IB Mathematics AA – Topic 1: Number & Algebra

\(i^2 = -1\) or equivalently \(i = \sqrt{-1}\)

Powers of i (Cyclic Pattern):

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = i^2 \cdot i = -1 \cdot i = -i\)
• \(i^4 = i^2 \cdot i^2 = (-1)(-1) = 1\)
• \(i^5 = i^4 \cdot i = 1 \cdot i = i\) (pattern repeats)

Key insight: Powers of \(i\) repeat every 4. To find \(i^n\), divide \(n\) by 4 and use the remainder.

Standard Cartesian Form:

• \(a\) is the real part: \(\text{Re}(z) = a\)
• \(b\) is the imaginary part: \(\text{Im}(z) = b\)
• Note: The imaginary part is just \(b\), not \(bi\)

Key Definitions:

• Complex Conjugate: \(\bar{z} = a - bi\)

  Change the sign of the imaginary part

• Modulus (Magnitude): \(|z| = \sqrt{a^2 + b^2}\)

  Distance from origin in the complex plane

• Argument (Angle): \(\arg(z) = \theta\) where \(\tan\theta = \frac{b}{a}\)

  Angle from positive real axis (in radians, \(-\pi < \theta \leq \pi\))

Addition & Subtraction:

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

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

Multiplication:

\((a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i\)

Remember: \(i^2 = -1\)

Division:

\(\displaystyle\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac+bd) + (bc-ad)i}{c^2 + d^2}\)

Multiply top and bottom by the conjugate of the denominator

⚠ Common Pitfalls:

• Don't forget \(i^2 = -1\) when multiplying complex numbers!
• The imaginary part is \(b\), not \(bi\). For \(z = 3 + 4i\), \(\text{Im}(z) = 4\), not \(4i\).
• When finding arguments, watch for quadrant—\(\tan^{-1}\) only gives values in \((-\frac{\pi}{2}, \frac{\pi}{2})\).
• Conjugate changes sign of imaginary part only: \(\overline{3+4i} = 3-4i\), not \(-3-4i\).
• \(|z|^2 = z\bar{z} = a^2 + b^2\) (useful property!)

Example 1: Basic Operations with Complex Numbers

Problem: Given \(z_1 = 3 + 4i\) and \(z_2 = 1 - 2i\), find:

(a) \(z_1 + z_2\)

(b) \(z_1 \cdot z_2\)

(c) \(\displaystyle\frac{z_1}{z_2}\)

(d) \(|z_1|\) and \(\arg(z_1)\)

Solution:

(a) \(z_1 + z_2 = (3+4i) + (1-2i)\)

\(= (3+1) + (4-2)i\)

\(= 4 + 2i\)

(b) \(z_1 \cdot z_2 = (3+4i)(1-2i)\)

\(= 3(1) + 3(-2i) + 4i(1) + 4i(-2i)\)

\(= 3 - 6i + 4i - 8i^2\)

\(= 3 - 2i - 8(-1)\)

\(= 3 - 2i + 8\)

\(= 11 - 2i\)

(c) \(\displaystyle\frac{z_1}{z_2} = \frac{3+4i}{1-2i}\)

Multiply by conjugate of denominator:

\(= \displaystyle\frac{(3+4i)(1+2i)}{(1-2i)(1+2i)}\)

Numerator: \((3+4i)(1+2i) = 3 + 6i + 4i + 8i^2 = 3 + 10i - 8 = -5 + 10i\)

Denominator: \((1-2i)(1+2i) = 1 - 4i^2 = 1 + 4 = 5\)

\(= \displaystyle\frac{-5 + 10i}{5} = -1 + 2i\)

\(= -1 + 2i\)

(d) Modulus: \(|z_1| = |3+4i| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25}\)

\(|z_1| = 5\)

Argument: \(\arg(z_1) = \tan^{-1}\left(\frac{4}{3}\right)\)

Since \(z_1\) is in the first quadrant (both \(a\) and \(b\) positive):

\(\arg(z_1) \approx 0.927\) radians or \(53.1°\)

Structure of the Argand Diagram:

• Horizontal axis (Real axis): Represents the real part \(a\)
• Vertical axis (Imaginary axis): Represents the imaginary part \(b\)
• Point representation: \(z = a + bi\) is plotted at coordinates \((a, b)\)
• Origin: Represents \(0 + 0i = 0\)

Key Geometric Interpretations:

• Modulus \(|z|\): Distance from the origin to the point \(z\)

  Length of the position vector

• Argument \(\arg(z)\): Angle from the positive real axis to the line joining origin to \(z\)

  Measured counterclockwise, in radians

• Complex conjugate \(\bar{z}\): Reflection of \(z\) across the real axis

  If \(z = a+bi\) is at \((a,b)\), then \(\bar{z} = a-bi\) is at \((a,-b)\)

• Addition: Vector addition (parallelogram law)

  \(z_1 + z_2\) is the diagonal of the parallelogram formed by \(z_1\) and \(z_2\)

Quadrant Rules for \(\arg(z)\):

Quadrant	Sign of \(a, b\)	Calculation
I (QI)	\(a > 0, b > 0\)	\(\theta = \tan^{-1}\left(\frac{b}{a}\right)\)
II (QII)	\(a < 0, b > 0\)	\(\theta = \pi + \tan^{-1}\left(\frac{b}{a}\right)\)
III (QIII)	\(a < 0, b < 0\)	\(\theta = -\pi + \tan^{-1}\left(\frac{b}{a}\right)\)
IV (QIV)	\(a > 0, b < 0\)	\(\theta = \tan^{-1}\left(\frac{b}{a}\right)\)

Note: Principal argument is in the range \((-\pi, \pi]\)

💡 Pro Tips for Argand Diagrams:

• Always sketch the complex number to identify the correct quadrant
• Use your calculator in complex mode to verify arguments
• Remember: \(|z_1 - z_2|\) represents the distance between \(z_1\) and \(z_2\)
• Geometric loci (circles, lines) are common IB exam questions

Example 2: Working with the Argand Diagram

Problem: Given \(z = -3 + 4i\):

(a) Plot \(z\) and \(\bar{z}\) on an Argand diagram

(b) Find \(|z|\) and \(\arg(z)\)

(c) Describe the locus of points satisfying \(|w - z| = 2\)

Solution:

(a) Plotting:

\(z = -3 + 4i\) is plotted at point \((-3, 4)\)

\(\bar{z} = -3 - 4i\) is plotted at point \((-3, -4)\)

Note: \(\bar{z}\) is the reflection of \(z\) across the real axis

(b) Modulus:

\(|z| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25}\)

\(|z| = 5\)

Argument (noting \(z\) is in Quadrant II):

Basic angle: \(\alpha = \tan^{-1}\left(\frac{4}{3}\right) \approx 0.927\) rad

Since \(z\) is in QII (negative real, positive imaginary):

\(\arg(z) = \pi - 0.927 \approx 2.214\) radians

\(\arg(z) \approx 2.21\) rad or \(126.9°\)

(c) Locus \(|w - z| = 2\):

The condition \(|w - z| = 2\) means the distance from \(w\) to \(z\) equals 2.

This is a circle with center at \(z = -3 + 4i\) and radius 2.

In Cartesian coordinates: center \((-3, 4)\), radius \(2\)

Polar Form:

Euler's Formula:

This is Euler's famous formula, connecting exponentials with trigonometry

Cartesian to Polar/Euler:

Given \(z = a + bi\):

• \(r = |z| = \sqrt{a^2 + b^2}\)
• \(\theta = \arg(z) = \tan^{-1}\left(\frac{b}{a}\right)\) (adjust for quadrant)
• Then: \(z = r\text{ cis }\theta = re^{i\theta}\)

Polar/Euler to Cartesian:

Given \(z = re^{i\theta}\) or \(z = r\text{ cis }\theta\):

• \(a = r\cos\theta\) (real part)
• \(b = r\sin\theta\) (imaginary part)
• Then: \(z = a + bi = r\cos\theta + ir\sin\theta\)

Multiplication:

\(z_1 \cdot z_2 = r_1e^{i\theta_1} \cdot r_2e^{i\theta_2} = r_1r_2e^{i(\theta_1+\theta_2)}\)

Multiply moduli, add arguments

Division:

\(\displaystyle\frac{z_1}{z_2} = \frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}} = \frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}\)

Divide moduli, subtract arguments

Powers:

\(z^n = (re^{i\theta})^n = r^ne^{in\theta}\)

Raise modulus to power \(n\), multiply argument by \(n\)

💡 When to Use Each Form:

• Cartesian form: Addition, subtraction, finding real/imaginary parts
• Polar/Euler form: Multiplication, division, powers, roots
• De Moivre's Theorem: Requires polar or Euler form

Example 3: Converting Between Forms

Problem:

(a) Convert \(z = 1 + \sqrt{3}i\) to polar form

(b) Convert \(w = 4e^{i\frac{\pi}{3}}\) to Cartesian form

(c) Find \(z \cdot w\) using polar form

Solution:

(a) Convert \(z = 1 + \sqrt{3}i\) to polar:

Modulus: \(r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2\)

Argument: \(\theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\)

(QI, so no adjustment needed)

\(z = 2e^{i\frac{\pi}{3}}\) or \(z = 2\text{ cis }\frac{\pi}{3}\)

(b) Convert \(w = 4e^{i\frac{\pi}{3}}\) to Cartesian:

Here \(r = 4\) and \(\theta = \frac{\pi}{3}\)

Real part: \(a = 4\cos\left(\frac{\pi}{3}\right) = 4 \times \frac{1}{2} = 2\)

Imaginary part: \(b = 4\sin\left(\frac{\pi}{3}\right) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}\)

\(w = 2 + 2\sqrt{3}i\)

(c) Multiply using polar form:

\(z = 2e^{i\frac{\pi}{3}}\) and \(w = 4e^{i\frac{\pi}{3}}\)

\(z \cdot w = 2 \times 4 \times e^{i(\frac{\pi}{3}+\frac{\pi}{3})}\)

\(= 8e^{i\frac{2\pi}{3}}\)

Convert to Cartesian:

\(= 8\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)\)

\(= 8\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)\)

\(= -4 + 4\sqrt{3}i\)

De Moivre's Theorem

In full form: \([r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))\)

Main Uses:

• Computing powers: Find \(z^n\) quickly for large \(n\)
• Finding roots: Find the \(n\)th roots of a complex number
• Trigonometric identities: Derive formulas for \(\cos(n\theta)\) and \(\sin(n\theta)\)
• Simplifying expressions: Work with powers of complex numbers elegantly

⚠ Important Notes:

• De Moivre's Theorem requires polar or Euler form—convert first if needed
• For negative \(n\), the theorem still works: \(z^{-n} = \frac{1}{z^n}\)
• When finding roots, remember there are \(n\) distinct \(n\)th roots
• Always check if your calculator is in radian or degree mode!

Example 4: Using De Moivre's Theorem for Powers

Problem: Use De Moivre's theorem to find \((1 + i)^{10}\), giving your answer in Cartesian form.

Solution:

Step 1: Convert \(z = 1 + i\) to polar form.

Modulus: \(r = \sqrt{1^2 + 1^2} = \sqrt{2}\)

Argument: \(\theta = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1) = \frac{\pi}{4}\)

So: \(z = \sqrt{2}e^{i\frac{\pi}{4}} = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)\)

Step 2: Apply De Moivre's Theorem with \(n = 10\).

\(z^{10} = \left(\sqrt{2}\right)^{10}\left(\cos\left(10 \cdot \frac{\pi}{4}\right) + i\sin\left(10 \cdot \frac{\pi}{4}\right)\right)\)

\(= 2^5\left(\cos\frac{10\pi}{4} + i\sin\frac{10\pi}{4}\right)\)

\(= 32\left(\cos\frac{5\pi}{2} + i\sin\frac{5\pi}{2}\right)\)

Step 3: Simplify the angle.

\(\frac{5\pi}{2} = 2\pi + \frac{\pi}{2}\) (equivalent to \(\frac{\pi}{2}\))

\(\cos\frac{\pi}{2} = 0\), \(\sin\frac{\pi}{2} = 1\)

Step 4: Calculate final answer.

\(z^{10} = 32(0 + i \cdot 1) = 32i\)

Answer: \((1+i)^{10} = 32i\)

Example 5: Deriving Trigonometric Identities (IB-Style)

Problem: Use De Moivre's theorem to express \(\cos(3\theta)\) in terms of \(\cos\theta\).

Solution:

Step 1: Apply De Moivre's theorem with \(n = 3\).

\((\cos\theta + i\sin\theta)^3 = \cos(3\theta) + i\sin(3\theta)\)

Step 2: Expand the left side using binomial theorem.

\((\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3\)

\(= \cos^3\theta + 3i\cos^2\theta\sin\theta + 3\cos\theta(i^2\sin^2\theta) + i^3\sin^3\theta\)

\(= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta\)

Grouping real and imaginary parts:

\(= (\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta)\)

Step 3: Equate real parts.

\(\cos(3\theta) = \cos^3\theta - 3\cos\theta\sin^2\theta\)

Using \(\sin^2\theta = 1 - \cos^2\theta\):

\(\cos(3\theta) = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta)\)

\(= \cos^3\theta - 3\cos\theta + 3\cos^3\theta\)

\(\cos(3\theta) = 4\cos^3\theta - 3\cos\theta\)

Formula for nth Roots:

Or in polar form: \(w_k = r^{\frac{1}{n}}\text{ cis }\left(\frac{\theta + 2\pi k}{n}\right)\)

Properties of nth Roots:

• All \(n\) roots lie on a circle of radius \(r^{\frac{1}{n}}\)
• The roots are evenly spaced at angular intervals of \(\frac{2\pi}{n}\)
• The roots form a regular \(n\)-sided polygon
• One root always lies at angle \(\frac{\theta}{n}\) from the positive real axis

💡 Special Case - Roots of Unity:

The \(n\)th roots of 1 (unity) are:

\(\omega_k = e^{i\frac{2\pi k}{n}} = \cos\left(\frac{2\pi k}{n}\right) + i\sin\left(\frac{2\pi k}{n}\right)\)

where \(k = 0, 1, 2, \ldots, n-1\)

These form a regular \(n\)-gon with vertices on the unit circle.

Example 6: Finding Cube Roots

Problem: Find all three cube roots of \(z = 8i\). Express your answers in Cartesian form and plot them on an Argand diagram.

Solution:

Step 1: Convert \(z = 8i\) to polar form.

\(z = 0 + 8i\)

Modulus: \(r = |8i| = 8\)

Argument: \(\theta = \frac{\pi}{2}\) (pointing straight up on imaginary axis)

So: \(z = 8e^{i\frac{\pi}{2}}\)

Step 2: Apply the nth root formula with \(n = 3\).

\(w_k = 8^{\frac{1}{3}}e^{i\left(\frac{\frac{\pi}{2} + 2\pi k}{3}\right)}\) for \(k = 0, 1, 2\)

\(= 2e^{i\left(\frac{\pi + 4\pi k}{6}\right)}\)

Step 3: Calculate each root.

For \(k = 0\):

\(w_0 = 2e^{i\frac{\pi}{6}} = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right)\)

\(w_0 = \sqrt{3} + i\)

For \(k = 1\):

\(w_1 = 2e^{i\frac{5\pi}{6}} = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) = 2\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right)\)

\(w_1 = -\sqrt{3} + i\)

For \(k = 2\):

\(w_2 = 2e^{i\frac{9\pi}{6}} = 2e^{i\frac{3\pi}{2}} = 2\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = 2(0 - i)\)

\(w_2 = -2i\)

Answer: The three cube roots are \(\sqrt{3} + i\), \(-\sqrt{3} + i\), and \(-2i\)

On an Argand diagram, these form an equilateral triangle with vertices on a circle of radius 2, centered at the origin.

Common Loci:

Complex Number Transformations:

• Translation: \(z \to z + w\) shifts by vector \(w\)
• Rotation: \(z \to ze^{i\theta}\) rotates by angle \(\theta\) about origin
• Scaling: \(z \to kz\) scales by factor \(k\) from origin
• Reflection: \(z \to \bar{z}\) reflects across real axis

Example 7: Geometric Loci (IB-Style)

Problem: Sketch and describe the locus of points satisfying:

(a) \(|z - 2i| = 3\)

(b) \(|z + 1| = |z - 3|\)

(c) \(\arg(z - 1 - i) = \frac{\pi}{4}\)

Solution:

(a) \(|z - 2i| = 3\)

This represents points at distance 3 from \(2i = 0 + 2i\)

Locus: Circle with center \((0, 2)\) and radius 3

The circle passes through points \((0, 5)\), \((3, 2)\), \((0, -1)\), and \((-3, 2)\)

(b) \(|z + 1| = |z - 3|\)

Rewrite: \(|z - (-1)| = |z - 3|\)

Points equidistant from \(-1\) and \(3\) (both on real axis)

Locus: Perpendicular bisector of segment from \(-1\) to \(3\)

This is the vertical line \(x = 1\) (or \(\text{Re}(z) = 1\))

(c) \(\arg(z - 1 - i) = \frac{\pi}{4}\)

This represents points such that the angle from \(1 + i\) to \(z\) is \(\frac{\pi}{4}\)

Locus: Ray (half-line) from \((1, 1)\) at angle \(\frac{\pi}{4}\) (45°)

This ray goes from \((1,1)\) through \((2,2)\), \((3,3)\), etc. (line \(y = x\) for \(x \geq 1\))

Example 8: Transformation Problem

Problem: The point \(z = 2 + i\) undergoes the following transformations in order:

• Rotation of \(\frac{\pi}{2}\) radians anticlockwise about the origin
• Translation by \(-3 + 2i\)

Find the final position of the point.

Solution:

Step 1: Rotation by \(\frac{\pi}{2}\).

Multiply by \(e^{i\frac{\pi}{2}}\):

\(z_1 = (2 + i)e^{i\frac{\pi}{2}} = (2 + i)\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)\)

\(= (2 + i)(0 + i) = (2 + i) \cdot i\)

\(= 2i + i^2 = 2i - 1\)

\(z_1 = -1 + 2i\)

Step 2: Translation by \(-3 + 2i\).

\(z_2 = z_1 + (-3 + 2i)\)

\(= (-1 + 2i) + (-3 + 2i)\)

\(= -4 + 4i\)

Final position: \(-4 + 4i\) or point \((-4, 4)\) in the Argand diagram

Form Selection:

• Cartesian: Addition, subtraction, finding real/imaginary parts
• Polar/Euler: Multiplication, division, powers, roots, De Moivre's theorem
• Always convert to the most convenient form for the operation!

Common Question Types:

• Operations: Add, multiply, divide complex numbers in appropriate form
• Powers: Use De Moivre's theorem (convert to polar first)
• Roots: Apply nth root formula, plot on Argand diagram
• Loci: Identify geometric shapes from complex equations
• Transformations: Apply rotations, translations geometrically

Calculator Tips:

• Set calculator to complex mode (rectangular or polar)
• Know how to convert between forms on your calculator
• Verify arguments are in correct range \((-\pi, \pi]\)
• Check if calculator is in radians or degrees!