Hi everyone! Read through the material below, watch the videos.

Lesson 21: Complex Numbers

Topic: This lesson covers Chapter 21: Complex numbers.

WeBWorK: There are four WeBWorK assignments on today’s material:Complex Numbers – Operations

Complex Numbers – Magnitude

Complex Numbers – Direction

Complex Numbers – Polar Form

**Question of the Day: **What is the square root of $-1$?

#### Lesson Notes (Notability – pdf):

This .pdf file contains most of the work from the videos in this lesson. It is provided for your reference.

## Review of Complex Numbers

How do we get the complex numbers? We start with the real numbers, and we throw in something that’s missing: the square root of $-1$.

**Definition 21.1**. We define the **imaginary unit** or **complex unit** to be:

$$i=\sqrt{-1}$$

The most important property of $i$ is: $\quad i^2=-1$

**Definition 21.2**. A complex number is a number of the form $a+bi$.

$a$ and $b$ are allowed to be any real numbers. $a$ is called the **real part** of $a+bi$, and $b$ is called the **imaginary part** of $a+bi$. The complex numbers are referred to as $\mathbb{C}$ (just as the real numbers are $\mathbb{R}$.

We can picture the complex number $a+bi$ as the point with coordinates $(a,b)$ in the *complex plane*.

**Example 21.3**. Perform the operation.

a) $(2-3 i)+(-6+4 i)$

b) $(3+5 i) \cdot(-7+i)$

c) $\frac{5+4 i}{3+2 i}$

*VIDEO: Review of Complex Numbers* – Example 21.3

## Polar form

Next, we will look at how we can describe a complex number slightly differently – instead of giving the $x$ and $y$ coordinates, we will give a *distance* $r$ (the modulus) and *angle* $\theta$ (the argument). We call this the **polar form** of a complex number.

Many amazing properties of complex numbers are revealed by looking at them in **polar form**! Let’s learn how to convert a complex number $a+bi$ into polar form, and back again.

**Definition 21.4**. Let $a+bi$ be a complex number. The **absolute value** of $a+bi$, denoted by $|a+bi|$, is the distance between the point $a+bi$ in the complex plane and the origin $(0,0)$. By the Pythagorean Theorem, we can calculate the absolute value of $a+bi$ as follows:

$$ |a+bi|=\sqrt{a^2+b^2}$$

**Definition 21.6**. Let $a+bi$ be a complex number. The coordinates in the plane can be expressed in terms of the absolute value, or **modulus**, $r=|a+bi|$ and the angle, or **argument**, $\theta$ formed with the positive real axis (the $x$-axis) as shown in the diagram:

As shown in the diagram, the coordinates $a$ and $b$ are given by:

$a=r\cdot\cos(\theta), \text{ and } b=r\cdot\sin(\theta)$

Substituting and factoring out $r$, we can use these to express $a+bi$ in **polar form:**

**Polar form**: $a+bi = r\left(\cos(\theta) + i\cdot\sin(\theta)\right)$

How do we find the modulus $r$ and the argument $\theta$?

Note that $r$ is given by the absolute value. For $\theta$, we note that $\frac{b}{a}=\frac{r \cdot \sin (\theta)}{r \cdot \cos (\theta)}=\frac{\sin (\theta)}{\cos (\theta)}=\tan (\theta)$. This leads to the following:

**Formulas for converting to polar form (finding the modulus $r$ and argument $\theta$):** $r=\sqrt{a^2+b^2}$, $\tan(\theta)=\frac{b}{a}$

With regards to the modulus $\theta$, we can certainly use the inverse tangent function $\arctan\left(\frac{b}{a}\right)$. However, we have to be a little careful: since the arctangent only gives angles in Quadrants I and II, we need to doublecheck the quadrant of $(a,b)$.

If $\arctan\left(\frac{b}{a}\right)$ is in the correct quadrant then $\theta=\arctan\left(\frac{b}{a}\right)$. If not, then we add $\pi$ radians or $180^\circ$ to obtain the angle in the opposing quadrant: $\theta=\arctan\left(\frac{b}{a}\right)+\pi$, or $\theta=\arctan\left(\frac{b}{a}\right)+180^\circ$. You’ll see this in action in the following example.

**Example 21.7**. Convert the complex number to polar form.

a) $2+3 i$

b) $-2-2 \sqrt{3} i$

c) $4-3 i$

d) $-4 i$

*VIDEO: Converting complex numbers to polar form – Example 21.7*

**Example 21.8**. Convert the number from polar form into the standard form $a+bi$

a) $3 \cdot\left(\cos \left(117^{\circ}\right)+i \sin \left(117^{\circ}\right)\right)$

b) $4 \cdot\left(\cos \left(\frac{5 \pi}{4}\right)+i \sin \left(\frac{5 \pi}{4}\right)\right)$

*VIDEO: Converting complex numbers from polar form into standard form – Example 21.8*

## Multiplication and division of complex numbers in polar form

Why is polar form useful? The primary reason is that it gives us a simple way to picture how multiplication and division work in the plane. The proposition below gives the formulas, which may look complicated – but the *idea* behind them is simple, and is captured in these two slogans:

**When we multiply complex numbers:** we multiply the $r$s and add the $\theta$s.**When we divide complex numbers:** we divide the $r$s and subtract the $\theta$s

**Proposition 21.9**. Let $r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right)$ and $r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right)$ be two complex numbers in polar form. Then, the product and quotient of these are given by

$r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right) \cdot r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right) \ =r_{1} r_{2} \cdot\left(\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right)$

$\frac{r_{1}\left(\cos \left(\theta_{1}\right)+i \sin \left(\theta_{1}\right)\right)}{r_{2}\left(\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right)} =\frac{r_{1}}{r_{2}} \cdot\left(\cos \left(\theta_{1}-\theta_{2}\right)+i \sin \left(\theta_{1}-\theta_{2}\right)\right)$

**Example 21.10**. Multiply or divide the complex numbers, and write your answer in polar and standard form.

a) $5\left(\cos \left(11^{\circ}\right)+i \sin \left(11^{\circ}\right)\right) \cdot 8\left(\cos \left(34^{\circ}\right)+i \sin \left(34^{\circ}\right)\right)$

b) $\quad 3\left(\cos \left(\frac{5 \pi}{8}\right)+i \sin \left(\frac{5 \pi}{8}\right)\right) \cdot 12\left(\cos \left(\frac{7 \pi}{8}\right)+i \sin \left(\frac{7 \pi}{8}\right)\right)$

c) $\frac{32\left(\cos \left(\frac{\pi}{4}\right)+i \sin \left(\frac{\pi}{4}\right)\right)}{8\left(\cos \left(\frac{7 \pi}{12}\right)+i \sin \left(\frac{7 \pi}{12}\right)\right)}$

d) $\frac{4\left(\cos \left(203^{\circ}\right)+i \sin \left(203^{\circ}\right)\right)}{6\left(\cos \left(74^{\circ}\right)+i \sin \left(74^{\circ}\right)\right)}$

e) INTUITIVE BONUS: Without doing any calculation or conversion, describe where in the complex plane to find the number obtained by multiplying $(5+2i)(-1+6i)$.

*VIDEO: Multiplication and division of complex numbers in polar form – Example 21.10 *

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