Although square roots of negative numbers such as \(\sqrt{-4}\) are not real numbers, they occur often in mathematics and its applications.
Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. René Descartes gave them the name imaginary numbers, which reflected the mistrust with which mathematicians regarded them at the time. Today, however, such numbers are well understood and used routinely by scientists and engineers.
We begin by defining a new number, \(i\text{,}\) whose square is \(-1\text{.}\)
Imaginary Unit.
We define the imaginary unit \(i\) by
\begin{equation*}
i^2=-1 ~~~\text{ or }~~~ i=\sqrt{-1}
\end{equation*}
Caution7.39.
The letter \(i\) used in this way is not a variable; it is the name of a specific number and hence is a constant.
The two solutions are \(1 + 2i\) and \(1 - 2i\text{.}\) These numbers are examples of complex numbers.
Complex Numbers.
A complex number can be written in the form \(a+bi\text{,}\) where \(a\) and \(b\) are real numbers.
Examples of complex numbers are
\begin{equation*}
3-5i\text{, }~~~2+\sqrt{7}i\text{, }~~~\frac{4-i}{3}\text{, }~~~6i\text{, }~~~\text{ and } -9
\end{equation*}
In a complex number \(a+bi\text{,}\)\(a\) is called the real part, and \(b\) is called the imaginary part. All real numbers are also complex numbers (with the imaginary part equal to zero). A complex number whose real part equals zero is called a pure imaginary number.
Example7.44.
Write the solutions to Example 7.37, \(\dfrac{2\pm\sqrt{-4}}{2}\text{,}\) as complex numbers.
Because \(\sqrt{-4}=\sqrt{-1}\sqrt{4}=2i\text{,}\) we have \(\dfrac{2\pm\sqrt{-4}}{2}=\dfrac{2\pm2i}{2}\text{,}\) or \(1\pm i\text{.}\) The solutions are \(1+i\) and \(1-i\text{.}\)
Checkpoint7.45.Practice 3.
Use extraction of roots to solve \((2x + 1)^2 + 9 = 0\text{.}\) Write your answers as complex numbers.
All the properties of real numbers listed in Algebra Skills Refresher Section A.13 are also true of complex numbers. We can carry out arithmetic operations with complex numbers.
Checkpoint7.46.QuickCheck 2.
Which statement about complex numbers is false?
A) A complex number has a real part and an imaginary part.
B) We can perform all four arithmetic operations on complex numbers.
C) Every quadratic equation has solutions in the complex numbers.
D) Complex numbers cannot be combined with real numbers.
You can verify that in general the following rule holds.
Product of Complex Numbers.
\begin{equation*}
(a + bi)(c + di) = (ac - bd) + (ad + bc)i
\end{equation*}
Caution7.51.
One property of real numbers that is not true of complex numbers is \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\text{.}\) This identity fails when \(a\) and \(b\) are both negative. For example, if \(a = b = -2\text{,}\) we have
We can avoid possible errors by writing square roots of negative numbers as imaginary numbers.
Checkpoint7.52.Pause and Reflect.
How is multiplying complex numbers similar to multiplying binomials?
SubsectionQuotients of Complex Numbers
To find the quotient of two complex numbers, we use the technique of rationalizing the denominator. (See Algebra Skills Refresher Section A.10.)
For example, consider the quotient
\begin{equation*}
\frac{3+4i}{2i}
\end{equation*}
Because \(i\) is really a radical (remember that \(i =\sqrt{-1})\text{,}\) we multiply the numerator and denominator of the quotient by \(\alert{i}\) to obtain
\begin{align*}
\frac{(3+4i)\cdot \alert{i}}{2i\cdot\alert{i}}\amp =\frac{3i+4i^2}{2i^2} \amp\amp \blert{\text{Apply the distributive law to the numerator.}}\\
\amp=\frac{3i-4}{-2}\amp\amp \blert{\text{Recall that }i^2=-1.}
\end{align*}
To write the quotient in the form \(a + bi\text{,}\) we divide \(-2\) into each term of the numerator to get
We multiply numerator and denominator by \(\alert{i}\text{.}\)
\begin{align*}
\frac{10-15i}{5i}\amp=\frac{(10-15i)\cdot \alert{i}}{5i\cdot \alert{i}}\amp\amp\\
\amp=\frac{10i-15i^2}{5i^2}\amp\amp \blert{\text{Replace }i^2 \text{ by } -1.}\\
\amp=\frac{10i+15}{-5}\amp\amp\\
\amp=\frac{10i}{-5}+\frac{15}{-5}\amp\amp \blert{\text{Divide }-5 \text{ into each term of numerator.}} \\
\amp=-2i-3
\end{align*}
If \(z = a + bi\) is any nonzero complex number, then the number \(\overline{z} = a - bi\) is called the complex conjugate of \(z\text{.}\) The product of a nonzero complex number and its conjugate is always a positive real number.
We use this fact to find the quotient of complex numbers. If the divisor has both a real and an imaginary part, we multiply numerator and denominator by the conjugate of the denominator.
Checkpoint7.55.QuickCheck 3.
What is the product of \(z=a+bi\) with its complex conjugate?
Explain how to divide one complex number by another.
SubsectionZeros of Polynomials
Because we can add, subtract, and multiply any two complex numbers, we can use a complex number as an input for a polynomial function. Thus, we can extend the domain of any polynomial to include all complex numbers.
Example7.59.
Evaluate the polynomial \(f(x) = x^2 - 2x + 2\) for \(x = 1 + i\text{,}\) then simplify.
In Chapter 6, we learned that irrational solutions of quadratic equations occur in conjugate pairs,
\begin{equation*}
x=\frac{-b}{2a}+\frac{\sqrt{b^2-4ac}}{2a} ~~\text{ and }~~ x=\frac{-b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}
\end{equation*}
If the discriminant \(D = b^2 - 4ac\) is negative, the two solutions are complex conjugates,
\begin{equation*}
z=\frac{-b}{2a}+\frac{i\sqrt{\abs{D}}}{2a} ~~\text{ and }~~ \overline{z}=\frac{-b}{2a}-\frac{i\sqrt{\abs{D}}}{2a}
\end{equation*}
Thus, if we know that \(z\) is a complex solution of a quadratic equation, we know that \(\overline{z}\) is the other solution. The quadratic equation with solutions \(z\) and \(\overline{z}\) is
The other solution of the equation is \(\overline{z} = 7 + 5i\text{,}\) and the equation is \((x-z)(x - \overline{z}) = 0\text{.}\) We expand the product to find
One of the most important results in mathematics is the fundamental theorem of algebra, which says that if we allow complex numbers as inputs, then every polynomial \(p(x)\) of degree \(n\ge 1\) has exactly \(n\) complex number zeros.
Fundamental Theorem of Algebra.
Let \(p(x)\) be a polynomial of degree \(n\ge 1\text{.}\) Then \(p(x)\) has exactly \(n\) complex zeros.
As a result, the factor theorem tells that every polynomial of degree \(n\) can be factored as the product of \(n\) linear terms. For example, although the graph of \(y = x^4 + 4\) shown at right has no \(x\)-intercepts, the fundamental theorem tells us that there are four complex solutions to \(x^4 + 4 = 0\text{,}\) and that \(x^4 + 4\) can be factored.
You can check that the four solutions to \(x^4 + 4 = 0\) are \(1 + i\text{,}\)\(-1 + i\text{,}\)\(-1 - i\) , and \(1 - i\text{.}\) For example, if \(x = 1 + i\text{,}\) then
Because each zero corresponds to a factor of the polynomial, the factored form of \(x^4 + 4\) is
\begin{equation*}
x^4 + 4 = \left[x - (1 + i )\right]\left[x - (-1 + i )\right]\left[x - (-1 - i )\right]\left[x - (1 - i )\right]
\end{equation*}
Note7.63.
The four solutions to \(x^4 + 4 = 0\) form two complex conjugate pairs, namely \(1\pm i\) and \(-1\pm i\text{.}\) In fact, for every polynomial with real coefficients, the nonreal zeros always occur in complex conjugate pairs.
Checkpoint7.64.QuickCheck 4.
If \(z=3+5i\) is a solution to a polynomial equation, then so is:
What can you say about the graph of a quadratic function whose zeros are complex numbers?
SubsectionGraphing Complex Numbers
Real numbers can be plotted on a number line, but to graph a complex number we use a plane, called the complex plane. In the complex plane, the real numbers lie on the horizontal or real axis, and pure imaginary numbers lie on the vertical or imaginary axis.
To plot a complex number a + bi, we move \(a\) units from the origin in the horizontal direction and \(b\) units in the vertical direction. The numbers \(2 + 3i\) and \(2 - 3i\) are plotted at left.
Example7.68.
Plot the numbers \(z\text{,}\)\(\overline{z}\text{,}\)\(-z\text{,}\) and \(-\overline{z}\) as points on the complex plane, for \(z = 2 - 2i\text{.}\)
If we draw an arrow from the origin to the point \(a + bi\) in the complex plane, we can see that multiplication by \(i\) corresponds to rotating a point around the origin by \(90\degree\) in the counterclockwise direction. For example, the figure at right shows the graphs of \(z = 2 + 3i\) and \(i z = 2i - 3\text{.}\)
SubsectionSection Summary
SubsubsectionVocabulary
Look up the definitions of new terms in the Glossary.
Imaginary unit
Imaginary number
Complex number
Imaginary axis
Complex conjugate
Real part
Imaginary part
Complex plane
Real axis
SubsubsectionCONCEPTS
The square root of a negative number is an imaginary number.
A complex number is the sum of a real number and an imaginary number.
We can perform the four arithmetic operations on complex numbers.
The product of a nonzero complex number and its conjugate is always a positive real number.
Fundamental Theorem of Algebra.
Let \(p(x)\) be a polynomial of degree \(n\ge 1\text{.}\) Then \(p(x)\) has exactly \(n\) complex zeros.
The nonreal zeros of a polynomial with real coefficients always occur in conjugate pairs.
We can graph complex numbers in the complex plane.
Multiplying a complex number by \(i\) rotates its graph by \(90\degree\) around the origin.
SubsubsectionSTUDY QUESTIONS
What are imaginary numbers, and why were they invented?
Explain how the complex conjugate is used in dividing complex numbers.
If one solution of a quadratic equation is \(3 + i\sqrt{2}\text{,}\) what is the other solution?
If \(P(x)\) is a polynomial of degree \(7\text{,}\) how many zeros does \(P(x)\) have? How many \(x\)-intercepts could its graph have? How many complex zeros could \(P(x)\) have?
SubsubsectionSKILLS
Practice each skill in the Homework problems listed.
Write and simplify complex numbers: #1–10
Perform arithmetic operations on complex numbers: #11–36
Evaluate polynomials at complex numbers, expand polynomials: #37–48
Find a polynomial with given zeros: #53–56, 59–62
Graph complex numbers: #63–70
ExercisesHomework 7.3
Exercise Group.
For Problems 1–6, write the complex number in the form \(a+bi\text{,}\) where \(a\) and \(b\) are real numbers.
1.
\(\sqrt{-25}-4 \)
2.
\(\sqrt{-9}+3 \)
3.
\(\dfrac{-8+\sqrt{-4}}{2} \)
4.
\(\dfrac{6-\sqrt{-36}}{2} \)
5.
\(\dfrac{-5-\sqrt{-2}}{6} \)
6.
\(\dfrac{7+\sqrt{-3}}{4} \)
Exercise Group.
For Problems 7–10, find the zeros of the quadratic polynomial. Write each in the form \(a+ bi\text{,}\) where \(a\) and \(b\) are real numbers.
For Problems 37–42, evaluate the polynomial for the given values of the variable.
37.
\(z^2+9\)
\(\displaystyle z=3i\)
\(\displaystyle z=-3i\)
38.
\(2y^2 - y - 2\)
\(\displaystyle y=2-i\)
\(\displaystyle y=-2-i\)
39.
\(x^2 - 2x + 2\)
\(\displaystyle x=1-i\)
\(\displaystyle x=1+i\)
40.
\(3w^2 +5\)
\(\displaystyle w=2i\)
\(\displaystyle w=-2i\)
41.
\(q^2 +4q + 13\)
\(\displaystyle q = -2 + 3i\)
\(\displaystyle q = -2 - 3i\)
42.
\(v^2 + 2v + 3\)
\(\displaystyle v = 1 + i\)
\(\displaystyle v=-1 + i\)
Exercise Group.
For Problems 43–48, expand each product of polynomials.
43.
\((2z + 7i )(2z - 7i )\)
44.
\((5w + 3i )(5w - 3i )\)
45.
\([x + (3 + i )][x + (3 - i )]\)
46.
\([s - (1 + 2i )][s - (1 - 2i )]\)
47.
\([v - (4 + i )][v - (4 - i )]\)
48.
\([Z + (2 + i )][Z + (2 - i )]\)
49.
For what values of \(x\) will \(\sqrt{x-5} \) be real? Imaginary?
50.
For what values of \(x\) will \(\sqrt{x+3} \) be real? Imaginary?
51.
Simplify.
\(\displaystyle i^6\)
\(\displaystyle i^{12}\)
\(\displaystyle i^{15}\)
\(\displaystyle i^{102}\)
52.
Express with a positive exponent and simplify.
\(\displaystyle i^{-1}\)
\(\displaystyle i^{-2}\)
\(\displaystyle i^{-3}\)
\(\displaystyle i^{-6}\)
Exercise Group.
In Problems 53–56,
Given one solution of a quadratic equation with rational coefficients, find the other solution.
Write a quadratic equation that has those solutions.
53.
\(2+\sqrt{5} \)
54.
\(3-\sqrt{2} \)
55.
\(4 - 3i \)
54.
\(5 + i \)
Exercise Group.
Every polynomial factors into a product of a constant and linear factors of the form \((x - a)\text{,}\) where \(a\) can be either real or complex. In Problems 57–58, how many linear factors are in the factored form of the given polynomial?
57.
\(\displaystyle x^4 - 2x^3 + 4x^2 + 8x - 6\)
\(\displaystyle 2x^5 - x^3 + 6x - 4\)
58.
\(\displaystyle x^6 - 6x\)
\(\displaystyle x^3 + 3x^2 - 2x + 1\)
Exercise Group.
For Problems 59–62, find a fourth-degree polynomial with real coefficients that has the given complex numbers as two of its zeros.
For Problems 63–66, plot each number and its complex conjugate in the complex plane. What is the geometric relationship between complex conjugates?
63.
\(z = -3 + 2i\)
64.
\(z = 4-3i\)
65.
\(z =\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i \)
66.
\(z =-\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}i \)
Exercise Group.
For Problems 59–62, simplify and plot each complex number as a point on the complex plane.
67.
\(1, i, i^2, i^3\) and \(i^4\)
68.
\(-1, -i, -i^2, -i^3\) and \(-i^4\)
69.
\(1+2i\) and \(i(1+2i)\)
70.
\(3-4i\) and \(i(3-4i)\)
Exercise Group.
Problems 71–72 show that multiplication by \(i\) results in a rotation of \(90\degree\text{.}\)
71.
Suppose that \(z = a + bi\) and that the real numbers \(a\) and \(b\) are both nonzero.
What is the slope of the segment in the complex plane joining the origin to \(z\text{?}\)
What is the slope of the segment in the complex plane joining the origin to \(zi\text{?}\)
What is the product of the slopes of the two segments from parts (a) and (b)? What can you conclude about the angle between the two segments?
72.
Suppose that \(z = a + bi\) and that \(a\) and \(b\) are both real numbers.
If \(a\ne 0\) and \(b=0\text{,}\) then what is the slope of the segment in the complex plane joining the origin to \(z\text{?}\) What is the slope of the segment joining the origin to \(iz\text{?}\)
If \(a=0\) and \(b \ne 0\text{,}\) then what is the slope of the segment in the complex plane joining the origin to \(z\text{?}\) What is the slope of the segment joining the origin to \(iz\text{?}\)
What can you conclude about the angle between the two segments from parts (a) and (b)?