14.4 The exponential form of a complex number

Consider the Maclaurin series for $\cos(x)$ , $i\sin(x)$ and $e^{ix}$ . What happens when we add $\cos(x)+i\sin(x)$ ? Well a fair amount of algebra to start with! After that, however, we do get an interesting result, though.

	$\displaystyle\cos(x)+i\sin(x)$	$\displaystyle=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\frac{ix}{1% !}-\frac{ix^{3}}{3!}+\frac{ix^{5}}{5!}-\frac{ix^{7}}{7!}+...$
		$\displaystyle=1+\frac{ix}{1!}-\frac{x^{2}}{2!}-\frac{ix^{3}}{3!}+\frac{x^{4}}{% 4!}+\frac{ix^{5}}{5!}-\frac{x^{6}}{6!}-\frac{ix^{7}}{7!}+...$
		$\displaystyle=1+\frac{ix}{1!}+\frac{(ix)^{2}}{2!}+\frac{(ix)^{3}}{3!}+\frac{(% ix)^{4}}{4!}+\frac{(ix)^{5}}{5!}+\frac{(ix)^{6}}{6!}+\frac{(ix)^{7}}{7!}+...$
		$\displaystyle=e^{ix}$

This means that ⁴⁴ 4 This is sometimes referred to as ”Euler’s formula”

e^{ix}=\cos(x)+i\sin(x)

(14.51)

Which provides a link between trigonometry and complex numbers! This turns out to be very useful in proving trig identities.

For any complex number, where $r$ is the modulus and $\theta$ is the argument, we have

re^{\theta i}=\cos(\theta)+i\sin(\theta)

(14.52)

We can use this to write other complex numbers, such as $1+i$ in the form $re^{i\theta}$ .