Polar and exponential form of complex numbers

IB Mathematics: Analysis and Approaches HL· 1.6 Complex numbers· 15 min read

1. Polar Form of Complex Numbers★★☆☆☆⏱ 5 min

📘 Definition

Polar Form

z = r(\cos\theta + i \sin\theta)

A representation of a non-zero complex number that uses its modulus $r$ (distance from the origin) and argument $θ$ (angle from the positive real axis)

Example:

$z = 1 + i$ has polar form $2 (cos \frac{π}{4} + i sin \frac{π}{4})$

To convert from Cartesian form $z = x + i y$ to polar form, calculate $r = x^{2} + y^{2}$ and $tan θ = \frac{y}{x}$ . You must always adjust $θ$ to match the quadrant of $z$ on the Argand diagram, as inverse tangent only returns values between $- \frac{π}{2}$ and $\frac{π}{2}$ .

📐 Worked Example

Convert $z = - 1 + 3 i$ to polar form

1
First calculate the modulus of $z$
$r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$
2
Find the reference angle for the argument
$\tan\alpha = \left|\frac{y}{x}\right| = \sqrt{3} \implies \alpha = \frac{\pi}{3}$
3
$z$ lies in the second quadrant, so adjust the argument
$\theta = \pi - \alpha = \frac{2\pi}{3}$
4
Substitute into the polar form formula
$z = 2\left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right)$

Exam tip:

Always plot your complex number on a rough Argand diagram to confirm the quadrant for the argument.

2. Multiplication, Division and De Moivre's Theorem★★☆☆☆⏱ 5 min

For two complex numbers in polar form $z_{1} = r_{1} (cos θ_{1} + i sin θ_{1})$ and $z_{2} = r_{2} (cos θ_{2} + i sin θ_{2})$ , the rules for multiplication and division are much simpler than in Cartesian form:

z_1 z_2 = r_1 r_2 \left(\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)\right) \\ \frac{z_1}{z_2} = \frac{r_1}{r_2} \left(\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)\right)

📘 Definition

De Moivre's Theorem

\left(r(\cos\theta + i\sin\theta)\right)^n

For any real number $n$ , the nth power of a complex number in polar form is equal to $r^{n} (cos n θ + i sin n θ)$

Example:

$(2 (cos \frac{π}{4} + i sin \frac{π}{4}))^{2} = 2 i$

📐 Worked Example

Given $z_{1} = 2 (cos \frac{π}{3} + i sin \frac{π}{3})$ and $z_{2} = 3 (cos \frac{π}{6} + i sin \frac{π}{6})$ , find $z_{1} z_{2}$ in polar form

1
Multiply the moduli of the two complex numbers
$r = r_1 r_2 = 2 \times 3 = 6$
2
Add the arguments of the two complex numbers
$\theta = \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2}$
3
Combine to get the product in polar form
$z_1 z_2 = 6\left(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}\right)$

3. Exponential Form of Complex Numbers★★★☆☆⏱ 4 min

📘 Definition

Exponential Form

z = r e^{i\theta}

A compact representation of a complex number derived from Euler's formula $e^{i θ} = cos θ + i sin θ$ , where $r$ is modulus and $θ$ is argument (in radians)

Example:

$z = 1 + i = 2 e^{i \frac{π}{4}}$

Exponential form follows all the standard exponent rules, making multiplication, division and powers very intuitive:

r_1 e^{i\theta_1} \times r_2 e^{i\theta_2} = r_1 r_2 e^{i(\theta_1 + \theta_2)} \\ \frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)} \\ \left(r e^{i\theta}\right)^n = r^n e^{i n\theta}

📐 Worked Example

Calculate $(2 e^{i \frac{π}{3}})^{3}$ and express the result in Cartesian form

1
Apply the power rule for exponential form
$(2 e^{i \pi/3})^3 = 2^3 (e^{i \pi/3})^3 = 8 e^{i \pi}$
2
Use Euler's formula to expand
$8 e^{i \pi} = 8 (\cos \pi + i \sin \pi)$
3
Evaluate and simplify to Cartesian form
$8(-1 + 0i) = -8$

4. Roots of Complex Numbers★★★★☆⏱ 6 min

Any non-zero complex number has exactly $n$ distinct $n$ th roots. These roots are equally spaced around a circle of radius $r^{1/ n}$ on the Argand diagram, with an angular separation of $\frac{2 π}{n}$ between consecutive roots.

📘 Definition

nth Root Formula

z_k = r^{1/n} e^{i \frac{\theta + 2k\pi}{n}}, \quad k = 0, 1, ..., n-1

Gives all $n$ distinct $n$ th roots of a complex number $z = r e^{i θ}$ by adding full rotations ( $2 k π$ ) to the original argument to generate all unique solutions.

📐 Worked Example

Find all cube roots of 1, leaving your answer in polar form

1
Write 1 in polar form, identify modulus and argument
$1 = 1(\cos 0 + i \sin 0), \quad r = 1, \theta = 0, n = 3$
2
Apply the nth root formula for polar form
$z_k = 1^{1/3} \left(\cos\left(\frac{0 + 2k\pi}{3}\right) + i \sin\left(\frac{0 + 2k\pi}{3}\right)\right)$
3
Substitute $k = 0, 1, 2$ to get all roots
$z_0 = \cos 0 + i \sin 0 = 1, \quad z_1 = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}, \quad z_2 = \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}$

Exam tip:

Always remember there are n distinct nth roots — if you only write one root, you will lose most of the marks for the question.

5. Common Pitfalls

Wrong move:

Forgetting to adjust the argument to the correct quadrant when converting from Cartesian to polar

Why:

Inverse tangent only returns values between $- \frac{π}{2}$ and $\frac{π}{2}$ , which is incorrect for complex numbers in the second and third quadrants

Correct move:

Plot the point on the Argand diagram to check the quadrant, then adjust the inverse tangent result by adding or subtracting $π$ as needed

Wrong move:

Using degrees instead of radians for the argument in exponential form

Why:

Euler's formula $e^{i θ} = cos θ + i sin θ$ only holds when $θ$ is measured in radians

Correct move:

Always convert any degree measure arguments to radians before writing exponential form

Wrong move:

Only giving one root when asked for nth roots of a complex number

Why:

The equation $z^{n} = a$ has $n$ distinct solutions for any non-zero complex $a$ , and exam markers require all solutions

Correct move:

Use the root formula for $k = 0, 1, ..., n - 1$ to list all $n$ roots

Wrong move:

Mixing up addition and subtraction of arguments for multiplication/division

Why:

Confusion between the geometric effects of multiplication and division

Correct move:

Remember: multiplication adds arguments (rotates the first number by the second's argument), division subtracts arguments

Wrong move:

Claiming $z = 0$ has a specific principal argument when converting to polar form

Why:

0 lies at the origin, so it has no defined direction or argument

Correct move:

Write 0 as $0 = 0 (cos θ + i sin θ)$ for any $θ$ — no specific argument is required

6. Quick Reference Cheatsheet

Operation	Polar Form	Exponential Form
Conversion from Cartesian	$r = x^{2} + y^{2}$ , $θ$ adjusted for quadrant: $z = r (cos θ + i sin θ)$	$r = x^{2} + y^{2}$ , $z = r e^{i θ}$ (θ in radians)
Multiplication	$z_{1} z_{2} = r_{1} r_{2} (cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2}))$	$z_{1} z_{2} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$
Division	$z_{1} / z_{2} = (r_{1} / r_{2}) (cos (θ_{1} - θ_{2}) + i sin (θ_{1} - θ_{2}))$	$z_{1} / z_{2} = (r_{1} / r_{2}) e^{i (θ_{1} - θ_{2})}$
nth Power	$z^{n} = r^{n} (cos n θ + i sin n θ)$	$z^{n} = r^{n} e^{in θ}$
nth Roots	$z_{k} = r^{1/ n} (cos (\frac{θ + 2 k π}{n}) + i sin (\frac{θ + 2 k π}{n})), k = 0.. n - 1$	$z_{k} = r^{1/ n} e^{i \frac{θ + 2 k π}{n}}, k = 0.. n - 1$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 1
Find all cube roots of a complex number
2023 · 2
Convert between forms, multiply complex numbers

What's Next

Mastering polar and exponential forms of complex numbers is essential for further topics in IB AA HL, including loci on the Argand diagram and solving higher-order polynomials with complex roots. These representations also underpin core applications in engineering and physics, from signal processing to alternating current circuit analysis. This topic forms the foundation for all advanced work with complex numbers in university-level mathematics and STEM.