# Polar and exponential form of complex numbers

> IB Mathematics: Analysis and Approaches HL · Number & Algebra
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u1-polar-and-exponential-form-of/

This sub-topic introduces polar and exponential representations of complex numbers, which simplify multiplication, division, and finding powers/roots compared to standard Cartesian form. We also cover their geometric interpretation on the Argand diagram.

**Prerequisites:** [Cartesian form of complex numbers](https://www.owlsprep.com/study/ib-math-aa-hl-u1-cartesian-form-complex-numbers/); Right-angled triangle trigonometry

## Learning objectives

- Convert between Cartesian, polar, and exponential forms of complex numbers
- Interpret polar and exponential forms geometrically on the Argand diagram
- Multiply, divide, and find powers/roots of complex numbers using polar and exponential forms
- Apply De Moivre's theorem to find all nth roots of a complex number

## Polar Form of Complex Numbers

**Polar Form** — A representation of a non-zero complex number that uses its modulus $r$ (distance from the origin) and argument $\theta$ (angle from the positive real axis)

*Notation:* z = r(\cos\theta + i \sin\theta)

*Example:* $z = 1 + i$ has polar form $\sqrt{2}\left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right)$

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

**Worked example:** Convert $z = -1 + \sqrt{3}i$ 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.

## Multiplication, Division and De Moivre's Theorem

For two complex numbers in polar form $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$, the rules for multiplication and division are much simpler than in Cartesian form:

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

**De Moivre's Theorem** — For any real number $n$, the nth power of a complex number in polar form is equal to $r^n(\cos n\theta + i\sin n\theta)$

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

*Example:* $\left(\sqrt{2}(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4})\right)^2 = 2i$

**Worked example:** Given $z_1 = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})$ and $z_2 = 3(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$, 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

   $$z_1 z_2 = 6\left(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}\right)$$

## Exponential Form of Complex Numbers

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

*Notation:* z = r e^{i\theta}

*Example:* $z = 1 + i = \sqrt{2} e^{i\frac{\pi}{4}}$

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

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

> **warning**
>
> Arguments in exponential form must always be in radians, never degrees. This is a common exam mistake that will cost you all marks for the question.

**Worked example:** Calculate $(2 e^{i \frac{\pi}{3}})^3$ 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

   $$8(-1 + 0i) = -8$$

## Roots of Complex Numbers

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 $\frac{2\pi}{n}$ between consecutive roots.

**nth Root Formula** — Gives all $n$ distinct $n$th roots of a complex number $z = r e^{i\theta}$ by adding full rotations ($2k\pi$) to the original argument to generate all unique solutions.

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

**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 command terms**

- **Find all roots** — You must list every distinct root, not just one *(For cube roots, you need 3 solutions)*

> **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.

## Common pitfalls

- **Wrong:** Forgetting to adjust the argument to the correct quadrant when converting from Cartesian to polar
  - Why it fails: Inverse tangent only returns values between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, which is incorrect for complex numbers in the second and third quadrants
  - Correct: Plot the point on the Argand diagram to check the quadrant, then adjust the inverse tangent result by adding or subtracting $\pi$ as needed
- **Wrong:** Using degrees instead of radians for the argument in exponential form
  - Why it fails: Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ only holds when $\theta$ is measured in radians
  - Correct: Always convert any degree measure arguments to radians before writing exponential form
- **Wrong:** Only giving one root when asked for nth roots of a complex number
  - Why it fails: The equation $z^n = a$ has $n$ distinct solutions for any non-zero complex $a$, and exam markers require all solutions
  - Correct: Use the root formula for $k = 0, 1, ..., n-1$ to list all $n$ roots
- **Wrong:** Mixing up addition and subtraction of arguments for multiplication/division
  - Why it fails: Confusion between the geometric effects of multiplication and division
  - Correct: Remember: multiplication adds arguments (rotates the first number by the second's argument), division subtracts arguments
- **Wrong:** Claiming $z = 0$ has a specific principal argument when converting to polar form
  - Why it fails: 0 lies at the origin, so it has no defined direction or argument
  - Correct: Write 0 as $0 = 0(\cos\theta + i\sin\theta)$ for any $\theta$ — no specific argument is required

## Cheatsheet

| Operation | Polar Form | Exponential Form |
| --- | --- | --- |
| Conversion from Cartesian | $r = \sqrt{x^2+y^2}$, $\theta$ adjusted for quadrant: $z = r(\cos\theta + i\sin\theta)$ | $r = \sqrt{x^2+y^2}$, $z = r e^{i\theta}$ (θ in radians) |
| Multiplication | $z_1z_2 = r_1r_2(\cos(\theta_1+\theta_2)+i \sin(\theta_1+\theta_2))$ | $z_1z_2 = r_1r_2 e^{i(\theta_1+\theta_2)}$ |
| Division | $z_1/z_2 = (r_1/r_2)(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2))$ | $z_1/z_2 = (r_1/r_2) e^{i(\theta_1-\theta_2)}$ |
| nth Power | $z^n = r^n(\cos n\theta + i \sin n\theta)$ | $z^n = r^n e^{i n\theta}$ |
| nth Roots | $z_k = r^{1/n}(\cos\left(\frac{\theta+2k\pi}{n}\right)+i \sin\left(\frac{\theta+2k\pi}{n}\right)), k=0..n-1$ | $z_k = r^{1/n} e^{i \frac{\theta+2k\pi}{n}}, k=0..n-1$ |

## 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.

- [De Moivre's theorem](https://www.owlsprep.com/study/ib-math-aa-hl-u1-de-moivre-s-theorem/)
- [Systems of linear equations](https://www.owlsprep.com/study/ib-math-aa-hl-u1-systems-of-linear-equations/)

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