Modeling, Functions, and Graphs

Section7.3Complex Numbers

SubsectionIntroduction

You know that not all quadratic equations have real solutions.
For example, the graph of
\begin{equation*} f(x) = x^2 - 2x + 2 \end{equation*}
has no $$x$$-intercepts (as shown at right), and the equation
\begin{equation*} x^2 - 2x + 2 = 0 \end{equation*}
has no real solutions.
We can still use completing the square or the quadratic formula to solve the equation.

Example7.37.

Solve the equation $$x^2 - 2x + 2 = 0$$ by using the quadratic formula.
Solution.
We substitute $$a = 1\text{,}$$ $$b =-2\text{,}$$ and $$c= 2$$ into the quadratic formula to get
\begin{equation*} x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(2)}}{2(1)}=\frac{2\pm\sqrt{-4}}{2} \end{equation*}
Because $$\sqrt{-4}$$ is not a real number, the equation $$x^2 - 2x + 2 = 0$$ has no real solutions.

Checkpoint7.38.Practice 1.

Solve the equation $$x^2 - 6x + 13 = 0$$ by using the quadratic formula.
$$x=$$
$$\frac{6+\sqrt{-16}}{2}, \frac{6-\sqrt{-16}}{2}$$
Solution.
$$x=\dfrac{6\pm\sqrt{-16} }{2}$$

SubsectionImaginary Numbers

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.

Checkpoint7.40.QuickCheck 1.

What is the imaginary unit?
• $$\displaystyle -1$$
• $$\displaystyle (-1)^2$$
• $$\displaystyle i^2$$
• $$\displaystyle \sqrt{-1}$$
$$\text{Choice 4}$$
Solution.
$$\sqrt{-1}$$ is the imaginary unit, denoted by $$i\text{.}$$
The square root of any negative number can be written as the product of a real number and $$i\text{.}$$ For example,
\begin{align*} \sqrt{-4}\amp=\sqrt{-1\cdot 4}\\ \amp= \sqrt{-1}\sqrt{4}=i\cdot2 \end{align*}
or $$\sqrt{-4}=2i\text{.}$$ Any number that is the product of $$i$$ and a real number is called an imaginary number.

Imaginary Numbers.

For $$a\gt 0\text{,}$$
\begin{equation*} \sqrt{-a}=\sqrt{-1}\cdot\sqrt{a}=i\sqrt{a} \end{equation*}
Examples of imaginary numbers are
\begin{equation*} 3i\text{, }~~~\frac{7}{8}i\text{, } ~~~-38i\text{, }~~~\text{ and }~~~ i\sqrt{5} \end{equation*}

Example7.41.

Write each radical as an imaginary number.
1. $$\displaystyle \sqrt{-25}$$
2. $$\displaystyle 2\sqrt{-3}$$
Solution.
1. \displaystyle \begin{aligned}[t]\\ \sqrt{-25}\amp=\sqrt{-1}\sqrt{25}\\ \amp=i\sqrt{25}=5i \end{aligned}
2. \displaystyle \begin{aligned}[t]\\ 2\sqrt{-3}\amp=2\sqrt{-1}\sqrt{3}\\ \amp=2i\sqrt{3} \end{aligned}

Checkpoint7.42.Practice 2.

Write each radical as an imaginary number. Use $$i$$ for the imaginary unit, and enter “sqrt(2)” for $$\sqrt{2}\text{.}$$
1. $$\sqrt{-18}=$$
2. $$-6\sqrt{-5}$$
$$3i\sqrt{2}$$
$$-6i\sqrt{5}$$
Solution.
1. $$\displaystyle 3i\sqrt{2}$$
2. $$\displaystyle -6i\sqrt{5}$$

Note7.43.

Every negative real number has two imaginary square roots, $$i\sqrt{a}$$ and $$-i\sqrt{a}\text{,}$$ because
\begin{equation*} \left(i\sqrt{a}\right)^2=i^2(\sqrt{a})^2=-a \end{equation*}
and
\begin{equation*} \left(-i\sqrt{a}\right)^2=(-i)^2(\sqrt{a})^2=-a \end{equation*}
For example, the two square roots of $$-9$$ are $$3i$$ and $$-3i\text{.}$$

SubsectionComplex Numbers

\begin{equation*} x^2 - 2x + 5 = 0 \end{equation*}
Using the quadratic formula to solve the equation, we find
\begin{equation*} x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(5)}}{2}=\frac{2\pm\sqrt{-16}}{2} \end{equation*}
If we now replace $$\sqrt{-16}$$ with $$4i\text{,}$$ we have
\begin{equation*} x=\frac{2\pm4i}{2}=1\pm2i \end{equation*}
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.
Solution.
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.
$$x=$$
$$\frac{-1}{2}+\frac{3}{2}i, \frac{-1}{2}-\frac{3}{2}i$$
Solution.
$$x=\dfrac{-1}{2}\pm\dfrac{3}{2}i$$

SubsectionArithmetic of 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.
$$\text{D) Complex ... real numbers.}$$
Solution.
“Complex numbers cannot be combined with real numbers” is a false statement.
We add and subtract complex numbers by combining their real and imaginary parts separately. For example,
\begin{align*} (4 + 5i ) + (2 - 3i ) \amp= (4 + 2) + (5 - 3)i\\ \amp= 6 + 2i \end{align*}

Sums and Differences of Complex Numbers.

\begin{equation*} (a + bi) + (c + di) = (a + c) + (b + d)i \end{equation*}
\begin{equation*} (a + bi) - (c + di) = (a - c) + (b - d)i \end{equation*}

Example7.47.

Subtract: $$(8 - 6i ) - (5 + 2i )\text{.}$$
Solution.
Combine the real and imaginary parts.
\begin{align*} (8 - 6i ) - (5 + 2i ) \amp= (8 - 5) + (-6 - 2)i\\ \amp = 3 + (-8)i = 3 - 8i \end{align*}

Checkpoint7.48.Practice 4.

Subtract: $$(-3 + 2i ) - (-3 - 2i )=$$
$$4i$$
Solution.
$$4i$$

SubsectionProducts of Complex Numbers

To find the product of two imaginary numbers, we use the fact that $$i^2=-1\text{.}$$ For example,
\begin{align*} (3i )\cdot(4i ) \amp= 3\cdot4 i^2\\ \amp= 12(-1) = -12 \end{align*}
To find the product of two complex numbers, we use the FOIL method, as if the numbers were binomials. For example,
\begin{equation*} (2 + 3i )(3 - 5i ) = 6 - 10i + 9i - 15i^2 \end{equation*}
Because $$i^2=-1\text{,}$$ the last term, $$-15i^2\text{,}$$ can be replaced by $$-15(-1)\text{,}$$ or $$15\text{,}$$ to obtain
\begin{equation*} 6 - 10i + 9i + 15 \end{equation*}
Finally, we combine the real parts and imaginary parts to obtain
\begin{equation*} (6 + 15) + (-10i + 9i ) = 21 - i \end{equation*}

Example7.49.

Multiply $$(7 - 4i )(-2 - i )\text{.}$$
Solution.
\begin{align*} (7 - 4i )(-2 - i ) \amp= -14 - 7i + 8i + 4i^2\amp\amp \blert{\text{Replace }i^2 \text{ by }-1.}\\ \amp = -14 - 7i + 8i - 4\amp\amp \blert{\text{Combine real parts and imaginary}}\\ \amp= -18 + i\amp\amp \blert{\text{parts.}} \end{align*}

Checkpoint7.50.Practice 5.

Multiply $$(-3 + 2i )(-3 - 2i )=$$
$$13$$
Solution.
$$13$$
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
\begin{equation*} \sqrt{ab}=\sqrt{(-2)(-2)}=\sqrt{4}=2 \end{equation*}
but
\begin{equation*} \sqrt{a}\cdot\sqrt{b}=\sqrt{-2}\cdot\sqrt{-2}=\sqrt{-1\cdot2}\cdot\sqrt{-1\cdot2} = i\sqrt{2}\cdot i\sqrt{2}=i^2(\sqrt{2})^2=-2 \end{equation*}
so in this case
\begin{equation*} \sqrt{ab}\ne\sqrt{a}\cdot\sqrt{b} \end{equation*}
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
\begin{equation*} \frac{3i}{-2}-\frac{4}{-2}=\frac{-3}{2}i+2=2+\frac{-3}{2}i \end{equation*}

Example7.53.

Divide $$~~\displaystyle{\frac{10-15i}{5i}}$$
Solution.
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*}
The quotient is $$-3 - 2i\text{.}$$

Checkpoint7.54.Practice 6.

Divide $$\displaystyle{\frac{8+9i}{3i}}=$$
$$3-\frac{8}{3}i$$
Solution.
$$3-\dfrac{8}{3}i$$
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.
\begin{equation*} z\overline{z} = (a + bi)(a - bi) = a^2 - b^2i^2 = a^2 - b^2(-1) = a^2 + b^2 \end{equation*}
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?
• $$\displaystyle a-bi$$
• $$\displaystyle a^2+b^2$$
• $$\displaystyle a^2-b^2$$
• $$\displaystyle ai+b$$
$$\text{Choice 2}$$
Solution.
$$a^2+b^2$$

Example7.56.

Divide $$~~\displaystyle{\frac{2+3i}{4-2i}}$$
Solution.
We multiply numerator and denominator by $$\alert{4 + 2i}\text{,}$$ the conjugate of the denominator.
\begin{align*} \frac{2+3i}{4-2i}\amp=\frac{(2+3i)(\alert{4 + 2i})}{(4-2i)(\alert{4 + 2i})} \amp\amp \blert{\text{Expand numerator and denominator.}}\\ \amp=\frac{8 + 4i + 12i + 6i^2}{16 + 8i - 8i - 4i^2} \amp\amp \blert{\text{Replace }i^2\text{ by } -1.}\\ \amp= \frac{8 + 16i - 6}{16 - (-4)} \amp\amp \blert{\text{Combine like terms.}}\\ \amp= \frac{2 + 16i}{20} \amp\amp \blert{\text{Divide 20 into each term of numerator.}}\\ \amp= \frac{2}{20}+ \frac{16i}{20}\\ \amp= \frac{1}{10}+ \frac{4}{5}i \end{align*}

Checkpoint7.57.Practice 7.

Write the quotient $$\displaystyle{\frac{4 - 2i}{1 + i}}$$ in the form $$a + bi\text{.}$$
$$1-3i$$
Solution.
$$1-3i$$

Checkpoint7.58.Pause and Reflect.

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.
Solution.
We substitute $$x = \alert{1 + i}$$ to find
\begin{align*} f(\alert{1 + i} ) \amp = (\alert{1 + i} )^2 - 2(\alert{1 + i} ) + 2\\ \amp = 1^2 + 2i + i^2 - 2 - 2i + 2\\ \amp = 1 + 2i + (-1) - 2 - 2i + 2\\ \amp =0 \end{align*}
Thus, $$f(1 + i ) = 0\text{,}$$ so $$1 + i$$ is a solution of $$x^2 - 2x + 2 = 0\text{.}$$

Checkpoint7.60.Practice 8.

If $$f(x) = x^2 - 6x + 13\text{,}$$ evaluate $$f(3 + 2i )=$$
$$0$$
Solution.
$$f(3 + 2i )=(3 + 2i)^2-6(3 + 2i)+13=0$$
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
\begin{align*} (x - z)(x - \overline{z}) \amp = 0\\ x^2-(z+ \overline{z}) +z\overline{z}\amp = 0 \end{align*}

Example7.61.

1. Let $$z = 7 - 5i\text{.}$$ Compute $$z\overline{z}\text{.}$$
2. Find a quadratic equation with one solution being $$z = 7 - 5i\text{.}$$
Solution.
1. The conjugate of $$z = 7 - 5i$$ is $$\overline{z} = 7 + 5i\text{,}$$ so
\begin{align*} z\overline{z} \amp= (7 - 5i )(7 + 5i )\\ \amp= 49 - 25i^2 \\ \amp= 49 + 25 \\ \amp= 74 \end{align*}
2. 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
\begin{align*} (x - z)(x - \overline{z}) \amp = x^2 - (z + \overline{z})x + z\overline{z}\\ \amp = x^2 - (7 - 5i + 7 + 5i )x + 74\\ \amp = x^2 - 14x + 74 \end{align*}
The equation is $$x^2 - 14x + 74 = 0\text{.}$$

Checkpoint7.62.Practice 9.

1. Let $$z = -3 + 4i\text{.}$$ Compute $$z\overline{z}=$$
2. Find a quadratic equation with one solution being $$z = -3 + 4i\text{.}$$
Answer: $$=0$$
$$25$$
$$x^{2}+6x+25$$
Solution.
1. $$\displaystyle 25$$
2. $$\displaystyle x^2 + 6x + 25 = 0$$
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
\begin{equation*} x^2 = (1 + i )^2 = 1 + 2i + i^2 = 2i \end{equation*}
and
\begin{equation*} x^4 = \left(x^2\right)^2 = (2i )^2 = -4, \end{equation*}
so $$x^4+4=(-4)+4=0\text{.}$$
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:
• $$\displaystyle 5+3i$$
• $$\displaystyle 9+25$$
• $$\displaystyle 3-5i$$
• $$\displaystyle 5-3i$$
$$\text{Choice 3}$$
Solution.
$$3-5i$$

Example7.65.

Find a fourth-degree polynomial with real coefficients, two of whose zeros are $$3i$$ and $$2 + i\text{.}$$
Solution.
The other two zeros are $$-3i$$ and $$2 - i\text{.}$$ The factored form of the polynomial is
\begin{equation*} (x - 3i )(x + 3i )[x - (2 + i )][x - (2 - i )] \end{equation*}
We multiply together the factors to find the polynomial. The product of $$(x - 3i )(x + 3i )$$ is $$x^2 + 9\text{,}$$ and
\begin{align*} [x - (2 + i )][x - (2 - i )] \amp = x^2 - (2 + i + 2 - i )x + (2 + i )(2 - i )\\ \amp = x^2 - 4x + 5 \end{align*}
Finally, we multiply these two partial products to find the polynomial we seek,
\begin{equation*} (x^2 + 9)(x^2 - 4x + 5) = x^4 - 4x^3 + 14x^2 - 36x + 45 \end{equation*}

Checkpoint7.66.Practice 10.

1. Find the zeros of the polynomial $$f(x) = x^4 + 15x^2 - 16\text{.}$$
Separate different solutions with a comma.
2. Write the polynomial in factored form.
$$f(x)=\big($$$$\big)\big($$$$\big)\big($$$$\big)\big($$$$\big)$$
$$1, -1, 4i, -4i$$
$$x-1$$
$$x+1$$
$$x-4i$$
$$x+4i$$
Solution.
1. $$\displaystyle \pm 1, ~\pm 4i$$
2. $$\displaystyle (x - 1)(x + 1)(x - 4i )(x + 4i )$$

Checkpoint7.67.Pause and Reflect.

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 $$z = 2 - 2i\text{.}$$
Solution.
• To plot $$z = 2 - 2i\text{,}$$ we move from the origin $$2$$ units to the right and $$2$$ units down.
• To plot $$\overline{z} = 2 + 2i\text{,}$$ we move from the origin $$2$$ units to the right and $$2$$ units up.
• To plot $$-z = -2 + 2i\text{,}$$ we move from the origin $$2$$ units to the left and $$2$$ units up.
• To plot $$-\overline{z} = -2 - 2i\text{,}$$ we move from the origin $$2$$ units to the left and $$2$$ units down.
All four points are plotted at right.

Checkpoint7.69.Practice 11.

Plot the following numbers as points on the complex plane.
• $$\displaystyle z = 1 + i$$
• $$\displaystyle i z = i + i^2$$
• $$\displaystyle i^2z = i^2 + i^3$$
• $$\displaystyle i^ 3z = i^3 + i^4$$
Solution.
A plot is below.

Note7.70.

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

1. The square root of a negative number is an imaginary number.
2. A complex number is the sum of a real number and an imaginary number.
3. We can perform the four arithmetic operations on complex numbers.
4. The product of a nonzero complex number and its conjugate is always a positive real number.
5. Fundamental Theorem of Algebra.
Let $$p(x)$$ be a polynomial of degree $$n\ge 1\text{.}$$ Then $$p(x)$$ has exactly $$n$$ complex zeros.
6. The nonreal zeros of a polynomial with real coefficients always occur in conjugate pairs.
7. We can graph complex numbers in the complex plane.
8. Multiplying a complex number by $$i$$ rotates its graph by $$90\degree$$ around the origin.

SubsubsectionSTUDY QUESTIONS

1. What are imaginary numbers, and why were they invented?
2. Simplify the following powers of $$i\text{:}$$
\begin{equation*} i^2, i^3, i^4, i^5, i^6, i^7, i^8 \end{equation*}
What do you notice?
3. Explain how the complex conjugate is used in dividing complex numbers.
4. If one solution of a quadratic equation is $$3 + i\sqrt{2}\text{,}$$ what is the other solution?
5. 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.
1. Write and simplify complex numbers: #1–10
2. Perform arithmetic operations on complex numbers: #11–36
3. Evaluate polynomials at complex numbers, expand polynomials: #37–48
4. Find a polynomial with given zeros: #53–56, 59–62
5. 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.
7.
$$x^2 + 6x + 13$$
8.
$$x^2 - 2x + 10$$
9.
$$3x^2 - x + 1$$
10.
$$5x^2 + 2x + 2$$

Exercise Group.

For Problems 11–14, add or subtract.
11.
$$(11 - 4i ) - (-2 - 8i )$$
12.
$$(7i - 2) + (6 - 4i )$$
13.
$$(2.1 + 5.6i ) + (-1.8i - 2.9)$$
14.
$$\left(\dfrac{1}{5}i-\dfrac{2}{5} \right)-\left(\dfrac{4}{5}-\dfrac{3}{5}i \right)$$

Exercise Group.

For Problems 15–24, multiply.
15.
$$5i (2 - 4i )$$
16.
$$-7i (-1 + 4i )$$
17.
$$(4 - i )(-6 + 7i )$$
18.
$$(2 - 3i )(2 - 3i )$$
19.
$$(7+i\sqrt{3} )^2$$
20.
$$(5-i\sqrt{2} )^2$$
21.
$$(7+i\sqrt{3} )(7-i\sqrt{3} )$$
22.
$$(5-i\sqrt{2} )(5+i\sqrt{2} )$$
23.
$$(1 - i )^3$$
24.
$$(2 + i )^3$$

Exercise Group.

For Problems 25–36, divide.
25.
$$\dfrac{12+3i}{-3i}$$
26.
$$\dfrac{12+4i}{8i}$$
27.
$$\dfrac{10 + 15i}{2+i}$$
28.
$$\dfrac{4-6i}{1-i}$$
29.
$$\dfrac{5i}{2-5i}$$
30.
$$\dfrac{-2i}{7+2i}$$
31.
$$\dfrac{\sqrt{3}}{\sqrt{3}+i}$$
32.
$$\dfrac{2\sqrt{2} }{1- i \sqrt{2} }$$
33.
$$\dfrac{1+i\sqrt{5}}{1-i\sqrt{5}}$$
34.
$$\dfrac{\sqrt{2}-i }{\sqrt{2}+i }$$
35.
$$\dfrac{3+2i}{2-3i}$$
36.
$$\dfrac{4-6i }{-3-2i }$$

Exercise Group.

For Problems 37–42, evaluate the polynomial for the given values of the variable.
37.
$$z^2+9$$
1. $$\displaystyle z=3i$$
2. $$\displaystyle z=-3i$$
38.
$$2y^2 - y - 2$$
1. $$\displaystyle y=2-i$$
2. $$\displaystyle y=-2-i$$
39.
$$x^2 - 2x + 2$$
1. $$\displaystyle x=1-i$$
2. $$\displaystyle x=1+i$$
40.
$$3w^2 +5$$
1. $$\displaystyle w=2i$$
2. $$\displaystyle w=-2i$$
41.
$$q^2 +4q + 13$$
1. $$\displaystyle q = -2 + 3i$$
2. $$\displaystyle q = -2 - 3i$$
42.
$$v^2 + 2v + 3$$
1. $$\displaystyle v = 1 + i$$
2. $$\displaystyle v=-1 + i$$

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.
1. $$\displaystyle i^6$$
2. $$\displaystyle i^{12}$$
3. $$\displaystyle i^{15}$$
4. $$\displaystyle i^{102}$$

52.

Express with a positive exponent and simplify.
1. $$\displaystyle i^{-1}$$
2. $$\displaystyle i^{-2}$$
3. $$\displaystyle i^{-3}$$
4. $$\displaystyle i^{-6}$$

Exercise Group.

In Problems 53–56,
1. Given one solution of a quadratic equation with rational coefficients, find the other solution.
2. 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.
1. $$\displaystyle x^4 - 2x^3 + 4x^2 + 8x - 6$$
2. $$\displaystyle 2x^5 - x^3 + 6x - 4$$
58.
1. $$\displaystyle x^6 - 6x$$
2. $$\displaystyle x^3 + 3x^2 - 2x + 1$$

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.
59.
$$1 + 3i, ~2 - i$$
60.
$$5 - 4i,~ -i$$
61.
$$\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i, ~ 3+2i$$
62.
$$-\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}i, ~ 4-i$$

Exercise Group.

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.
1. What is the slope of the segment in the complex plane joining the origin to $$z\text{?}$$
2. What is the slope of the segment in the complex plane joining the origin to $$zi\text{?}$$
3. 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.
1. 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 $$iz\text{?}$$
2. 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 $$iz\text{?}$$
3. What can you conclude about the angle between the two segments from parts (a) and (b)?