Perkalian dan pembagian dalam bentuk polar

Multiplying and Dividing Complex Numbers in Polar form

Since $re^{i\theta}$ and $r(\cos\theta+i\sin\theta)$ are both equal. We will look at multiplication and division of complex numbers in the form $re^{i\theta}$.

Lets say there are two complex numbers $z_1$, $z_2$ with magnitude $r_1$, $r_2$ and argument $\theta_1$, $\theta_2$ respectively.

Then, the two complex numbers can be expressed as: $r_1e^{i\theta_1}$ and $r_2e^{i\theta_2}$

So lets multiply them together!

$$\begin{aligned} z_1z_2&=\left(r_1e^{i\theta_1}\right)\left(r_2e^{i\theta_2}\right) \\&= r_1r_2e^{i\theta_1+i\theta_2} \\&= r_1r_2e^{i(\theta_1+\theta_2)} \end{aligned}$$

Alright, now lets try dividing the two complex numbers!

$$\begin{aligned} \frac{z_1}{z_2}&=\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}} \\&= \frac{r_1}{r_2}e^{i\theta_1-i\theta_2} \\&= \frac{r_1}{r_2}e^{i(\theta_1-\theta_2)} \end{aligned}$$

This holds even when we express the complex numbers in the form of $r(\cos\theta+i\sin\theta)$, but the proof of that is a bit more tedious.

$$\textbf{Multiplying and Dividing Complex Numbers in Polar Form}$$

When you are multiplying and dividing complex numbers in rectangular form, there is a lot of FOILing and algebraic manipulation, which may lead you to wonder whether there is a more efficient way to go about solving these problems. Luckily, we are also able to multiply and divide complex numbers in polar form, which can be a lot easier!

$$\textbf{Multiplying Complex Numbers in Polar Form}$$

When you multiply two complex numbers in polar form, $z_1=r_1(\text{cos}(\theta_1)+i\text{sin}(\theta_1))$ and $z_2=r_2(\text{cos}(\theta_2)+i\text{sin}(\theta_2)),$ you can use the following formula to solve for their product:

$$z_1z_2 = r_1r_2\big(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\big)$$

Let's do an example to practice using the formula!

Let's use our formula, $z_1z_2 = r_1r_2\big(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\big),$ to solve this problem. $$\begin{align} z_1z_2&=\bigg(\sqrt{2}(\text{cos}(\frac{\pi}{6})+ i\text{sin}(\frac{\pi}{6}))\bigg)\bigg(2\sqrt{2}(\text{cos}(\frac{\pi}{2})+ i\text{sin}(\frac{\pi}{2}))\bigg) \\ \\ &= (\sqrt{2})(2\sqrt{2})\bigg(\text{cos}(\frac{\pi}{6} + \frac{\pi}{2})+ i\text{sin}(\frac{\pi}{6} + \frac{\pi}{2})\bigg)\text{ by our formula}\\ \\ &= 4\bigg(\text{cos}(\frac{2\pi}{12} + \frac{6\pi}{12})+ i\text{sin}(\frac{2\pi}{12} + \frac{6\pi}{12})\bigg)\\ \\ &= 4\bigg(\text{cos}(\frac{8\pi}{12})+ i\text{sin}(\frac{8\pi}{12})\bigg)\\ \\ &= 4\bigg(\text{cos}(\frac{2\pi}{3})+ i\text{sin}(\frac{2\pi}{3})\bigg)\\ \\ &= 4\bigg(-\frac{1}{2}+ i(\frac{\sqrt{3}}{2})\bigg)\\ \\ &= -\frac{4}{2}+ i(\frac{4\sqrt{3}}{2})\\ \\ &= -2+ 2i\sqrt{3}\\ \end{align}$$

This method is a lot easier than converting complex numbers in polar form to complex numbers in rectangular form, and then using FOILing to find their product! We can also use a similar formula when we are dividing complex numbers in polar form.

$$\textbf{Dividing Complex Numbers in Polar Form}$$

When you divide two complex numbers in polar form, $z_1=r_1(\text{cos}(\theta_1)+i\text{sin}(\theta_1))$ and $z_2=r_2(\text{cos}(\theta_2)+i\text{sin}(\theta_2)),$ you can use the following formula so solve for their product, as long as $z_2 \neq 0:$

$$\frac{z_1}{z_2} = \frac{r_1}{r_2}\big(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\big)$$

Let's do another example to practice using this new formula!

Let's use our formula, $\frac{z_1}{z_2} = \frac{r_1}{r_2}\big(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\big),$ to solve this problem. $\big($First, it's important to notice that $z_2 \neq 0,$ which is why we can use this formula. If $z_2$ was $0,$ then $r_2$ would be $0,$ and we would not be able to calculate $\frac{r_1}{r_2}.\big)$ $$\begin{align} \frac{z_1}{z_2}&=\frac{\bigg(\sqrt{2}(\text{cos}(\frac{\pi}{6})+ i\text{sin}(\frac{\pi}{6}))\bigg)}{\bigg(2\sqrt{2}(\text{cos}(\frac{\pi}{2})+ i\text{sin}(\frac{\pi}{2}))\bigg)} \\ \\ &= \frac{\sqrt{2}}{2\sqrt{2}}\bigg(\text{cos}(\frac{\pi}{6} - \frac{\pi}{2})+ i\text{sin}(\frac{\pi}{6} - \frac{\pi}{2})\bigg)\text{ by our formula}\\ \\ &= \frac{1}{2}\bigg(\text{cos}(\frac{2\pi}{12} - \frac{6\pi}{12})+ i\text{sin}(\frac{2\pi}{12} - \frac{6\pi}{12})\bigg)\\ \\ &= \frac{1}{2}\bigg(\text{cos}(\frac{-4\pi}{12})+ i\text{sin}(\frac{-4\pi}{12})\bigg)\\ \\ &= \frac{1}{2}\bigg(\text{cos}(\frac{-\pi}{3})+ i\text{sin}(\frac{\pi}{3})\bigg)\\ \\ &= \frac{1}{2}\bigg(\frac{1}{2}+ i(-\frac{\sqrt{3}}{2})\bigg)\\ \\ &= \frac{1}{2} \cdot \frac{1}{2}+ i(\frac{1}{2}\cdot -\frac{\sqrt{3}}{2})\\ \\ &= \frac{1}{4} + \frac{-\sqrt{3}}{4}i \end{align}$$

Like with multiplying complex numbers, this method is a lot easier than converting two complex numbers from polar form to rectangular form, and then using FOILing and complex conjugation to find their quotient!

Multiplying and Dividing Complex Numbers in Polar Form

by Euler's Academy

In this video by Euler's Academy, you will learn how to multiply and divide complex numbers when they're written in polar form.

Summary

Try the first example on your own, then watch the video to make sure you understood correctly.

What is $\frac{w_1}{w_2}$?

$$w_1 = 60(\cos(30^{\circ})+ i \sin (30^{\circ}))$$ $$w_2 = 10(\cos(45^{\circ})+ i \sin (45^{\circ}))$$

In the previous lesson, we learned about the exponential form of a complex number.

Multiplying and Dividing Complex Numbers in Polar Form

Multiplying and dividing complex numbers in polar form is much easier than doing it in rectangular form. In rectangular form, you have to distribute and simplify using $i^2=-1$, but in polar form, you just have to work with the modulus and the angle.

A complex number in polar form looks like: $$z = r \text{cis} \theta$$

, which means:

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

Let's talk through what this looks like. Let $z_1 = r_1 \text{cis} \theta_1$ and $z_2 = r_2 \text{cis} \theta_2$.

Multiplication

To multiply two complex numbers in polar form, you multiply the $r$ values and add the angles.

$$z_1 \cdot z_1 = (r_1 \cdot r_2)\text{cis}(\theta_1 + \theta_2)$$

Division

To divide two complex numbers in polar form, you divide the $r$ values and subtract the angles.

$$\frac{z_1}{z_2} = \frac{r_1}{r_2} \text{cis} (\theta_1 - \theta_2)$$

This method makes it easier to work with rotations and magnitudes. You don't need to deal with real and imaginary parts! Multiplying stretches and rotates the complex number. Dividing shrinks and reverses the rotation.

In the next two lessons, you'll learn about roots of unity and De Moivre's Theorem.