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