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Section 7.3 Complex Numbers

Subsection Introduction

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.

parabola without x-intercepts

We can still use completing the square or the quadratic formula to solve the equation.

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.

Solve the equation \(x^2 - 6x + 13 = 0\) by using the quadratic formula.

Answer

\(x=\dfrac{6\pm\sqrt{-16} }{2} \)

Subsection Imaginary 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*}
Caution 7.25.

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 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*}

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. \(\displaystyle \begin{aligned}[t]\\ 2\sqrt{-3}\amp=2\sqrt{-1}\sqrt{3}\\ \amp=2i\sqrt{3} \end{aligned}\)

Write each radical as an imaginary number.

  1. \(\displaystyle \sqrt{-18}\)
  2. \(\displaystyle -6\sqrt{-5}\)
Answer

  1. \(3i\sqrt{2} \)

  2. \(-6i\sqrt{5} \)

Note 7.28.

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 \(-3i\text{.}\)

Subsection Complex Numbers

Consider the quadratic equation

\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.

Write the solutions to Example 6.24, \(\dfrac{2\pm\sqrt{-4}}{2}\text{,}\) 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 \(1-i\text{.}\)

Use extraction of roots to solve \((2x + 1)^2 + 9 = 0\text{.}\) Write your answers as complex numbers.

Answer

\(x=\dfrac{-1}{2}\pm\dfrac{3}{2}i \)

Subsection Arithmetic 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.

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*}

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*}

Subtract: \((-3 + 2i ) - (-3 - 2i )\text{.}\)

Answer

\(4i\)

Subsection Products 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*}

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*}

Multiply \((-3 + 2i )(-3 - 2i )\text{.}\)

Answer

\(13\)

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*}
Caution 7.35.

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(i\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.

Subsection Quotients 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*}

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 \(-3 - 2i\text{.}\)

Divide \(~~\displaystyle{\frac{8+9i}{3i}}\)

Answer

\(3-\dfrac{8}{3}i \)

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.

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*}

Write the quotient \(~\displaystyle{\frac{4 - 2i}{1 + i}}~\) in the form \(a + bi\text{.}\)

Answer

\(1-3i\)

Subsection Zeros 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.

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 \(x^2 - 2x + 2 = 0\text{.}\)

If \(f(x) = x^2 - 6x + 13\text{,}\) evaluate \(f(3 + 2i )\text{.}\)

Answer

\(0\)

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*}
  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 - z) \amp = x^2 - (z + z)x + z\overline{z}\\ \amp = x^2 - (7 - 5i + 7 + 5i )x + 74\\ \amp = x^2 - 14x + 74 \end{align*}

    The equation is \(x^2 - 14x + 74 = 0\text{.}\)

  1. Let \(z = -3 + 4i\text{.}\) Compute \(z\overline{z}\text{.}\)
  2. Find a quadratic equation with one solution being \(z = -3 + 4i\text{.}\)
Answer

  1. \(25\)

  2. \(x^2 + 6x + 25 = 0\)

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.

quartic with no x-intercepts

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*}
Note 7.44.

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.

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*}

  1. Find the zeros of the polynomial \(f(x) = x^4 + 15x^2 - 16\text{.}\)

  2. Write the polynomial in factored form.

Answer

  1. \(\pm 1, ~\pm 4i\)

  2. \((x - 1)(x + 1)(x - 4i )(x + 4i )\)

Subsection Graphing 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.

plots of complex conjugates

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.

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.

figure of comlex numbers arranged symmetrically with respect to the axes

Plot the following numbers as points on the complex plane.

  • \(\displaystyle z = 1 + i\)
  • \(\displaystyle i z = i + i^2\)
  • \(\displaystyle i^2z = i^2 + i^3\)
  • \(\displaystyle i^ 3z = i^3 + i^4\)
Answer
points in complex plane
Note 7.49.

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 \(i z = 2i - 3\text{.}\)

figure of comlex number multiplied by i causing a rotation of 90 degrees

Subsection Section Summary

Subsubsection Vocabulary

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

Subsubsection CONCEPTS

  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.

Subsubsection STUDY 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?

Subsubsection SKILLS

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

Exercises Homework 7.3

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.

\(\dfrac{7+\sqrt{-3}}{4} \)

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.

\(5x^2 + 2x + 2\)

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.

\(\left(\dfrac{1}{5}i-\dfrac{2}{5} \right)-\left(\dfrac{4}{5}-\dfrac{3}{5}i \right) \)

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.

\((2 + i )^3 \)

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.

\(\dfrac{4-6i }{-3-2i } \)

For Problems 37–42, evaluate the polynomial for the given values of the variable.

37.

\(z^2+9\)

  1. \(z=3i\)

  2. \(z=-3i\)

38.

\(2y^2 - y - 2\)

  1. \(y=2-i\)

  2. \(y=-2-i\)

39.

\(x^2 - 2x + 2\)

  1. \(x=1-i\)

  2. \(x=1+i\)

40.

\(3w^2 +5\)

  1. \(w=2i\)

  2. \(w=-2i\)

41.

\(q^2 +4q + 13\)

  1. \(q = -2 + 3i\)

  2. \(q = -2 - 3i\)

42.

\(v^2 + 2v + 3\)

  1. \(v = 1 + i\)

  2. \(v=-1 + i\)

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. \(i^6\)

  2. \(i^{12}\)

  3. \(i^{15}\)

  4. \(i^{102}\)

52.

Express with a positive exponent and simplify.

  1. \(i^{-1}\)

  2. \(i^{-2}\)

  3. \(i^{-3}\)

  4. \(i^{-6}\)

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.

\(5 + i \)

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. \(x^4 - 2x^3 + 4x^2 + 8x - 6\)

  2. \(2x^5 - x^3 + 6x - 4\)

58.

  1. \(x^6 - 6x\)

  2. \(x^3 + 3x^2 - 2x + 1\)

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.

\(-\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}i, ~ 4-i \)

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.

\(z =-\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}i \)

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 \(i(3-4i)\)

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)?