# Multiplication and division in polar form

When multiplying two complex numbers $$z$$ and $$w$$, because $$|zw| = |z||w|$$, the modulus of the new complex number is just the product of the moduli.

We also saw that $$\arg(zw) = \arg(z) + \arg(w)$$, so the argument of a product of two complex numbers is just the sum of the arguments.

Theorem. Let $$z_1 = r_1 (\cos{\theta_1} + i \sin{\theta_1}) = r_1 e^{i \theta_1}$$ and $$z_2 = r_2 (\cos{\theta_2} + i \sin{\theta_2}) = r_2 e^{i \theta_2}$$. Then $$z_1 z_2 = r_1 r_2 (\cos(\theta_1+\theta_2) + i \sin(\theta_1+\theta_2)) = r_1 r_2 e^{i(\theta_1 + \theta_2)}.$$

In other words, when multiplying complex numbers in polar form, you “multiply the mods and add the args”. You have a similar statement for division.

Theorem. Let $$z_1 = r_1 (\cos{\theta_1} + i \sin{\theta_1}) = r_1 e^{i \theta_1}$$ and $$z_2 = r_2 (\cos{\theta_2} + i \sin{\theta_2}) = r_2 e^{i \theta_2}$$. Then $$\frac{z_1}{z_2} = \frac{r_1}{r_2} (\cos(\theta_1-\theta_2) + i \sin(\theta_1-\theta_2)) = r_1 r_2 e^{i(\theta_1 + \theta_2)}.$$

Notice that if you use the exponential form, this looks like you’re simply using index laws!

Worked exercise: Let $$z_1 = \sqrt{2}+\sqrt{2} i$$ and $$z_2 = \sqrt{3}+i$$. Write $$z_1$$ and $$z_2$$ in polar form, and hence find the polar form of $$z_1 z_2$$.
Solution: Plotting $$z_1$$ and $$z_2$$ on the complex plane will allow us to see that $$z_1 = 2 e^{i \frac{\pi}{4}}$$ and $$z_2 = 2 e^{i \frac{\pi}{6}}$$. Hence, $$z_1 z_2 = 4 e^{i \left(\frac{\pi}{4}+\frac{\pi}{6}\right)} = 4 e^{\frac{5\pi i}{12}}.$$

## Problems

1. Evaluate the following in polar form.
1. $$\left(3\cos\left( \frac{\pi}{4}\right) + 3i \sin\left( \frac{\pi}{4}\right)\right) \left(\sqrt{2}\cos\left( \frac{\pi}{6}\right) + \sqrt{2}i \sin\left( \frac{\pi}{6}\right)\right)$$
2. $$\displaystyle \frac{1+\sqrt{3} i}{1-i}$$
3. $$(\sqrt{3}-i)(\sqrt{2}+\sqrt{2}i)$$
4. $$\displaystyle 2e^{\frac{i\pi}{3}} \div \sqrt{2} e^{\frac{\pi i}{12}}$$
5. $$(-1-i)(1+i)$$
6. $$\left( \frac{\sqrt{3}}{2}-\frac{1}{2}i\right)(3-3i)$$