

eiθ := cos θ + i sin θ. It's the exponential form of a phasor.

See proof

Sine and cosine as exponentials

cos⁡θ=eiθ+e−iθ2,sin⁡θ=eiθ−e−iθ2i\operatorname{cos}\theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2},\qquad \operatorname{sin}\theta = \dfrac{e^{i\theta} - e^{-i\theta}}{2i}

Adding and subtracting eiθ and e−iθ isolates cos θ and sin θ.

See proof

Properties of eiθ

(eiθ)n=einθ,ei(θ+ω)=eiθeiω(e^{i\theta})^n = e^{in\theta},\qquad e^{i(\theta + \omega)} = e^{i\theta}e^{i\omega}

It behaves like an ordinary power: exponents add and multiply.

See proof
10 · Roots

n-th roots

In ℂ, every nonzero number has exactly n distinct n-th roots, spread out like the vertices of a regular polygon.

n-th roots

ρθn={(ρn)θ+2πkn:k=0,…,n−1}\sqrt[n]{\rho_\theta} = \left\{ \left(\sqrt[n]{\rho}\right)_{\frac{\theta + 2\pi k}{n}} : k = 0, \ldots, n - 1 \right\}

Every z ≠ 0 has n n-th roots, with modulus ⁿ√ρ and angles (θ + 2πk)/n.

³√(2 + i): ≈ 1.292 + 0.201i · ≈ −0.82 + 1.018i · ≈ −0.472 − 1.22i
See proof

Example: the cube roots of 1

13={1, −1±3 i2}\sqrt[3]{1} = \left\{ 1,\ \dfrac{-1 \pm \sqrt{3}\,i}{2} \right\}

∛1 = {1, ω, ω²} with ω = −1/2 + (√3/2)i.

See proof

Example: z³ − 8 = 0

z3−8=0⇒z∈{2, −1±3 i}z^3 - 8 = 0 \quad\Rightarrow\quad z \in \{ 2,\ -1 \pm \sqrt{3}\,i \}

The solutions are 2 and −1 ± √3 i: the cube roots of 1 times 2.

See proof
11 · Exponential and logarithm

ez and log z in ℂ

The exponential extends to any complex exponent; its inverse, the logarithm, exists even for negative numbers, but it has infinitely many values.

Complex exponential ez

ez≔eRe⁡z eiIm⁡ze^z \coloneqq e^{\operatorname{Re} z}\, e^{i \operatorname{Im} z}

ea+bi = ea (cos b + i sin b): modulus ea and angle b.

e^(2 + i) = ≈ 3.992 + 6.218i, |e^z| = e^2 = 7.389
See proof

Complex logarithm

log⁡z=ln⁡∣z∣+iarg⁡z,Log⁡z=ln⁡∣z∣+iArg⁡z\log z = \ln|z| + i\arg z,\qquad \operatorname{Log} z = \ln|z| + i\operatorname{Arg} z

log z = ln|z| + i arg z. It has infinitely many values, 2πi apart.

Log(2 + i) = ≈ 0.805 + 0.464i
See proof
12 · Applications

What are complex numbers for?

ElectricityIn AC circuits, voltages and currents are phasors. The impedance Z = R + iX packs resistance and reactance into a single number.
Signals and soundThe Fourier transform breaks any signal into waves eiωt. That's how MP3, JPEG and 5G work.
Graphics and gamesRotating a point by θ is multiplying it by eiθ. Quaternions, their 3D version, rotate cameras and characters.
EquationsFundamental theorem of algebra: every polynomial of degree n has exactly n roots in ℂ (counting repeats).
Quantum physicsA particle's wave function is complex; probabilities come from its modulus squared |ψ|².
FractalsThe Mandelbrot set comes from repeating z → z² + c with complex numbers and checking whether the result escapes.
Proof

Axiomatic definition: ℂ = (ℝ², +, ·)

(a,b)⋅(c,d)=(ac−bd, ad+bc)(a, b) \cdot (c, d) = (ac - bd,\ ad + bc)
  1. C=R2={(a,b):a,b∈R}\mathbb{C} = \mathbb{R}^2 = \{(a, b) : a, b \in \mathbb{R}\}

    A complex number is an ordered pair of real numbers (a, b). The first is the real part and the second the imaginary part.

  2. (a,b)+(c,d)=(a+c, b+d)(a, b) + (c, d) = (a + c,\ b + d)

    Addition is done component by component, just like with vectors.

  3. (a+bi)(c+di)=ac+adi+bci+bd⋅i2(a + bi)(c + di) = ac + adi + bci + bd \cdot i^2

    To guess what the product should be, we multiply (a + bi)(c + di) like two binomials: every term by every term (distributive property).

  4. (a+bi)(c+di)=ac+bd⋅(−1)+(ad+bc)i=(ac−bd)+(ad+bc)i(a + bi)(c + di) = ac + \hl{bd \cdot (-1)} + (ad + bc)i = (ac - bd) + (ad + bc)i

    We use i² = −1, so bd·i² = −bd. Grouping: what has no i is the real part (ac − bd); what has i, the imaginary part (ad + bc).

  5. (a,b)⋅(c,d)≔(ac−bd, ad+bc)(a, b) \cdot (c, d) \coloneqq (ac - bd,\ ad + bc)

    That rule is exactly what's taken as the definition of the product of pairs. The symbol := means “is defined as”.

  6. i=(0,1)  →  i2=(0⋅0−1⋅1, 0⋅1+1⋅0)=(−1,0)=−1i = (0, 1) \;\to\; i^2 = (0 \cdot 0 - 1 \cdot 1,\ 0 \cdot 1 + 1 \cdot 0) = (-1, 0) = -1

    With this definition, i = (0, 1), and multiplying it by itself gives (−1, 0), which is the real number −1. So i² = −1 is no longer an “impossible number” but a consequence.

What it's for

Complex numbers are built from real numbers alone, no magic: they are points of the plane with a special way of multiplying.

Proof

Powers of i

i0=1,i1=i,i2=−1,i3=−i,i4=1i^0 = 1,\quad i^1 = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1
  1. i2=−1i^2 = -1

    Start from the definition: i squared is −1.

  2. i3=i2⋅i=−ii^3 = i^2 \cdot i = -i

    For i³ multiply by i once more: (−1)·i = −i.

  3. i4=i2⋅i2=(−1)(−1)=1i^4 = i^2 \cdot i^2 = (-1)(-1) = 1

    i⁴ = i²·i² = (−1)(−1) = 1. We're back to the start.

  4. in=ir,r=remainder(n÷4)i^n = i^r,\quad r = \text{remainder}(n \div 4)

    Since every four powers we're back to 1, to compute iⁿ we only need the remainder r of dividing n by 4 (r = 0, 1, 2 or 3).

  5. i27=i4⋅6+3=(i4)6⋅i3=−ii^{27} = i^{4 \cdot 6 + 3} = (i^4)^6 \cdot i^3 = -i

    Example: 27 = 4·6 + 3, remainder 3. Since i⁴ = 1, (i⁴)⁶ = 1, leaving i³ = −i.

What it's for

Multiplying by i rotates 90° counterclockwise; four 90° turns make a full turn. That's why the cycle has length 4.

Proof

ℂ is a field

z≠0  ⇒  z−1=a−bia2+b2z \neq 0 \;\Rightarrow\; z^{-1} = \dfrac{a - bi}{a^2 + b^2}
  1. z+w=w+z,zw=wzz + w = w + z,\quad zw = wz

    Addition and multiplication are commutative (order doesn't matter) and associative (grouping doesn't matter). They inherit this from ℝ.

  2. z(w+v)=zw+zvz(w + v) = zw + zv

    The distributive property holds: multiplying by a sum is adding the products.

  3. 0=0+0i,1=1+0i0 = 0 + 0i,\quad 1 = 1 + 0i

    There are identity elements: 0 = 0 + 0i changes nothing when added, and 1 = 1 + 0i changes nothing when multiplied.

  4. −z=−a−bi-z = -a - bi

    Every z has an opposite −z, which added to z gives 0.

  5. 1a+bi=1a+bi⋅a−bia−bi=a−bia2+b2\dfrac{1}{a + bi} = \dfrac{1}{a + bi} \cdot \dfrac{\hl{a - bi}}{\hl{a - bi}} = \dfrac{a - bi}{a^2 + b^2}

    Inverse: to make the denominator real, multiply top and bottom by a − bi (the conjugate). The bottom becomes a² + b², which is positive if z ≠ 0.

  6. (1+i)−1=1−i2=12−12i(1 + i)^{-1} = \dfrac{1 - i}{2} = \dfrac{1}{2} - \dfrac{1}{2}i

    Example: the inverse of 1 + i is (1 − i)/2. Check: (1 + i)(1 − i)/2 = (1 − i²)/2 = 2/2 = 1.

What it's for

ℂ being a field means you can solve equations with complex numbers exactly like with real ones.

Proof

Is 2 < i? No: ℂ can't be ordered

2<i ?No2 < i\,?\quad \text{No}
  1. x>0, y>0  ⇒  xy>0x > 0,\ y > 0 \;\Rightarrow\; xy > 0

    A “compatible” order (like the one on ℝ) must make the product of two positives positive.

  2. x≠0  ⇒  x2>0x \neq 0 \;\Rightarrow\; x^2 > 0

    Consequence: the square of any nonzero number is positive (if x < 0, then −x > 0 and x² = (−x)² > 0).

  3. i≠0  ⇒  i2>0i \neq 0 \;\Rightarrow\; i^2 > 0

    Suppose such an order exists on ℂ. Since i ≠ 0, we'd need i² > 0.

  4. i2=−1  ⇒  −1>0i^2 = -1 \;\Rightarrow\; -1 > 0

    But i² = −1, so we'd get −1 > 0.

  5. 1=12>01 = 1^2 > 0

    On the other hand, 1 = 1² is also a square, so 1 > 0.

  6. 0=(−1)+1>0(contradiction)0 = (-1) + 1 > 0\quad (\text{contradiction})

    Adding two positives (−1 and 1) must give a positive number: 0 > 0. That's impossible, so no such order exists.

What it's for

In ℂ you never say one number is bigger than another. What you can compare is their moduli |z| and |w|, which are real numbers.

Proof

Phasor: a complex number of modulus 1

∣z∣=1  ⇔  a2+b2=1  ⇔  z=cos⁡θ+isin⁡θ|z| = 1 \;\Leftrightarrow\; a^2 + b^2 = 1 \;\Leftrightarrow\; z = \operatorname{cos}\theta + i\operatorname{sin}\theta
  1. z=a+bi,∣z∣=a2+b2z = a + bi,\quad |z| = \sqrt{a^2 + b^2}

    A phasor is a complex z with |z| = 1. Recall that |z| = √(a² + b²).

  2. ∣z∣=1  ⇔  a2+b2=1|z| = 1 \;\Leftrightarrow\; a^2 + b^2 = 1

    Squaring, |z| = 1 means a² + b² = 1, the equation of the circle of radius 1 centred at the origin.

  3. a=cos⁡θ,b=sin⁡θa = \operatorname{cos}\theta,\quad b = \operatorname{sin}\theta

    Every point on that circle has coordinates (cos θ, sin θ), as on the trigonometric circle.

  4. z=cos⁡θ+isin⁡θ=1θ=eiθz = \operatorname{cos}\theta + i\operatorname{sin}\theta = 1_\theta = e^{i\theta}

    So every phasor can be written z = cos θ + i sin θ, which is 1θ in polar form and eiθ in exponential form.

What it's for

Multiplying by a phasor rotates without changing size. In electricity they represent sinusoidal waves.

Proof

The conjugate of the conjugate

z‾‾=z\overline{\overline{z}} = z
  1. z=a+bi  ⇒  z‾=a−biz = a + bi \;\Rightarrow\; \overline{z} = a - bi

    Conjugating flips the sign of the imaginary part: a + bi becomes a − bi.

  2. z‾‾=a+(−b)i‾=a−(−b)i\overline{\overline{z}} = \overline{a + (-b)i} = a - (-b)i

    Conjugate again: the imaginary part is now −b, and we flip its sign once more.

  3. z‾‾=a+bi=z\overline{\overline{z}} = a + bi = z

    −(−b) = b, so we get back a + bi = z.

What it's for

Geometrically, conjugating is reflecting across the real axis; reflecting twice leaves the point where it was.

Proof

Conjugate of a sum

z+w‾=z‾+w‾\overline{z + w} = \overline{z} + \overline{w}
  1. z=a+bi,w=c+diz = a + bi,\quad w = c + di

    Write both numbers in rectangular form.

  2. z+w=(a+c)+(b+d)iz + w = (a + c) + (b + d)i

    Add real parts to real parts and imaginary parts to imaginary parts.

  3. z+w‾=(a+c)−(b+d)i\overline{z + w} = (a + c) - (b + d)i

    Conjugate the result: flip the sign of the imaginary part.

  4. z+w‾=(a−bi)+(c−di)=z‾+w‾\overline{z + w} = (a - bi) + (c - di) = \overline{z} + \overline{w}

    Split it up: it's the sum of a − bi and c − di, that is, z̄ + w̄.

What it's for

It lets you conjugate long expressions term by term, for example when simplifying fractions.

Proof

Conjugate of a product

z⋅w‾=z‾⋅w‾⇒zn‾=z‾ n\overline{z \cdot w} = \overline{z} \cdot \overline{w} \quad\Rightarrow\quad \overline{z^n} = \overline{z}^{\,n}
  1. zw=(ac−bd)+(ad+bc)izw = (ac - bd) + (ad + bc)i

    Product in rectangular form (definition of the product).

  2. zw‾=(ac−bd)−(ad+bc)i\overline{zw} = (ac - bd) - (ad + bc)i

    Conjugate it: only the sign of the imaginary part changes.

  3. z‾⋅w‾=(a−bi)(c−di)=ac−adi−bci+bd⋅i2\overline{z} \cdot \overline{w} = (a - bi)(c - di) = ac - adi - bci + bd \cdot i^2

    Now multiply the conjugates (a − bi)(c − di) term by term.

  4. z‾⋅w‾=(ac−bd)−(ad+bc)i\overline{z} \cdot \overline{w} = (ac - bd) - (ad + bc)i

    With i² = −1 we get (ac − bd) − (ad + bc)i: the same as in step 2.

  5. zn‾=z⋅z⋯z‾=z‾⋅z‾⋯z‾=z‾ n\overline{z^n} = \overline{z \cdot z \cdots z} = \overline{z} \cdot \overline{z} \cdots \overline{z} = \overline{z}^{\,n}

    A power is a repeated product, so applying the rule n times: conj(zⁿ) = (z̄)ⁿ.

What it's for

If a polynomial with real coefficients has a complex root z, it also has the root z̄: complex roots come in pairs.

Proof

The modulus is ≥ 0 and only 0 at z = 0

∣z∣=a2+b2≥0|z| = \sqrt{a^2 + b^2} \ge 0
  1. ∣z∣=Re⁡(z)2+Im⁡(z)2|z| = \sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}

    The modulus is the distance from (a, b) to the origin: by Pythagoras, the hypotenuse with legs a and b.

  2. a2≥0, b2≥0  ⇒  a2+b2≥0  ⇒  ∣z∣≥0a^2 \ge 0,\ b^2 \ge 0 \;\Rightarrow\; a^2 + b^2 \ge 0 \;\Rightarrow\; |z| \ge 0

    a² and b² are squares, so they aren't negative; neither is their sum, nor its square root.

  3. ∣z∣=0  ⇔  a2+b2=0  ⇔  a=b=0  ⇔  z=0|z| = 0 \;\Leftrightarrow\; a^2 + b^2 = 0 \;\Leftrightarrow\; a = b = 0 \;\Leftrightarrow\; z = 0

    A sum of squares is 0 only if both squares are 0, that is, a = 0 and b = 0.

  4. ∣3+4i∣=9+16=25=5|3 + 4i| = \sqrt{9 + 16} = \sqrt{25} = 5

    Example: |3 + 4i| = √(9 + 16) = √25 = 5 (the 3-4-5 triangle).

What it's for

The modulus measures the size of a complex number. It's what gets compared instead of the numbers themselves.

Proof

Triangle inequality

∣z+w∣≤∣z∣+∣w∣|z + w| \le |z| + |w|
  1. z+wz + w

    z + w is the diagonal of the parallelogram formed by the arrows z and w (parallelogram rule).

  2. 0  →  z  →  z+w0 \;\to\; z \;\to\; z + w

    To get from 0 to z + w we can go through z: first the arrow z (length |z|), then a copy of w (length |w|). That makes a triangle.

  3. ∣z+w∣≤∣z∣+∣w∣|z + w| \le |z| + |w|

    In a triangle, one side is never longer than the other two combined: the direct path is the shortest.

  4. ∣z+w∣=∣z∣+∣w∣  ⇔  θz=θw|z + w| = |z| + |w| \;\Leftrightarrow\; \theta_z = \theta_w

    Equality only happens when z and w point the same way: the triangle flattens into a segment.

  5. ∣3+4i∣=5  ≤  ∣3∣+∣4i∣=7|3 + 4i| = 5 \;\le\; |3| + |4i| = 7

    Example: |3 + 4i| = 5, while |3| + |4i| = 3 + 4 = 7.

What it's for

It gives bounds: if you know how big z and w are, you know at most how big z + w is.

Proof

Modulus of a product

∣zw∣=∣z∣⋅∣w∣⇒∣zn∣=∣z∣n|zw| = |z| \cdot |w| \quad\Rightarrow\quad |z^n| = |z|^n
  1. ∣zw∣2=(ac−bd)2+(ad+bc)2|zw|^2 = (ac - bd)^2 + (ad + bc)^2

    Apply the definition of modulus to the product zw = (ac − bd) + (ad + bc)i.

  2. ∣zw∣2=a2c2−2abcd+b2d2+a2d2+2abcd+b2c2|zw|^2 = a^2c^2 \hl{{}- 2abcd} + b^2d^2 + a^2d^2 \hl{{}+ 2abcd} + b^2c^2

    Expand both squares. The terms −2abcd and +2abcd cancel.

  3. ∣zw∣2=a2(c2+d2)+b2(c2+d2)=(a2+b2)(c2+d2)|zw|^2 = a^2(c^2 + d^2) + b^2(c^2 + d^2) = (a^2 + b^2)(c^2 + d^2)

    Factor: a² multiplies (c² + d²) and so does b².

  4. ∣zw∣2=∣z∣2⋅∣w∣2⇒∣zw∣=∣z∣⋅∣w∣|zw|^2 = |z|^2 \cdot |w|^2 \quad\Rightarrow\quad |zw| = |z| \cdot |w|

    That's |z|²·|w|². Since moduli are ≥ 0, we can drop the squares.

  5. ∣zn∣=∣z⋅z⋯z∣=∣z∣n|z^n| = |z \cdot z \cdots z| = |z|^n

    A power is a repeated product, so |zⁿ| = |z|ⁿ.

What it's for

It's half of the polar product rule: sizes multiply (the other half: angles add).

Proof

A number and its conjugate have the same size

∣z∣=∣z‾∣|z| = |\overline{z}|
  1. z‾=a−bi\overline{z} = a - bi

    The conjugate of a + bi is a − bi.

  2. ∣z‾∣=a2+(−b)2=a2+b2=∣z∣|\overline{z}| = \sqrt{a^2 + (-b)^2} = \sqrt{a^2 + b^2} = |z|

    When squaring, (−b)² = b², so the modulus is the same.

What it's for

z and z̄ are at the same distance from the origin, with opposite angles θ and −θ.

Proof

z · z̄ = |z|²

z⋅z‾=∣z∣2z \cdot \overline{z} = |z|^2
  1. z⋅z‾=(a+bi)(a−bi)z \cdot \overline{z} = (a + bi)(a - bi)

    Multiply z = a + bi by its conjugate a − bi.

  2. z⋅z‾=a2−abi+abi−b2i2z \cdot \overline{z} = a^2 \hl{{}- abi + abi} - b^2i^2

    Expand: the terms −abi and +abi cancel.

  3. z⋅z‾=a2−b2(−1)=a2+b2z \cdot \overline{z} = a^2 - b^2(-1) = a^2 + b^2

    Since i² = −1, −b²i² = +b².

  4. z⋅z‾=∣z∣2∈R, ≥0z \cdot \overline{z} = |z|^2 \in \mathbb{R},\ \ge 0

    We're left with a² + b², the modulus squared: a real, non-negative number.

What it's for

It's the trick for dividing: multiplying by the conjugate turns a complex number into a real one.

Proof

Division: z/w = z·w̄ / |w|²

zw=z w‾∣w∣2,1w=w‾∣w∣2(w≠0)\dfrac{z}{w} = \dfrac{z\,\overline{w}}{|w|^2},\qquad \dfrac{1}{w} = \dfrac{\overline{w}}{|w|^2}\quad (w \neq 0)
  1. zw=zw⋅w‾w‾\dfrac{z}{w} = \dfrac{z}{w} \cdot \dfrac{\hl{\overline{w}}}{\hl{\overline{w}}}

    Multiply numerator and denominator by w̄. That's multiplying by w̄/w̄ = 1, so the value doesn't change.

  2. zw=z w‾w w‾=z w‾∣w∣2\dfrac{z}{w} = \dfrac{z\,\overline{w}}{w\,\overline{w}} = \dfrac{z\,\overline{w}}{|w|^2}

    The bottom becomes w·w̄ = |w|², a real number. Dividing by a real number is easy: divide each part.

  3. z=1  ⇒  1w=w‾∣w∣2z = 1 \;\Rightarrow\; \dfrac{1}{w} = \dfrac{\overline{w}}{|w|^2}

    Special case z = 1: the inverse of w is w̄/|w|².

  4. 2+3i1+2i=(2+3i)(1−2i)12+22\dfrac{2 + 3i}{1 + 2i} = \dfrac{(2 + 3i)(1 - 2i)}{1^2 + 2^2}

    Example: (2 + 3i)/(1 + 2i). The conjugate of 1 + 2i is 1 − 2i and |1 + 2i|² = 1² + 2² = 5.

  5. 2+3i1+2i=2−4i+3i−6i25=8−i5=85−15i\dfrac{2 + 3i}{1 + 2i} = \dfrac{2 - 4i + 3i - 6i^2}{5} = \dfrac{8 - i}{5} = \dfrac{8}{5} - \dfrac{1}{5}i

    Expand: 2 − 4i + 3i − 6i² = 2 + 6 − i = 8 − i. Divide each part by 5.

What it's for

That's how you divide in rectangular form. In polar form it's even easier: divide the moduli and subtract the angles.

Proof

From rectangular to polar (and the arctan trap)

ρ=a2+b2,θ=Arg⁡z∈(−π,π]\rho = \sqrt{a^2 + b^2},\qquad \theta = \operatorname{Arg} z \in (-\pi, \pi]
  1. cos⁡θ=aρ,sin⁡θ=bρ\operatorname{cos}\theta = \dfrac{a}{\rho},\quad \operatorname{sin}\theta = \dfrac{b}{\rho}

    In the triangle with legs a (horizontal) and b (vertical) and hypotenuse ρ, sine is opposite/hypotenuse and cosine is adjacent/hypotenuse.

  2. tan⁡θ=ba  ⇒  θ=arctan⁡ba\operatorname{tan}\theta = \dfrac{b}{a} \;\Rightarrow\; \theta = \arctan\dfrac{b}{a}

    Dividing, tan θ = b/a, so θ = arctan(b/a)… but only sometimes.

  3. a>0:θ=arctan⁡baa > 0:\quad \theta = \arctan\dfrac{b}{a}

    arctan only returns angles between −π/2 and π/2 (quadrants I and IV), which are the ones with a > 0. There the formula is right.

  4. a<0, b≥0: θ=arctan⁡ba+π;a<0, b<0: θ=arctan⁡ba−πa < 0,\ b \ge 0:\ \theta = \arctan\dfrac{b}{a} + \pi;\qquad a < 0,\ b < 0:\ \theta = \arctan\dfrac{b}{a} - \pi

    If a < 0 the point is on the left: b/a gives the same value for z and for −z (the opposite point), so we must correct by adding π (if b ≥ 0) or subtracting π (if b < 0).

  5. a=0:θ=±π2a = 0:\quad \theta = \pm\dfrac{\pi}{2}

    If a = 0, b/a doesn't exist: the point is on the imaginary axis, θ = π/2 if b > 0 and −π/2 if b < 0.

  6. arg⁡z={θ+2πk:k∈Z}\arg z = \{\theta + 2\pi k : k \in \mathbb{Z}\}

    arg (lowercase) is the set of all valid angles; Arg (capital) is the principal one, the only one in (−π, π].

What it's for

Careful with calculators: arctan((−1)/(−1)) = π/4, but −1 − i is in the third quadrant and its Arg is −3π/4.

Proof

From polar to rectangular

a=ρcos⁡θ,b=ρsin⁡θ⇒z=ρ(cos⁡θ+isin⁡θ)a = \rho\operatorname{cos}\theta,\quad b = \rho\operatorname{sin}\theta \quad\Rightarrow\quad z = \rho(\operatorname{cos}\theta + i\operatorname{sin}\theta)
  1. cos⁡θ=aρ  ⇒  a=ρcos⁡θ\operatorname{cos}\theta = \dfrac{a}{\rho} \;\Rightarrow\; a = \rho\operatorname{cos}\theta

    From cos θ = a/ρ, solve for a by multiplying by ρ.

  2. sin⁡θ=bρ  ⇒  b=ρsin⁡θ\operatorname{sin}\theta = \dfrac{b}{\rho} \;\Rightarrow\; b = \rho\operatorname{sin}\theta

    From sin θ = b/ρ, solve for b the same way.

  3. z=a+bi=ρcos⁡θ+iρsin⁡θ=ρ(cos⁡θ+isin⁡θ)z = a + bi = \rho\operatorname{cos}\theta + i\rho\operatorname{sin}\theta = \rho(\operatorname{cos}\theta + i\operatorname{sin}\theta)

    Substitute into a + bi and factor out ρ: that's the trigonometric form ρ(cos θ + i sin θ).

  4. 2π/3=2(12+32i)=1+3 i2_{\pi/3} = 2\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right) = 1 + \sqrt{3}\,i

    Example: 2π/3. Since cos(π/3) = 1/2 and sin(π/3) = √3/2, we get 1 + √3 i.

What it's for

Rectangular form is best for adding and subtracting; polar form for multiplying, dividing, powers and roots.

Proof

Product in polar form: multiply moduli, add angles

(ρ1)θ1⋅(ρ2)θ2=(ρ1ρ2)θ1+θ2(\rho_1)_{\theta_1} \cdot (\rho_2)_{\theta_2} = (\rho_1\rho_2)_{\theta_1 + \theta_2}
  1. z=ρ1(cos⁡α+isin⁡α),w=ρ2(cos⁡β+isin⁡β)z = \rho_1(\operatorname{cos}\alpha + i\operatorname{sin}\alpha),\quad w = \rho_2(\operatorname{cos}\beta + i\operatorname{sin}\beta)

    Write both numbers in trigonometric form, with angles α and β.

  2. zw=ρ1ρ2(cos⁡α+isin⁡α)(cos⁡β+isin⁡β)zw = \rho_1\rho_2(\operatorname{cos}\alpha + i\operatorname{sin}\alpha)(\operatorname{cos}\beta + i\operatorname{sin}\beta)

    The moduli ρ₁ and ρ₂ are real numbers: move them to the front.

  3. zw=ρ1ρ2[(cos⁡α cos⁡β−sin⁡α sin⁡β)+i(sin⁡α cos⁡β+cos⁡α sin⁡β)]zw = \rho_1\rho_2[(\operatorname{cos}\alpha\,\operatorname{cos}\beta - \operatorname{sin}\alpha\,\operatorname{sin}\beta) + i(\operatorname{sin}\alpha\,\operatorname{cos}\beta + \operatorname{cos}\alpha\,\operatorname{sin}\beta)]

    Multiply the two brackets like binomials, with i² = −1, and group real and imaginary parts.

  4. zw=ρ1ρ2[cos⁡(α+β)+i sin⁡(α+β)]zw = \rho_1\rho_2[\hl{\operatorname{cos}(\alpha + \beta)} + i\,\hl{\operatorname{sin}(\alpha + \beta)}]

    They are exactly the angle-sum formulas for cosine and sine! (the ones in the α ± β tab).

  5. zw=(ρ1ρ2)α+βzw = (\rho_1\rho_2)_{\alpha + \beta}

    The result is a complex number with modulus ρ₁ρ₂ and angle α + β.

  6. α+β∉(−π,π]  ⇒  ±2π\alpha + \beta \notin (-\pi, \pi] \;\Rightarrow\; \pm 2\pi

    If α + β falls outside (−π, π], add or subtract 2π to get the principal argument: it's the same point.

What it's for

Multiplying = multiplying sizes and adding turns. That's why complex numbers are perfect for describing rotations.

Proof

Quotient in polar form: divide moduli, subtract angles

(ρ1)θ1(ρ2)θ2=(ρ1ρ2)θ1−θ2\dfrac{(\rho_1)_{\theta_1}}{(\rho_2)_{\theta_2}} = \left(\dfrac{\rho_1}{\rho_2}\right)_{\theta_1 - \theta_2}
  1. q=zw  ⇔  q⋅w=zq = \dfrac{z}{w} \;\Leftrightarrow\; q \cdot w = z

    Dividing means finding the number q that, multiplied by w, gives z.

  2. q=rφ  ⇒  rφ⋅(ρ2)θ2=(rρ2)φ+θ2q = r_\varphi \;\Rightarrow\; r_\varphi \cdot (\rho_2)_{\theta_2} = (r\rho_2)_{\varphi + \theta_2}

    Write q = rφ and use the product rule: moduli multiply, angles add.

  3. (rρ2)φ+θ2=(ρ1)θ1  ⇒  rρ2=ρ1,φ+θ2=θ1(r\rho_2)_{\varphi + \theta_2} = (\rho_1)_{\theta_1} \;\Rightarrow\; r\rho_2 = \rho_1,\quad \varphi + \theta_2 = \theta_1

    Two complex numbers in polar form are equal if they have the same modulus and the same angle (up to full turns of 2π).

  4. r=ρ1ρ2,φ=θ1−θ2r = \dfrac{\rho_1}{\rho_2},\qquad \varphi = \theta_1 - \theta_2

    Solving: r = ρ₁/ρ₂ and φ = θ₁ − θ₂.

What it's for

Dividing undoes the turn: sizes divide and angles subtract.

Proof

Conjugate and inverse in polar form

ρθ‾=ρ−θ,(ρθ)−1=(1ρ)−θ\overline{\rho_\theta} = \rho_{-\theta},\qquad (\rho_\theta)^{-1} = \left(\dfrac{1}{\rho}\right)_{-\theta}
  1. ρθ=ρcos⁡θ+iρsin⁡θ\rho_\theta = \rho\operatorname{cos}\theta + i\rho\operatorname{sin}\theta

    Write ρθ in trigonometric form.

  2. ρθ‾=ρcos⁡θ−iρsin⁡θ=ρcos⁡(−θ)+iρsin⁡(−θ)\overline{\rho_\theta} = \rho\operatorname{cos}\theta - i\rho\operatorname{sin}\theta = \rho\operatorname{cos}(-\theta) + i\rho\operatorname{sin}(-\theta)

    Conjugating flips the sign of the imaginary part. Since cos(−θ) = cos θ and sin(−θ) = −sin θ, that's the same as replacing θ with −θ.

  3. ρθ‾=ρ−θ\overline{\rho_\theta} = \rho_{-\theta}

    Same modulus, opposite angle: ρ−θ.

  4. 1ρθ=10ρθ=(1ρ)0−θ=(1ρ)−θ\dfrac{1}{\rho_\theta} = \dfrac{1_0}{\rho_\theta} = \left(\dfrac{1}{\rho}\right)_{0 - \theta} = \left(\dfrac{1}{\rho}\right)_{-\theta}

    For the inverse, write 1 = 1₀ (modulus 1, angle 0) and apply the quotient rule.

What it's for

The inverse points the same way as the conjugate, but with size 1/ρ: if z is far from the origin, 1/z is close to it.

Proof

De Moivre's theorem

(cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ(\operatorname{cos}\theta + i\operatorname{sin}\theta)^n = \operatorname{cos} n\theta + i\operatorname{sin} n\theta
  1. (ρθ)n=ρθ⋅ρθ⋯ρθ(\rho_\theta)^n = \rho_\theta \cdot \rho_\theta \cdots \rho_\theta

    Raising to n means multiplying the number by itself n times.

  2. (ρθ)n=(ρ⋅ρ⋯ρ)θ+θ+⋯+θ=(ρn)nθ(\rho_\theta)^n = (\rho \cdot \rho \cdots \rho)_{\theta + \theta + \cdots + \theta} = (\rho^n)_{n\theta}

    By the product rule, the moduli multiply (ρⁿ) and the angles add (θ + θ + … = nθ).

  3. ρ=1:(1θ)n=1nθ\rho = 1:\quad (1_\theta)^n = 1_{n\theta}

    If the modulus is 1, we get 1 to the n, which is 1: only the angle changes.

  4. (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ(\operatorname{cos}\theta + i\operatorname{sin}\theta)^n = \operatorname{cos} n\theta + i\operatorname{sin} n\theta

    Written in trigonometric form, that's De Moivre's theorem.

  5. n=2: cos⁡2θ−sin⁡2θ+2isin⁡θ cos⁡θ=cos⁡2θ+isin⁡2θn = 2:\ \operatorname{cos}^2\theta - \operatorname{sin}^2\theta + 2i\operatorname{sin}\theta\,\operatorname{cos}\theta = \operatorname{cos} 2\theta + i\operatorname{sin} 2\theta

    Example n = 2: expanding the square on the left and matching real and imaginary parts gives cos 2θ = cos²θ − sin²θ and sin 2θ = 2 sin θ cos θ.

  6. (1+i)8=(2π/4)8=162π=16(1 + i)^8 = \left(\sqrt{2}_{\pi/4}\right)^8 = 16_{2\pi} = 16

    Example: 1 + i = √2π/4. To the 8th: (√2)⁸ = 16 and 8·π/4 = 2π, a full turn: the result is 16.

What it's for

Compute large powers without expanding binomials, and derive multiple-angle formulas (cos 3θ, sin 4θ…).

Proof

Multiplying by 1β rotates by β

(x+yi)⋅1β  =  rotate β(x + yi) \cdot 1_\beta \;=\; \text{rotate}\ \beta
  1. 1⋅(cos⁡β+isin⁡β)=cos⁡β+isin⁡β1 \cdot (\operatorname{cos}\beta + i\operatorname{sin}\beta) = \operatorname{cos}\beta + i\operatorname{sin}\beta

    The point 1 = (1, 0) becomes cos β + i sin β = (cos β, sin β).

  2. i⋅(cos⁡β+isin⁡β)=−sin⁡β+icos⁡βi \cdot (\operatorname{cos}\beta + i\operatorname{sin}\beta) = -\operatorname{sin}\beta + i\operatorname{cos}\beta

    The point i = (0, 1) becomes i cos β + i² sin β = −sin β + i cos β = (−sin β, cos β).

  3. (1,0)→(cos⁡β,sin⁡β),(0,1)→(−sin⁡β,cos⁡β)(1, 0) \to (\operatorname{cos}\beta, \operatorname{sin}\beta),\qquad (0, 1) \to (-\operatorname{sin}\beta, \operatorname{cos}\beta)

    Both axes rotate by the same angle β, so the whole plane rotates by β.

  4. β=90°:(1,0)→(0,1),(0,1)→(−1,0)\beta = 90\degree:\quad (1, 0) \to (0, 1),\qquad (0, 1) \to (-1, 0)

    With β = 90°, multiplying by 190° = i takes the real axis to the imaginary one: a 90° turn.

  5. 130°⋅190°=1120°=−12+32i1_{30\degree} \cdot 1_{90\degree} = 1_{120\degree} = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i

    Example from your notes: rotating 30° and then 90° is rotating 120°, with sin 120° = √3/2 and cos 120° = −1/2.

What it's for

It's the basis of rotations in computer graphics, robotics and animation.

Proof

Euler's formula

eiθ≔cos⁡θ+isin⁡θ=1θe^{i\theta} \coloneqq \operatorname{cos}\theta + i\operatorname{sin}\theta = 1_\theta
  1. eiθ≔cos⁡θ+isin⁡θ(θ in rad)e^{i\theta} \coloneqq \operatorname{cos}\theta + i\operatorname{sin}\theta\quad (\theta\ \text{in}\ \text{rad})

    It's a definition: it extends powers of e to imaginary exponents. θ is always in radians.

  2. ∣eiθ∣=cos⁡2θ+sin⁡2θ=1=1|e^{i\theta}| = \sqrt{\operatorname{cos}^2\theta + \operatorname{sin}^2\theta} = \sqrt{1} = 1

    Its modulus is 1 by Pythagoras (sin²θ + cos²θ = 1): eiθ always lies on the unit circle.

  3. Arg⁡(eiθ)=θ(θ∈(−π,π])\operatorname{Arg}(e^{i\theta}) = \theta\quad (\theta \in (-\pi, \pi])

    Its argument is θ (if θ is in (−π, π]; otherwise θ plus or minus full turns of 2π).

  4. ρθ=ρ(cos⁡θ+isin⁡θ)=ρeiθ\rho_\theta = \rho(\operatorname{cos}\theta + i\operatorname{sin}\theta) = \rho e^{i\theta}

    Any complex number in polar form can be written ρ eiθ: the exponential form.

  5. eiπ=cos⁡π+isin⁡π=−1⇒eiπ+1=0e^{i\pi} = \operatorname{cos}\pi + i\operatorname{sin}\pi = -1 \quad\Rightarrow\quad e^{i\pi} + 1 = 0

    With θ = π we get eiπ = −1, Euler's identity, which puts e, i, π, 1 and 0 into a single formula.

What it's for

With exponential notation, working with complex numbers is working with powers: eiα·eiβ = ei(α+β).

Proof

Sine and cosine as exponentials

cos⁡θ=eiθ+e−iθ2,sin⁡θ=eiθ−e−iθ2i\operatorname{cos}\theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2},\qquad \operatorname{sin}\theta = \dfrac{e^{i\theta} - e^{-i\theta}}{2i}
  1. eiθ=cos⁡θ+isin⁡θe^{i\theta} = \operatorname{cos}\theta + i\operatorname{sin}\theta

    Euler's formula with θ.

  2. e−iθ=cos⁡(−θ)+isin⁡(−θ)=cos⁡θ−isin⁡θe^{-i\theta} = \operatorname{cos}(-\theta) + i\operatorname{sin}(-\theta) = \operatorname{cos}\theta - i\operatorname{sin}\theta

    Euler's formula with −θ. Since cosine is even and sine is odd, we get cos θ − i sin θ: the conjugate of eiθ.

  3. eiθ+e−iθ=2cos⁡θ+isin⁡θ−isin⁡θe^{i\theta} + e^{-i\theta} = 2\operatorname{cos}\theta \hl{{}+ i\operatorname{sin}\theta - i\operatorname{sin}\theta}

    Add them: the sines cancel, leaving 2 cos θ.

  4. cos⁡θ=eiθ+e−iθ2\operatorname{cos}\theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2}

    Divide by 2.

  5. eiθ−e−iθ=2isin⁡θ+cos⁡θ−cos⁡θe^{i\theta} - e^{-i\theta} = 2i\operatorname{sin}\theta \hl{{}+ \operatorname{cos}\theta - \operatorname{cos}\theta}

    Subtract them: the cosines cancel, leaving 2i sin θ.

  6. sin⁡θ=eiθ−e−iθ2i\operatorname{sin}\theta = \dfrac{e^{i\theta} - e^{-i\theta}}{2i}

    Divide by 2i.

What it's for

It turns products of sines and cosines into sums of exponentials, which are much easier to integrate and differentiate.

Proof

Properties of eiθ

(eiθ)n=einθ,ei(θ+ω)=eiθeiω(e^{i\theta})^n = e^{in\theta},\qquad e^{i(\theta + \omega)} = e^{i\theta}e^{i\omega}
  1. eiθ⋅eiω=1θ⋅1ω=1θ+ω=ei(θ+ω)e^{i\theta} \cdot e^{i\omega} = 1_\theta \cdot 1_\omega = 1_{\theta + \omega} = e^{i(\theta + \omega)}

    In polar form, eiθ = 1θ. Multiplying adds the angles: that's adding exponents.

  2. (eiθ)n=(1θ)n=1nθ=einθ(e^{i\theta})^n = (1_\theta)^n = 1_{n\theta} = e^{in\theta}

    Raising to n multiplies the angle by n (De Moivre): that's multiplying the exponent by n.

  3. (ρeiθ)(reiω)=ρr ei(θ+ω)(\rho e^{i\theta})(r e^{i\omega}) = \rho r\, e^{i(\theta + \omega)}

    With any moduli: the moduli multiply and the angles add.

What it's for

The exponent rules work just like with real numbers: ea·eb = ea+b and (ea)ⁿ = ena.

Proof

n-th roots

ρθn={(ρn)θ+2πkn:k=0,…,n−1}\sqrt[n]{\rho_\theta} = \left\{ \left(\sqrt[n]{\rho}\right)_{\frac{\theta + 2\pi k}{n}} : k = 0, \ldots, n - 1 \right\}
  1. w=rφ,wn=z=ρθw = r_\varphi,\quad w^n = z = \rho_\theta

    We look for every w = rφ with wⁿ = z.

  2. wn=(rn)nφw^n = (r^n)_{n\varphi}

    By De Moivre, wⁿ has modulus rⁿ and angle nφ.

  3. rn=ρ  ⇒  r=ρnr^n = \rho \;\Rightarrow\; r = \sqrt[n]{\rho}

    Match moduli: rⁿ = ρ, so r is the real (positive) n-th root of ρ.

  4. nφ=θ+2πk  ⇒  φ=θ+2πknn\varphi = \theta + \hl{2\pi k} \;\Rightarrow\; \varphi = \dfrac{\theta + 2\pi k}{n}

    Match angles, but angles may differ by full turns: nφ = θ + 2πk. Solve for φ.

  5. k=0,1,…,n−1k = 0, 1, \ldots, n - 1

    With k = n we're back at the angle for k = 0 (a full turn), so k = 0, 1, …, n − 1 is enough: there are n distinct roots.

  6. φk+1−φk=2πn\varphi_{k+1} - \varphi_k = \dfrac{2\pi}{n}

    Consecutive roots are 2π/n apart: they form a regular n-sided polygon centred at the origin.

What it's for

Solve equations like zⁿ = c. In ℝ, x³ = 8 has only one solution; in ℂ it has three.

Proof

Example: the cube roots of 1

13={1, −1±3 i2}\sqrt[3]{1} = \left\{ 1,\ \dfrac{-1 \pm \sqrt{3}\,i}{2} \right\}
  1. z=1:ρ=1, θ=0⇒z=10z = 1:\quad \rho = 1,\ \theta = 0 \quad\Rightarrow\quad z = 1_0

    Write 1 in polar form: modulus 1 and angle 0.

  2. 103={10+2πk3:k=0,1,2}\sqrt[3]{1_0} = \left\{ 1_{\frac{0 + 2\pi k}{3}} : k = 0, 1, 2 \right\}

    Apply the formula with n = 3. The real root of 1 is 1, so all of them have modulus 1.

  3. k=0:10=1k = 0:\quad 1_0 = 1

    k = 0: angle 0, the real root 1.

  4. k=1:12π/3=cos⁡2π3+isin⁡2π3=−12+32i=ωk = 1:\quad 1_{2\pi/3} = \operatorname{cos}\dfrac{2\pi}{3} + i\operatorname{sin}\dfrac{2\pi}{3} = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i = \omega

    k = 1: angle 2π/3 = 120°. cos 120° = −1/2 and sin 120° = √3/2. This root is called ω.

  5. k=2:14π/3=−12−32i=ω‾=ω2k = 2:\quad 1_{4\pi/3} = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i = \overline{\omega} = \omega^2

    k = 2: angle 4π/3 = 240°. It's the conjugate of ω, and also ω² (two 120° turns).

  6. k=3:12π=10k = 3:\quad 1_{2\pi} = 1_0

    k = 3 would give 2π = 360°: back to 1. That's why there are only 3 roots.

What it's for

The cube roots of unity form an equilateral triangle and show up when solving cubic equations.

Proof

Example: z³ − 8 = 0

z3−8=0⇒z∈{2, −1±3 i}z^3 - 8 = 0 \quad\Rightarrow\quad z \in \{ 2,\ -1 \pm \sqrt{3}\,i \}
  1. z3−8=0  ⇒  z3=8  ⇒  z∈83z^3 - 8 = 0 \;\Rightarrow\; z^3 = 8 \;\Rightarrow\; z \in \sqrt[3]{8}

    Isolate z: z is a cube root of 8.

  2. z3−8=0  ⇒  (z2)3−1=0(2⋅2⋅2=8)z^3 - 8 = 0 \;\Rightarrow\; \left(\dfrac{z}{2}\right)^3 - 1 = 0\quad (2 \cdot 2 \cdot 2 = 8)

    Trick: since 8 = 2³, divide the equation by 8 to get (z/2)³ = 1.

  3. z2∈13={1, −1±3 i2}\dfrac{z}{2} \in \sqrt[3]{1} = \left\{ 1,\ \dfrac{-1 \pm \sqrt{3}\,i}{2} \right\}

    So z/2 is a cube root of 1, which we already know.

  4. z∈{2, −1±3 i}z \in \{ 2,\ -1 \pm \sqrt{3}\,i \}

    Multiplying by 2: z = 2, z = −1 + √3 i and z = −1 − √3 i.

  5. 80:83=2,φ=0, 2π3, 4π38_0:\quad \sqrt[3]{8} = 2,\quad \varphi = 0,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3}

    With the direct formula: 8 = 8₀, its real cube root is 2 and the angles are 0, 2π/3 and 4π/3. Same result.

What it's for

In ℝ this equation only has the solution 2. In ℂ all three appear, on an equilateral triangle of radius 2.

Proof

Complex exponential ez

ez≔eRe⁡z eiIm⁡ze^z \coloneqq e^{\operatorname{Re} z}\, e^{i \operatorname{Im} z}
  1. z=a+bi  ⇒  ez=ea+bi=eaebiz = a + bi \;\Rightarrow\; e^z = e^{a + bi} = e^a e^{bi}

    It's defined so that ex+y = ex · ey still holds.

  2. ez=ea(cos⁡b+isin⁡b)=(ea)be^z = e^a(\operatorname{cos} b + i\operatorname{sin} b) = (e^a)_b

    ea is a positive real and ebi is a phasor with angle b: the result is already in polar form.

  3. ∣ez∣=eRe⁡z|e^z| = e^{\operatorname{Re} z}

    The modulus of ez depends only on the real part of z.

  4. arg⁡(ez)=Im⁡z+2πk\arg(e^z) = \operatorname{Im} z + 2\pi k

    The angle of ez is the imaginary part of z (plus full turns).

  5. ez+2πi=ez⋅e2πi=ez⋅1=eze^{z + 2\pi i} = e^z \cdot e^{2\pi i} = e^z \cdot 1 = e^z

    Adding 2πi adds a full turn and changes nothing: ez is periodic with period 2πi.

  6. e1+iπ/2=e⋅eiπ/2=e⋅ie^{1 + i\pi/2} = e \cdot e^{i\pi/2} = e \cdot i

    Example: e1 + iπ/2 = e · eiπ/2 = e · i.

What it's for

It describes oscillations that grow or decay: the real part controls size and the imaginary part the turning (RLC circuits, springs).

Proof

Complex logarithm

log⁡z=ln⁡∣z∣+iarg⁡z,Log⁡z=ln⁡∣z∣+iArg⁡z\log z = \ln|z| + i\arg z,\qquad \operatorname{Log} z = \ln|z| + i\operatorname{Arg} z
  1. ew=z,w=x+yie^w = z,\quad w = x + yi

    The logarithm of z is the number w you must put in the exponent to get z: ew = z.

  2. ew=(ex)y=ρθe^w = (e^x)_y = \rho_\theta

    Write w = x + yi. Then ew has modulus ex and angle y; z has modulus ρ and angle θ.

  3. ex=ρ  ⇒  x=ln⁡ρ=ln⁡∣z∣e^x = \rho \;\Rightarrow\; x = \ln\rho = \ln|z|

    Match moduli: ex = ρ, so x is the natural (real) logarithm of |z|.

  4. y=θ+2πk  ⇒  log⁡z=ln⁡∣z∣+i(θ+2πk)y = \theta + 2\pi k \;\Rightarrow\; \log z = \ln|z| + i(\theta + 2\pi k)

    Match angles up to full turns: y = θ + 2πk. There are infinitely many logarithms, one for each k.

  5. Log⁡z=ln⁡∣z∣+iArg⁡z\operatorname{Log} z = \ln|z| + i\operatorname{Arg} z

    The principal logarithm Log z uses the principal argument Arg z ∈ (−π, π] (k = 0).

  6. Log⁡(−1)=ln⁡1+iπ=iπ\operatorname{Log}(-1) = \ln 1 + i\pi = i\pi

    Example: −1 has modulus 1 and angle π, so Log(−1) = ln 1 + iπ = iπ. In ℂ, negative numbers do have logarithms!

What it's for

It lets you define complex powers as zw = ew·log z and solve exponential equations in ℂ.

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