Exponential Form (AQA A-Level Further Maths): Revision Notes

Exponential Form

Understanding exponential form

Exponential form is a powerful way to represent complex numbers using the exponential function. This form connects complex numbers with trigonometric functions through one of mathematics' most elegant formulas.

A complex number $z = a + bi$ can be written in modulus-argument form as:

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

where $r = |z|$ (the modulus) and $\theta = \arg z$ (the argument).

This representation can be further simplified using exponential form, which we'll explore in this note.

Deriving Euler's formula

To understand exponential form, we start with the Maclaurin series expansions for $\cos\theta$, $\sin\theta$, and $e^x$:

$\cos\theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + ...$

$\sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + ...$

Now, consider the expression $z = r\left[\left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - ...\right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - ...\right)\right]$

We can rearrange this by grouping terms:

$z = r\left(1 + i\theta - \frac{\theta^2}{2!} - \frac{\theta^3 i}{3!} + \frac{\theta^4}{4!} + \frac{\theta^5 i}{5!} - ...\right)$

Notice that this can be rewritten as:

$z = r\left(1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + ...\right)$

This works because $(i\theta)^2 = -\theta^2$, $(i\theta)^3 = -i\theta^3$, $(i\theta)^4 = \theta^4$, and so on.

The series we've obtained is the Maclaurin expansion of $e^{i\theta}$. This leads us to Euler's formula.

Euler's formula

Exponential form definition

Using Euler's formula, we can write any complex number in exponential form:

$z = re^{i\theta}$

where:

• $r = |z|$ is the modulus of $z$
• $\theta = \arg z$ is the argument of $z$, with $-\pi < \theta \leq \pi$

This is the exponential form (or Euler form) of a complex number.

Converting from exponential form to Cartesian form

To convert from exponential form $re^{i\theta}$ to Cartesian form $a + bi$, use Euler's formula. The key step is to expand using the relationship $e^{i\theta} = \cos\theta + i\sin\theta$, then evaluate the trigonometric functions.

Converting from Cartesian form to exponential form

To convert from Cartesian form $a + bi$ to exponential form $re^{i\theta}$, follow these steps:

1. Calculate the modulus: $r = |z| = \sqrt{a^2 + b^2}$
2. Sketch an Argand diagram to visualize the complex number
3. Find the argument: Use $\tan^{-1}\left(\frac{b}{a}\right)$ and adjust based on the quadrant
4. Ensure the argument is in the correct range: $-\pi < \theta \leq \pi$
5. Write in exponential form: $z = re^{i\theta}$

Trigonometric functions in exponential form

Euler's formula can be rearranged to express trigonometric functions in terms of exponentials. These formulas are incredibly useful for proving trigonometric identities.

Key formulas

$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$

$\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$

These follow directly from Euler's formula and its conjugate:

• $e^{i\theta} = \cos\theta + i\sin\theta$
• $e^{-i\theta} = \cos(-\theta) + i\sin(-\theta) = \cos\theta - i\sin\theta$

Adding these gives: $e^{i\theta} + e^{-i\theta} = 2\cos\theta$

Subtracting gives: $e^{i\theta} - e^{-i\theta} = 2i\sin\theta$

Strategy for proving trigonometric identities

Working with products and quotients in exponential form

When complex numbers are in exponential form, multiplication and division become straightforward:

Multiplication: If $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$, then:

$z_1z_2 = r_1r_2e^{i(\theta_1 + \theta_2)}$

Division: If $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$, then:

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

Summary