Exponential Form & Multiplying/Dividing Complex Numbers

The Maclaurin series for $\cos\theta$, $\sin\theta$ and $e^x$ (which you'll meet properly in Chapter 2) are:

$$\cos\theta=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots,\quad \sin\theta=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots,\quad e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

Substituting $x=i\theta$ into the series for $e^x$ and using $i^2=-1,\,i^3=-i,\,i^4=1$ to simplify each power:

$$e^{i\theta}=1+i\theta-\frac{\theta^2}{2!}-\frac{i\theta^3}{3!}+\frac{\theta^4}{4!}+\cdots=\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots\right)+i\left(\theta-\frac{\theta^3}{3!}+\cdots\right)$$

The bracketed real part is exactly the series for $\cos\theta$, and the bracketed imaginary part is exactly the series for $\sin\theta$. This gives Euler's relation:

$$e^{i\theta}=\cos\theta+i\sin\theta$$

Combining Euler's relation with the modulus-argument form gives the exponential form of a complex number:

$$z=re^{i\theta},$$

$$\text{where } r=|z|>0 \text{ and } \theta=\arg z \text{ satisfies } -\pi<\theta\leq\pi$$

As with the principal argument, $\theta$ is always chosen in this range unless a question says otherwise.

Exponential form turns the modulus-argument multiplication/division rules into ordinary laws of indices. If $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$:

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

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

In words: moduli multiply (or divide) as normal, while arguments simply add (or subtract) — the exponent behaves exactly like a real-number exponent, because $e^{i\theta_1}\cdot e^{i\theta_2}=e^{i(\theta_1+\theta_2)}$ is inherited directly from the ordinary law of indices $e^ae^b=e^{a+b}$, extended to complex exponents.