Complex Numbers HL

IB Mathematics: Applications & Interpretation · Topic 1: Number & Algebra

JMaths

www.jmaths.xyz

1

Introduction & Cartesian Form

The imaginary unit \( i \) is defined by \( i^2 = -1 \). A complex number in Cartesian form is \( z = a + bi \), where \( a = \operatorname{Re}(z) \) and \( b = \operatorname{Im}(z) \).

Addition / Subtraction

\( (a+bi) \pm (c+di) = (a\pm c) + (b\pm d)i \)

Multiplication

\( (a+bi)(c+di) = (ac-bd) + (ad+bc)i \)

2

Conjugate, Modulus & Argument

Complex conjugate

\( z^{*} = a - bi \)

\( z \times z^{*} = a^2 + b^2 \) (always real)

Modulus

\( |z| = \sqrt{a^2 + b^2} \)

Distance from origin on Argand diagram

Argument

\( \arg(z) = \theta \), reference angle \( \beta = \arctan\left|\dfrac{b}{a}\right| \)

By quadrant (principal value \( -\pi < \theta \leq \pi \)): Q1: \( \theta = \beta \); Q2: \( \theta = \pi - \beta \); Q3: \( \theta = -(\pi - \beta) \); Q4: \( \theta = -\beta \). Plot the point first.

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GDC first: use abs(z) for \( |z| \) and angle(z) / Arg for \( \arg(z) \) — the calculator handles the quadrant automatically (set RADIAN mode, see §7).

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Division: To divide by a complex number, multiply top and bottom by the conjugate of the denominator: \( \dfrac{a+bi}{c+di} \times \dfrac{c-di}{c-di} \).

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Common error: \( \arctan(b/a) \) only gives the correct angle for Q1 and Q4. For \( z = -1 + i \), \( \arctan(-1) = -45 \)° but the actual argument is 135° (Q2). Always plot the point first.

3

Argand Diagram

The Argand diagram represents complex numbers as points in a plane: horizontal axis = Real, vertical axis = Imaginary.

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Key interpretations: \( |z_1 - z_2| = \) distance between two points. \( |z - w| = r \) describes a circle centre \( w \), radius \( r \). \( \arg(z - w) = \theta \) describes a half-line from \( w \).

Complex Numbers HL

IB Mathematics: Applications & Interpretation · Topic 1: Number & Algebra

JMaths

www.jmaths.xyz

4

Polar & Exponential Form

Polar (modulus-argument) form

\( z = r\operatorname{cis}\theta = r(\cos \theta + i \sin \theta ) \)

where \( r = |z| \) and \( \theta = \arg(z) \)

Exponential (Euler) form

\( z = r e^{i\theta} \)

Same \( r \) and \( \theta \) (in radians): \( r\operatorname{cis}\theta = r(\cos\theta + i\sin\theta) = re^{i\theta} \). Products/powers are easiest here: \( (re^{i\theta})^n = r^n e^{in\theta} \).

Converting between forms

Cartesian \( \to \) Polar / Exponential: \( r = \sqrt{a^2+b^2} \), \( \theta = \arg(z) \) (radians, adjust for quadrant — see §2)
Polar / Exponential \( \to \) Cartesian: \( a = r \cos \theta \), \( b = r \sin \theta \)

So \( a + bi = r\operatorname{cis}\theta = re^{i\theta} \) all describe the same number.

Worked Example

Write \( z = -1 + \sqrt{3}\, i \) in polar and exponential form.

\( r = \sqrt{1 + 3} = 2 \)

Reference \( \beta = \arctan\left|\dfrac{\sqrt{3}}{-1}\right| = \dfrac{\pi}{3} \); point is in Q2, so \( \theta = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3} \)

Answer: \( z = 2\operatorname{cis}\dfrac{2\pi}{3} = 2e^{i\,2\pi/3} \)

GDC: abs(-1+√3 i)=2, angle(-1+√3 i)=2π/3 (radian mode).

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Common error: taking \( \theta = \arctan(b/a) \) without checking the quadrant. Here \( \arctan(-\sqrt{3}) = -\dfrac{\pi}{3} \), which is Q4 — you must add \( \pi \) to land in Q2.

5

Complex Roots of a Quadratic

When the discriminant is negative

\( ax^2+bx+c=0 \) with real \( a,b,c \) and \( b^2-4ac < 0 \):
\( x = \dfrac{-b \pm i\sqrt{4ac-b^2}}{2a} \)

The two roots are a conjugate pair \( p \pm qi \).

Worked Example

Solve \( x^2 - 2x + 5 = 0 \).

Discriminant: \( b^2-4ac = 4 - 20 = -16 < 0 \) ⇒ complex roots.

\( x = \dfrac{2 \pm \sqrt{-16}}{2} = \dfrac{2 \pm 4i}{2} \)

Answer: \( x = 1 \pm 2i \) (conjugate pair)

GDC: a complex-mode equation solver / polyRoots returns both roots directly.

6

Multiplication & Division in Polar Form

Multiplication

\( r_1\operatorname{cis}\theta_1 \times r_2\operatorname{cis}\theta_2 = r_1 r_2 \operatorname{cis}(\theta_1+\theta_2) \)

Multiply moduli, add arguments

Division

\( \dfrac{r_1\operatorname{cis}\theta_1}{r_2\operatorname{cis}\theta_2} = \dfrac{r_1}{r_2}\operatorname{cis}(\theta_1-\theta_2) \)

Divide moduli, subtract arguments

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Geometric meaning: Multiplying by \( r\operatorname{cis}\theta \) scales by \( r \) and rotates by \( \theta \). Multiplying by \( \operatorname{cis}\left(\dfrac{\pi}{2}\right) \) is a rotation of 90° anticlockwise.

Complex Numbers HL

IB Mathematics: Applications & Interpretation · Topic 1: Number & Algebra

JMaths

www.jmaths.xyz

7

De Moivre’s Theorem

De Moivre’s theorem (given in formula booklet)

\( (r\operatorname{cis}\theta)^n = r^n \operatorname{cis}(n\theta) \)

Equivalently: \( [r(\cos\theta + i\sin\theta)]^{n} = r^{n}(\cos n\theta + i\sin n\theta) \). In exponential form: \( (re^{i\theta})^n = r^n e^{in\theta} \).

Worked Example

Find \( (1 + i)^8 \).

Convert to polar: \( |1+i| = \sqrt{2} \), \( \arg = \dfrac{\pi}{4} \). So \( 1+i = \sqrt{2}\operatorname{cis}\dfrac{\pi}{4} \).

\( \left(\sqrt{2}\operatorname{cis}\dfrac{\pi}{4}\right)^{8} = (\sqrt{2})^{8}\operatorname{cis}\left(8 \times \dfrac{\pi}{4}\right) = 16\operatorname{cis}(2\pi) = 16(1 + 0i) \)

Answer: \( (1 + i)^8 = 16 \)

GDC: in a+bi mode type (1+i)^8 directly → 16.

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Common error: Applying De Moivre’s theorem to Cartesian form directly. You MUST convert to polar form first: \( (a+bi)^{n} \neq a^{n} + (bi)^{n} \).

8

Adding Sinusoidal Functions

Complex numbers can be used to combine sinusoidal functions of the same frequency into a single sinusoidal function.

Combining sinusoids

\( A_1 \sin(\omega t + \phi_1) + A_2 \sin(\omega t + \phi_2) = R \sin(\omega t + \alpha) \)

Represent each term as a complex number \( A\operatorname{cis}\phi \), add them, then read off \( R \) and \( \alpha \).

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Method: Write \( z_1 = A_1\operatorname{cis}\phi_1 \) and \( z_2 = A_2\operatorname{cis}\phi_2 \). Then \( z_1 + z_2 = R\operatorname{cis}\alpha \) where \( R = |z_1+z_2| \) and \( \alpha = \arg(z_1+z_2) \).

Worked Example

Write \( 3\sin(\omega t) + 4\sin\left(\omega t + \dfrac{\pi}{2}\right) \) as \( R\sin(\omega t + \alpha) \).

\( z_1 = 3\operatorname{cis}0 = 3 + 0i \), \( z_2 = 4\operatorname{cis}\dfrac{\pi}{2} = 0 + 4i \)

\( z_1 + z_2 = 3 + 4i \Rightarrow R = |3+4i| = \sqrt{9+16} = 5 \)

\( \alpha = \arg(3+4i) = \arctan\dfrac{4}{3} \approx 0.927 \) rad

Answer: \( 5\sin(\omega t + 0.927) \)

GDC: abs(3+4i)=5, angle(3+4i)=0.927 (radian mode).

GDC: Complex Numbers

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Set RADIAN mode first (all models): the IB wants arguments in radians. In degree mode \( \arg(-1+\sqrt{3}\,i) \) returns 120° instead of \( \dfrac{2\pi}{3} \). Check the angle/mode setting before reading any arg / angle / Arg output.

TI-84 Plus CE

[MODE] → set RADIAN and a+b\( i \) (or re^(\( \theta i) \))
Enter \( i \) with [2ND][.]
abs( for modulus, angle( for argument
Powers: (1+i)^8 [ENTER]
Radian mode — else angle returns degrees.

TI-Nspire CX II

Enter \( i \) from the keypad (or symbol palette)
abs(z) modulus, angle(z) argument
conj(z) conjugate real(z), imag(z) parts
Polar: type \( r\cdot e^{(i\cdot\theta)} \) or use ▸Polar
Set Document Settings → Angle = Radian before reading angle.

Casio fx-CG50

[MENU] → Run-Matrix [SHIFT][0] for \( i \)
[OPTN] → COMPLEX → Abs, Arg, Conj
Form: [SHIFT][MENU] → Complex Mode → a+bi or r∠\( \theta \)
Same SET UP screen: set Angle = Rad before Arg.

9

Exam Traps & Key Reminders

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Powers of \( i \): \( i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, \) then repeats. For \( i^{n} \), find \( n \) mod 4.

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Argument in radians. The IB expects arguments in radians (usually in terms of \( \pi \)), not degrees, unless otherwise stated.

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Formula booklet: De Moivre’s theorem, modulus, and polar/Cartesian conversion are given. The multiplication/division rules for polar form follow directly from De Moivre’s theorem.

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Complex Numbers HL

Complex Numbers HL

Complex Numbers HL