5.1: Complex Numbers

Section Preview

Note to the Instructor (click to expand)

This required topic is needed for students to work with imaginary solutions of quadratic equations (coming up in the next section).

Core Prerequisites (i.e., "Understanding the Lesson"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are required to understand the core concepts in this lesson.

(click to expand)

• Sets and Numbers
  • Properties of Real Numbers

Operational Prerequisites (i.e., "Completing the Homework"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are needed to complete the homework.

(click to expand)

• Simplifying Expressions
  • Laws of Exponents — Product and Quotient Rules
  • Simplifying Products of Radical Expressions
  • Rationalizing Denominators and Numerators; Simplifying Quotients of Radical Expressions

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

• Express square roots of negative numbers as multiples of \(i\).
• Plot complex numbers on the complex plane.
• Add and subtract complex numbers.
• Multiply and divide complex numbers.

The study of Mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. For example, we still have no solution to equations such as \(x^2 +4=0\). Our best guesses might be \(2\) or \(-2\). But if we test \(2\) in this equation, it does not work. If we test \(-2\), it does not work. If we want to have a solution for this equation, we will have to go farther than we have so far. After all, to this point we have described the square root of a negative number as undefined. Fortunately, there is another system of numbers that provides solutions to problems such as these. In this section, we will explore this number system and how to work within it.

Expressing Square Roots of Negative Numbers as Multiples of \(i\)

We know how to find the square root of any positive real number. In a similar way, we can find the square root of a negative number. The difference is that the root is not real. We begin our exploration of this new beast by defining the unit imaginary number.

Definition: Unit Imaginary Number

The unit imaginary number, denoted as \(i\), is defined as the number whose square root is \( -1 \). That is,\[\sqrt{ −1 } = i. \nonumber \]

The unit imaginary number is often referred to as the imaginary unit. This definition leads directly to the following theorem.

Theorem: The Square of the Imaginary Unit

\[ i^2 = -1 \nonumber \]

Proof

\[ i^2 = (i)^2 =  \left( \sqrt{-1} \right)^2 = -1 \nonumber \]

We can write the square root of any negative number as a multiple of \(i\). Consider the square root of \(-25\).\[\sqrt{-25} =\sqrt{25 \cdot (-1)} =\sqrt{25} \sqrt{-1} =5i. \nonumber\]We use \(5i\) and not \(-5i\) because the principal root of \(25\) is the positive root.

By defining the imaginary unit, we have access to an entirely new type of number—the complex number.

Definition: Complex Number

A complex number is a number that can be written in the form \(a+bi\) where \(a\) is called the real part and \(b\) is called the imaginary part of the complex number. If \(b=0\), then \(a+bi\) is simply a real number. If \(a=0\) and \(b \neq 0 \), the complex number is called an imaginary number. The form, \( a + bi \) is called the standard form of the complex number.

Complex numbers are a combination of real and imaginary numbers. For example, \(5+2i\) is a complex number. The real part is \(5\), and it happens that \(5\) is a real number. The imaginary part is \(2\), and \(2i\) is called an imaginary number.

Figure \(\PageIndex{1}\)

Recall, when a positive real number is squared, the result is a positive real number and when a negative real number is squared, again, the result is a positive real number. Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative real number!

If we were to gather all of the complex numbers in existence, we would have a completely new number system.

Definition: Complex Number System

The set of all numbers of the form \( a + bi \) is called the complex number system (symbolically denoted as \( \mathbb{C} \)).

Example \(\PageIndex{1}\): Expressing an Imaginary Number in Standard Form

Express \(\sqrt{-9}\) in standard form.

Solution

\[\sqrt{-9} =\sqrt{9} \sqrt{-1} =3i \nonumber\]In standard form, this is \(0+3i\).

Checkpoint \(\PageIndex{1}\)

Express \(\sqrt{-24}\) in standard form.

Plotting a Complex Number on the Complex Plane

We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number we need to address the two components of the number. Two components means we need two number lines—one for the real part and one for the imaginary part. Thus is borne the complex plane.

Definition: Complex Plane

The complex plane is a coordinate system in which the horizontal axis represents the real part of a complex number and the vertical axis represents the imaginary part. In this sense, we define the horizontal axis to be the real axis and the vertical axis to be the complex axis. Complex numbers are the points on the plane, expressed as ordered pairs \((a,b)\), where \(a\) represents the coordinate for the horizontal axis and \(b\) represents the coordinate for the vertical axis.

Let's consider the number \(-2+3i\). The real part of the complex number is \(-2\) and the imaginary part is \(3\). We plot the ordered pair \((-2,3)\) to represent the complex number \(-2+3i\) as shown in Figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\)

Example \(\PageIndex{2}\): Plotting a Complex Number on the Complex Plane

Plot the complex number \(3-4i\) on the complex plane.

Solution

The real part of the complex number is \(3\), and the imaginary part is \(-4\). We plot the ordered pair \((3,-4)\) as shown in Figure \(\PageIndex{3}\).

Figure \(\PageIndex{3}\)

Interactive Element: A Complex Number as a Point

The following Interactive Element might help you visualize how a complex number \(a+bi\) corresponds to a single point in the complex plane.

Interact: Drag the red point anywhere in the plane.

Observation: The horizontal position gives the real part and the vertical position gives the imaginary part, so \(a+bi\) is exactly the point \((a,b)\). Land on the horizontal axis and the number is purely real; land on the vertical axis and it is purely imaginary.

Checkpoint \(\PageIndex{2}\)

Plot the complex number \(-4-i\) on the complex plane.

Adding and Subtracting Complex Numbers

Just as with real numbers, we can perform arithmetic operations on complex numbers; however, we need to define what it means to add and subtract complex numbers.

Definition: Addition and Subtraction of Complex Numbers

Given two complex numbers, \( z_1 = a + bi \) and \( z_2 = c + di \), we define addition and subtraction as follows:

Addition:\[ z_1 + z_2 = ( a+bi )+( c+di )=( a+c )+( b+d )i. \nonumber \]Subtraction:\[z_1 - z_2 = ( a+bi )−( c+di )=( a−c )+( b−d )i. \nonumber \]

That is, to add or subtract complex numbers, we combine the real parts and combine the imaginary parts.

Example \(\PageIndex{3}\): Adding Complex Numbers

Add \(3-4i\) and \(2+5i\).

Solution

We add the real parts and add the imaginary parts.\[\begin{array}{rclcl}
(a + bi) + (c + di) & = & (a + c) + (b + d)i & \quad & \left( \text{regrouping} \right) \\[6pt]
& = & (3 + 2) + (-4 + 5)i & \quad & \left( \text{substituting} \right) \\[6pt]
& = & 5 + i & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber\]

Checkpoint \(\PageIndex{3}\)

Subtract \(2+5i\) from \(3-4i\).

Multiplying Complex Numbers

When defining multiplication of complex numbers, we borrow heavily from binomial products.

Definition: Multiplication of Complex Numbers

Given two complex numbers, \( z_1 = a + bi \) and \( z_2 = c + di \), we define multiplication as follows:\[ \begin{array}{rcl}
z_1 \cdot z_2 & = & (z_1)(z_2) \\[6pt]
& = & ( a+bi )( c+di ) \\[6pt]
& = & ac + ad i + bc i + bd i^2 \\[6pt]
& = & ac +(ad + bc)i + bc (-1) \\[6pt]
& = & (ac - bd) + (ad + bc)i \\[6pt]
\end{array} \nonumber \]

While it might be tempting, you should never memorize the result of this definition. Instead, treat multiplying complex numbers like multiplying binomials.

Let's begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,

Figure \(\PageIndex{4}\)

Example \(\PageIndex{4}\): Multiplying a Complex Number by a Real Number

Find the product \(4(2+5i)\).

Solution

Distribute the 4.\[\begin{array}{rclcl}
4(2 + 5i) & = & (4 \cdot 2) + (4 \cdot 5i) & \quad & \left( \text{distributing} \right) \\[6pt]
& = & 8 + 20i & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber\]

Checkpoint \(\PageIndex{4}\)

Find the product \(-4(2+6i)\).

Example \(\PageIndex{5}\): Multiplying a Complex Number by a Complex Number

Multiply \((4+3i)(2-5i)\).

Solution

\[\begin{array}{rclcl}
(4+3i)(2-5i) & = & (4 \cdot 2) - (4 \cdot 5i) + (3i \cdot 2) - (3i \cdot 5i) & \quad & \left( \text{distributing} \right) \\[6pt]
& = & 8 - 20i + 6i - 15 i^2 & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & 8 - 14i - 15 (-1) & \quad & \left( \text{combining like terms, and }i^2 = -1 \right) \\[6pt]
& = & 8 - 14i + 15 & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & 23 - 14i & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber\]

Checkpoint \(\PageIndex{5}\)

Multiply \((3-4i)(2+3i)\).

Dividing Complex Numbers

When building a definition for division of complex numbers, we must come to some agreement concerning allowable forms of complex numbers. At this time, a complex number in standard form must look like \(a + b i\), where both \(a\) and \(b\) are real numbers. Thus, it is safe to claim that \(a\) and \(b\), if fractions, cannot have complex numbers in their denominators. Said differently, we require that any fraction must have a real-number denominator. This is so important, it bears repeating.

Any fraction must have a real-valued denominator.

Therefore, when dividing a complex number by another complex number, we need to find a number by which we can multiply the numerator and the denominator that will force the denominator to be a pure real number. That is, given \(z_1 = a + bi\) and \(z_2 = c + di\), we need to find a number, let's call this \(k\) for now, such that\[\dfrac{z_1}{z_2} = \dfrac{a + bi}{c + di} = \dfrac{(a + bi)}{(c + di)} \cdot \dfrac{k}{k} = \dfrac{k(a + bi)}{k(c + di)}, \nonumber\]where \(k(c + di)\) is a pure real number.

A great choice for \(k\) is \(c - di\). This is because\[k(c + di) = (c - di)(c + di) = c^2 + cdi - cdi - d^2 i^2 = c^2 + \cancel{cdi} - \cancel{cdi} - d^2(-1) = c^2 + d^2. \nonumber\]Since both \(c\) and \(d\) are real numbers, this result is a real number. Our choice for \(k\) has a special name (and is incredibly important as we move through the rest of Mathematics).

Definition: Complex Conjugate

The complex conjugate of \( z = a + bi \) is defined to be \( \bar{z} = a - bi \). Likewise, the complex conjugate of \( z = a - bi \) is defined to be \( \bar{z} = a + bi \).

Prior to this definition, we proved the following theorem.

Theorem: Products of Complex Conjugates

Let \( z = a + bi \), so that its complex conjugate is \( \bar{z} = a - bi \). Then\[ z \cdot \bar{z} = a^2 + b^2. \nonumber \]

Proof

Let \( z = a + bi \). Then\[ \begin{array}{rcl} z \cdot \bar{z}& = & (a + bi)(a - bi) \\[6pt]  & = & a^2 - abi + abi - b^2 i^2 \\[6pt]  & = & a^2 - \cancel{abi} + \cancel{abi} - b^2 (-1) \\[6pt]  & = & a^2 + b^2 \\[6pt] \end{array} \nonumber \]

We now have everything we need to create a definition for division of complex numbers that will result in denominators being real numbers.

Definition: Division of Complex Numbers

Given two complex numbers, \( z_1 = a + bi \) and \( z_2 = c + di \), we define division as follows:\[ \begin{array}{rcl}
z_1 \div z_2 & = & \dfrac{z_1}{z_2} \\[6pt]
& = & \dfrac{a+bi}{c+di} \\[6pt]
& = & \dfrac{a+bi}{c+di} \cdot \dfrac{c - di}{c - di} \\[6pt]
& = & \dfrac{(a+bi)(c - di)}{(c+di)(c - di)} \\[6pt]
& = & \dfrac{ac -ad i + bc i - bd i^2}{c^2+ d^2} \\[6pt]
& = & \dfrac{ac + (-ad + bc) i - bd (-1)}{c^2+ d^2} \\[6pt]
& = & \dfrac{(ac + bd) + (bc - ad) i}{c^2+ d^2} \\[6pt]
& = & \dfrac{ac + bd}{c^2 + d^2} + \dfrac{bc - ad}{c^2 + d^2} i \\[6pt]
& = & r + si \\[6pt]
\end{array} \nonumber \]where \( r \) and \( s \) are real numbers.

As was mentioned with multiplication of complex numbers, it is best not to memorize this resulting formula. You should be able to just "do the math."

Example \(\PageIndex{6}\): Finding Complex Conjugates

Find the complex conjugate of each number.

1. \(2+i\sqrt{5}\)
2. \(-\frac{1}{2} i\)

Solution

1. The number is already in the form \(a+bi\). The complex conjugate is \(a-bi\), or \(2-i\sqrt{5}\).
2. We can rewrite this number in the form \(a+bi\) as \(0- \frac{1}{2} i\). The complex conjugate is \(a-bi\), or \(0+ \frac{1}{2} i\). This can be written simply as \(\frac{1}{2} i\).

Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by \(i\).

Example \(\PageIndex{7}\): Dividing Complex Numbers

Divide \((2+5i)\) by \((4-i)\).

Solution

We begin by writing the problem as a fraction:\[\dfrac{2+5i}{4-i}. \nonumber\]Then we multiply the numerator and denominator by the complex conjugate of the denominator:\[\dfrac{(2+5i)}{(4-i)} \cdot \dfrac{(4+i)}{(4+i)}. \nonumber\]To multiply two complex numbers, we expand the product as we would with polynomials.\[\begin{array}{rclcl}
\dfrac{(2+5i)}{(4-i)} \cdot \dfrac{(4+i)}{(4+i)} & = & \dfrac{8 + 2i + 20i + 5i^2}{(4)^2 + (-1)^2} & \quad & \left( \text{distributing and using the fact that }(a + bi)(a - bi) = a^2 + b^2 \right) \\[6pt]
& = & \dfrac{8 + 22i - 5}{16 + 1} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{3 + 22i}{17} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{3}{17} + \dfrac{22}{17}i & \quad & \left( \text{separating the real and imaginary parts} \right) \\[6pt]
\end{array} \nonumber\]Note that this expresses the quotient in standard form.

Example \(\PageIndex{8}\): Substituting a Complex Number into a Polynomial Function

Let \(f(x)= x^2 -5x+2\). Evaluate \(f(3+i)\).

Solution

\[\begin{array}{rclcl}
f(3 + i) & = & (3 + i)^2 - 5(3 + i) + 2 & \quad & \left( \text{substituting} \right) \\[6pt]
& = & (9 + 6i + i^2) - 15 - 5 i + 2 & \quad & \left( \text{distributing} \right) \\[6pt]
& = & 9 + 6i + (-1) - 15 - 5 i + 2 & \quad & \left( i^2 = -1 \right) \\[6pt]
& = & -5 + i & \quad & \left( \text{combining like terms} \right) \\[6pt]
\end{array} \nonumber\]

Checkpoint \(\PageIndex{8}\)

Let \(f(x)=2 x^2 -3x\). Evaluate \(f(8-i)\).

Example \(\PageIndex{9}\): Substituting an Imaginary Number in a Rational Function

Let \(f(x)= \frac{2+x}{x+3}\). Evaluate \(f(10i)\).

Solution

\[\begin{array}{rclcl}
f(10i) & = & \dfrac{2 + 10i}{10i + 3} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{2 + 10i}{3 + 10i} & \quad & \left( \text{rewriting the denominator in standard form} \right) \\[6pt]
& = & \dfrac{(2 + 10i)}{(3 + 10i)} \cdot \dfrac{(3 - 10i)}{(3 - 10i)} & \quad & \left( \text{multiplying by the conjugate of the denominator} \right) \\[6pt]
& = & \dfrac{6 - 20i + 30i - 100i^2}{(3)^2 + (10)^2} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{6 - 20i + 30i - 100(-1)}{(3)^2 + (10)^2} & \quad & \left( i^2 = -1 \right) \\[6pt]
& = & \dfrac{106 + 10i}{109} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{106}{109} + \dfrac{10}{109}i & \quad & \left( \text{separating the real and imaginary parts} \right) \\[6pt]
\end{array} \nonumber\]

Checkpoint \(\PageIndex{9}\)

Let \(f(x)= \frac{x+1}{x-4}\). Evaluate \(f(-i)\).

Simplifying Powers of \(i\)

The powers of \(i\) are cyclic. Let's look at what happens when we raise \(i\) to increasing powers.\[\begin{array}{rcl}
i^1 & = & i \\[6pt]
i^2 & = & -1 \\[6pt]
i^3 & = & i^2 \cdot i =-1 \cdot i =-i \\[6pt]
i^4 & = & i^3 \cdot i = -i \cdot i =- i^2 = -(-1) = 1 \\[6pt]
i^5 & = & i^4 \cdot i =1 \cdot i = i \\[6pt]
\end{array} \nonumber\]We can see that when we get to the fifth power of \(i\), it is equal to the first power. As we continue to multiply \(i\) by itself for increasing powers, we will see a cycle of 4. Let's examine the next 4 powers of \(i\).\[\begin{array}{rcl}
i^6 & = & i^5 \cdot i = i \cdot i = i^2 = -1 \\[6pt]
i^7 & = & i^6 \cdot i = i^2 \cdot i = i^3 = -i \\[6pt]
i^8 & = & i^7 \cdot i =i^3 \cdot i = i^4 = 1 \\[6pt]
i^9 & = & i^8 \cdot i = i^4 \cdot i = i^5 = i \\[6pt]
\end{array} \nonumber\]

Theorem: Powers of the Imaginary Unit

Let \( p \in \mathbb{W}\) (the Whole Number system) and \( r \) be the remainder when \( p \) is divided by 4. Then\[ i^p = i^r. \nonumber \]

Proof

Let \( p \in \mathbb{W} \) and \( r \) be the remainder when \( p \) is divided by 4. Then \( r \) is either 0, 1, 2, or 3. The Divison Algorithm tells us that\[ \dfrac{p}{4} = q + \dfrac{r}{4}, \nonumber \]where \( q \) is the quotient and \( r \) is the remainder. Multiplying both sides by 4, we get\[ p = 4q + r. \nonumber \]Hence,\[ \begin{array}{rclcl}
i^p & = & i^{4q + r} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & i^{4q} \cdot i^r & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
& = & \left(i^4\right)^q \cdot i^r & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
& = & \left(1\right)^q \cdot i^r & \quad & \left( i^4 = 1 \right) \\[6pt]
& = & 1 \cdot i^r &  & \\[6pt]
& = & i^r &  & \\[6pt]
\end{array} \nonumber \]Therefore, all whole-numbered powers of \( i \) can be simplified to\[ \begin{array}{rcl}
i^0 & = & 1 \\[6pt]
i^1 & = i \\[6pt]
i^2 & = & -1 \\[6pt]
i^3 & = & i^2 \cdot i = (-1) \cdot i = -i \\[6pt]
\end{array} \nonumber \]

Example \(\PageIndex{10}\): Simplifying Powers of \(i\)

Evaluate \(i^{35}\).

Solution

Since \(i^4 =1\), we can simplify the problem by factoring out as many factors of \(i^4\) as possible. To do so, first determine how many times 4 goes into 35: \(35=4 \cdot 8+3\). Therefore,\[i^{35} = i^{4 \cdot 8+3} = i^{4 \cdot 8} \cdot i^3 = (i^4)^8 \cdot i^3 = 1^8 \cdot i^3 = i^3 = -i. \nonumber\]

Q&A

Can we write \(i^{35}\) in other helpful ways?

As we saw in Example \(\PageIndex{10}\), we reduced \(i^{35}\) to \(i^3\) by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of \(i^{35}\) may be more useful. Table \(\PageIndex{1}\) shows some other possible factorizations.

Table \(\PageIndex{1}\)

Factorization of \(i^{35}\)	\(i^{34} \cdot i\)	\(i^{33} \cdot i^{2}\)	\(i^{31} \cdot i^4\)	\(i^{19} \cdot i^{16}\)
Reduced form	\((i^2)^{17} \cdot i\)	\(i^{33} \cdot (-1)\)	\(i^{31} \cdot 1\)	\(i^{19} \cdot (i^4)^4\)
Simplified form	\((-1)^{17} \cdot i\)	\(- i^{33}\)	\(i^{31}\)	\(i^{19}\)

Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.

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Success in College: Study Time

Believe it or not, there is a direct link between the number of units a class is worth and how many hours you should be studying (outside of class). For each hour of class time, the California state education code suggests at least two hours of studying outside of class. If your math background is weak, you may need to spend more time studying.

At Cosumnes River College, each unit of a math course corresponds to one hour of in-class time per week. Thus, a five-unit math course requires five hours of in-class interaction per week and ten hours of study time outside of class per week.

One way to study is to do what you're doing now - read! Class time is an excellent opportunity to introduce new concepts and see how they fit with those previously learned; however, there is usually not enough time to address details with the depth that a textbook can. In this way, your professor and the textbook should be considered as a combined primary source of information in your course.

Here are some questions to consider regarding how you study.

1. Do you study at least two hours for every one hour of class time? If not, why?
2. How many hours do you dedicate to TV, YouTube, and social media every week (on average)?
3. How many hours do you dedicate to leisure activities per week (e.g., hanging out with friends, going to the movies, gaming)?
4. Would it be too difficult to trade a few hours for study time?

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MyOpenMath Advice: Reading Solutions to Exercises

Some problems have complete worked-out solutions. If your feedback has a comment like, "Please see the solutions," then you should probably look through these solutions. You can see solutions by going into the Grades area, finding the assignment, and clicking "Show Detailed Solution" on an exercise with a solution. Solutions are generally not shown until after the due date has passed.

I usually leave a comment of the form, "Please see the written solutions," if there were too many mistakes for me to cover by typing or if I could not follow your logic.

- Professor Simpson, Cosumnes River College