Tag: complex numbers

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

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 $-1$?

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

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) $-4 i$

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

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

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 $(5+2i)(-1+6i)$.

That’s it for today! Give the WeBWorK a try.