Section 1.4 Complex Numbers: Quadratic Equations with Complex Solutions

Before Starting this Section, Review

  1. 1 Properties of real numbers (Section P.1 , page 12)

  2. 2 Special products (Section P.3 , pages 34 and 37)

  3. 3 Factoring (Section P.4 )

Objectives

  1. 1 Define complex numbers.

  2. 2 Add and subtract complex numbers.

  3. 3 Multiply complex numbers.

  4. 4 Divide complex numbers.

  5. 5 Solve quadratic equations having complex solutions.

Alternating Current Circuits

In the early days of the study of alternate current (AC) circuits, scientists concluded that AC circuits were somehow different from the battery-powered direct current (DC) circuits. However, both types of circuits obey the same physical and mathematical laws. In 1893, Charles Steinmetz provided the breakthrough in understanding AC circuits. He explained that in AC circuits, the voltage, current, and resistance (called impedance in AC circuits) were not scalars, but alternate in direction, and possessed frequency and phase shift that must be taken into account. He advocated the use of the polar form of complex numbers to provide a convenient method of symbolically denoting the magnitude, frequency, and phase shift simultaneously for the AC circuit quantities voltage, current, and impedance. Steinmetz eventually became known as “the wizard who generated electricity from the square root of minus one.” In Example 7, we use complex numbers to compute the total impedance in an AC circuit.

Complex Numbers

  1. 1 Define complex numbers.

Because the square of a real number is nonnegative (that is, x2≥0x2≥0 for any real number x), the equation x2=−1x2=−1 has no solution in the set of real numbers. To solve such equations, we extend the real numbers to a larger set called the set of complex numbers. We first introduce a new number i whose square is −1.−1.

With the introduction of the number i, the equation x2=−1x2=−1 has two solutions, x=±√−1=±i.x=±−1−−−√=±i.

A complex number z written in the form a+bia+bi is said to be in standard form. Where convenient, we use the form a+iba+ib as equivalent to a+bi.a+bi. A complex number with a=0a=0 and b≠0,b≠0, written as bi, is called a pure imaginary number. If b=0,b=0, then the complex number a+bi=aa+bi=a is a real number. So real numbers form a subset of the complex numbers (with imaginary part 0). Figure 1.5 shows how various sets of numbers are contained within larger sets.

Figure 1.5

Complex numbers

We can express the square root of any negative number as a product of a real number and i.

We usually write i√bib√ rather than √bib√i because √bib√i is easily confused with √bi.bi−−√.

Side Note

Here is a visual reminder of the parts of a complex number.

a+bi↑↑RealpartImaginarypart
a↑⏐Realpart+bi↑⏐Imaginarypart

Example 1 Identifying the Real and Imaginary Parts of a Complex Number

Identify the real and imaginary parts of each complex number.

  1. 2+5i2+5i

  2. 7−12i7−12i

  3. 3i.

  4. −9−9

  5. 0

  6. 3+√−253+−25−−−−√

Solution

To identify the real and imaginary parts, we express each number in the standard form a+bi.a+bi.

  1. 2+5i2+5i is already written as a+bi;a+bi; real part 2, imaginary part 5

  2. 7−12i=7+(−12)i;7−12i=7+(−12)i; real part 7, imaginary part −12−12

  3. 3i=0+3i;3i=0+3i; real part 0, imaginary part 3

  4. −9=−9+0i;−9=−9+0i; real part −9,−9, imaginary part 0

  5. 0=0+0i;0=0+0i; real part 0, imaginary part 0

  6. 3+√−25=3+(√25)i=3+5i;3+−25−−−−√=3+(25−−√)i=3+5i; real part 3, imaginary part 5

Practice Problem 1

  1. Identify the real and imaginary parts of each complex number.

    1. −1+2i−1+2i

    2. −13−6i−13−6i

    3. 8

Addition and Subtraction

  1. 2 Add and subtract complex numbers.

The standard form of a complex number a+bia+bi is similar to the binomial form a+bx.a+bx. The rules for equality, addition, and subtraction preserve the familiar rules used for these binomials.

To add or subtract complex numbers, add or subtract the real parts and add or subtract the imaginary parts. Write the sum or difference in standard form, a+bi.a+bi.

Example 2 Adding and Subtracting Complex Numbers

Write the sum or difference of two complex numbers in standard form.

  1. (3+7i)+(2−4i)(3+7i)+(2−4i)

  2. (5+9i)−(6−8i)(5+9i)−(6−8i)

  3. (2+√−9)−(−2+√−4)(2+−9−−−√)−(−2+−4−−−√)

Solution

  1. (3+7i)+(2−4i)=(3+2)+[7+(−4)]i=5+3i(3+7i)+(2−4i)=(3+2)+[7+(−4)]i=5+3i

  2. (5+9i)−(6−8i)=(5−6)+[9−(−8)]i=−1+17i(5+9i)−(6−8i)=(5−6)+[9−(−8)]i=−1+17i

  3. (2+√−9)−(−2+√−4)=(2+3i)−(−2+2i)√9=3i, √−4=2i=(2−(−2))+(3−2)i=4+i(2+−9−−−√)−(−2+−4−−−√)===(2+3i)−(−2+2i)(2−(−2))+(3−2)i4+i9–√=3i, −4−−−√=2i

Practice Problem 2

  1. Write the following complex numbers in standard form.

    1. (1−4i)+(3+2i)(1−4i)+(3+2i)

    2. (4+3i)−(5−i)(4+3i)−(5−i)

    3. (3−√−9)−(5−√−64)(3−−9−−−√)−(5−−64−−−−√)

Multiplying Complex Numbers

  1. 3 Multiply complex numbers.

We multiply complex numbers by using FOIL (as we do with binomials; see page 35) and then replace i2i2 with −1.−1. For example,

FOIL(2+5i)(4+3i)=2⋅4+2⋅3i+5i⋅4+5i⋅3i=8+6i+20i+15i2=8+26i+15(−1)Replace i2 with −1.=(8−15)+26i=−7+26i
(2+5i)(4+3i)=====FOIL2⋅4+2⋅3i+5i⋅4+5i⋅3i8+6i+20i+15i28+26i+15(−1)(8−15)+26i−7+26iReplace i2 with −1.

Example 3 Multiplying Complex Numbers

Write the following products in standard form.

  1. (3−5i)(2+7i)(3−5i)(2+7i)

  2. −2i(5−9i)−2i(5−9i)

Solution

  1. FOIL(3−5i)(2+7i)=6+21i−10i−35i2=6+11i+35Because i2=−1, −35i2=35.=41+11iCombine terms.(3−5i)(2+7i)===FOIL6+21i−10i−35i26+11i+3541+11iBecause i2=−1, −35i2=35.Combine terms.

  2. −2i(5−9i)=−10i+18i2Distributive property=−10i−18Because i2=−1, 18i2=−18.=−18−10i−2i(5−9i)===−10i+18i2−10i−18−18−10iDistributive propertyBecause i2=−1, 18i2=−18.

Practice Problem 3

  1. Write the following products in standard form.

    1. (2−6i)(1+4i)(2−6i)(1+4i)

    2. −3i(7−5i)−3i(7−5i)

Warning

Recall from algebra that if a and b are positive real numbers, then

√a√b=√ab.
a−−√b√=ab−−√.

However, this property is not true for all complex numbers. For example,

√−9 √−9=(3i)(3i)=9i2=9(−1)=−9,
−9−−−√ −9−−−√=(3i)(3i)=9i2=9(−1)=−9,

but

√(−9)(−9)= √81=9.
(−9)(−9)−−−−−−−−√= 81−−√=9.

Thus,

√−9 √−9≠√(−9)(−9).
−9−−−√ −9−−−√≠(−9)(−9)−−−−−−−−√.

When performing multiplication (or division) involving square roots of negative numbers (say, √−b−b−−−√ with b>0),b>0), always write √−b=i√b−b−−−√=ib−−√ before performing the operation.

Example 4 Multiplication Involving Square Roots of Negative Numbers

Perform the indicated operation and write the result in standard form.

  1. √−2√−8−2−−−√−8−−−√

  2. √−3(2+√−3)−3−−−√(2+−3−−−√)

  3. (−2+√−3)2(−2+−3−−−√)2

  4. (3+√−2)(1+√−32)(3+−2−−−√)(1+−32−−−−√)

Solution

  1. √−2√−8=i√2⋅i√8=i2√2⋅8=i2√16=(−1)(4)=−4−2−−−√−8−−−√=i2–√⋅i8–√=i22⋅8−−−−√=i216−−√=(−1)(4)=−4

  2. √−3(2+√−3)=i√3(2+i√3)√−3=i√3=2i√3+i2√9Distributive property=2i√3+(−1)(3)i2=−1,√9=3=−3+2i√3Standard form−3−−−√(2+−3−−−√)====i3–√(2+i3–√)2i3–√+i29–√2i3–√+(−1)(3)−3+2i3–√−3−−−√=i3–√Distributive propertyi2=−1,9–√=3Standard form

  3. (−2+√−3)2=(−2+i√3)2√−3=i√3=(−2)2+2(−2)(i√3)+(i√3)2(a+b)2=a2+2ab+b2=4−4i√3+3i2(i√3)2=(i√3)(i√3)=3i2=4−4i√3+3(−1)i2=−1=1−4i√3Simplify.(−2+−3−−−√)2=====(−2+i3–√)2(−2)2+2(−2)(i3–√)+(i3–√)24−4i3–√+3i24−4i3–√+3(−1)1−4i3–√−3−−−√=i3–√(a+b)2=a2+2ab+b2(i3–√)2==(i3–√)(i3–√)3i2i2=−1Simplify.

  4. (3+√−2)(1+√−32)=(3+i√2)(1+i√32)√−2=i√2, √−32=i√32=3+3i√32+i√2+(i√2)(i√32)FOIL=3+3i(4√2)+i√2+i2√64√32=√16⋅2=4√2=3+12i√2+i√2+(−1)(8)i2=−1=−5+13√2iAdd.(3+−2−−−√)(1+−32−−−−√)=====(3+i2–√)(1+i32−−√)3+3i32−−√+i2–√+(i2–√)(i32−−√)3+3i(42–√)+i2–√+i264−−√3+12i2–√+i2–√+(−1)(8)−5+132–√i−2−−−√=i2–√, −32−−−−√=i32−−√FOIL32−−√=16⋅2−−−−√=42–√i2=−1Add.

Practice Problem 4

  1. Perform the indicated operation and write the result in standard form.

    1. (−3+√−4)2(−3+−4−−−√)2

    2. (5+√−2)(4+√−8)(5+−2−−−√)(4+−8−−−√)

Complex Conjugates and Division

  1. 4 Divide complex numbers.

To perform the division of complex numbers, it is helpful to learn about the conjugate of a complex number.

Note that the conjugate ˉzz¯ of a complex number has the same real part as z does, but the sign of the imaginary part is changed. For example, ¯2+7i=2−7i2+7i¯¯¯¯¯¯¯¯¯=2−7i and ¯5−3i=¯5+(−3)i=5−(−3)i=5+3i.5−3i¯¯¯¯¯¯¯¯¯=5+(−3)i¯¯¯¯¯¯¯¯¯¯¯¯¯¯=5−(−3)i=5+3i.

Example 5 Multiplying a Complex Number by Its Conjugate

Find the product zˉzzz¯ for each complex number z.

  1. z=2+5iz=2+5i

  2. z=1−3iz=1−3i

Solution

  1. If z=2+5i,z=2+5i, then ˉz=2−5i.z¯=2−5i.

    zˉz=(2+5i)(2−5i)=22−(5i)2Difference of squares=4−25i2(5i)2=52i2=25i2=4−(−25)Because i2=−1,25i2=−25.=29Simplify.
    zz¯=(2+5i)(2−5i)====22−(5i)24−25i24−(−25)29Difference of squares(5i)2=52i2=25i2Because i2=−1,25i2=−25.Simplify.

  2. If z=1−3i,z=1−3i, then ˉz=1+3i.z¯=1+3i.

    zˉz=(1−3i)(1+3i)=12−(3i)2Difference of squares=1−9i2(3i)2=32i2=9i2=1−(−9)Because i2=−1, 9i2=−9.=10Simplify.
    zz¯=(1−3i)(1+3i)====12−(3i)21−9i21−(−9)10Difference of squares(3i)2=32i2=9i2Because i2=−1, 9i2=−9.Simplify.

Practice Problem 5

  1. Find the product zˉzzz¯ for each complex number z.

    1. 1+6i1+6i

    2. −2i−2i

The results in Example 5 correctly suggest the following theorem.

Side Note

Notice that the product of a complex number and its conjugate is always a real number.

To write the reciprocal of a nonzero complex number or the quotient of two complex numbers in the form a+bi,a+bi, multiply the numerator and denominator by the conjugate of the denominator. By the Complex Conjugate Product Theorem, the resulting denominator is a real number.

Example 6 Dividing Complex Numbers

Write the following quotients in standard form.

  1. 12+i12+i

  2. 4+√−252−√−94+−25−−−−√2−−9−−−√

Solution

  1. The denominator is 2+i,2+i, so its conjugate is 2−i.2−i.

    12+i=1(2−i)(2+i)(2−i)Multiply numerator anddenominator by 2−i.=2−i22+12(2+i)(2−i)=22+12=2−i5Simplify.=25+−15iStandard form
    12+i====1(2−i)(2+i)(2−i)2−i22+122−i525+−15iMultiply numerator anddenominator by 2−i.(2+i)(2−i)=22+12Simplify.Standard form

  2. We write √−25=5i−25−−−−√=5i and √−9=3i−9−−−√=3i so that 4+√−252−√−9=4+5i2−3i.4+−25−−−−√2−−9−−−√=4+5i2−3i.

    4+5i2−3i=(4+5i)(2+3i)(2−3i)(2+3i)Multiply numerator anddenominator by 2+3i.=8+12i+10i+15i222+32Use FOIL in the numerator;(2−3i)(2+3i)=22+32.=−7+22i1315i2=−15, 8−15=−7=−713+2213iStandard form
    4+5i2−3i====(4+5i)(2+3i)(2−3i)(2+3i)8+12i+10i+15i222+32−7+22i13−713+2213iMultiply numerator anddenominator by 2+3i.Use FOIL in the numerator;(2−3i)(2+3i)=22+32.15i2=−15, 8−15=−7Standard form

Practice Problem 6

  1. Write the following quotients in standard form.

    1. 21−i21−i

    2. −3i4+√−25−3i4+−25−−−−√

Example 7 Using Complex Numbers in AC Circuits

In a parallel circuit, the total impedance ZtZt is given by

Zt=Z1Z2Z1+Z2.
Zt=Z1Z2Z1+Z2.

Find ZtZt assuming that Z1=2+3iZ1=2+3i ohms and Z2=3−4iZ2=3−4i ohms.

Solution

We first calculate Z1+Z2Z1+Z2 and Z1Z2.Z1Z2.

Z1+Z2=(2+3i)+(3−4i)=5−iZ1Z2=(2+3i)(3−4i)=6−8i+9i−12i2FOIL=18+iSet i2=−1 and simplify.Zt=Z1Z2Z1+Z2=18+i5−i=(18+i)(5+i)(5−i)(5+i)Multiply numerator and denominator byconjugate of the denominator.=90+18i+5i+i252+12Use FOIL in the numerator;(5−i)(5+i)=52+12.=89+23i26Set i2=−1 and simplify.=8926+2326iStandard form
Z1+Z2Z1Z2Zt=Z1Z2Z1+Z2=========(2+3i)+(3−4i)=5−i(2+3i)(3−4i)6−8i+9i−12i218+i18+i5−i(18+i)(5+i)(5−i)(5+i)90+18i+5i+i252+1289+23i268926+2326iFOILSet i2=−1 and simplify.Multiply numerator and denominator byconjugate of the denominator.Use FOIL in the numerator;(5−i)(5+i)=52+12.Set i2=−1 and simplify.Standard form

Practice Problem 7

  1. Repeat Example 7 assuming that Z1=1+2i and Z2=2−3i.

Quadratic Equations with Complex Solutions

  1. 5 Solve quadratic equations having complex solutions.

The methods of solving quadratic equations introduced in the previous section are also applicable to solving quadratic equations with complex solutions.

Example 8 Solving Quadratic Equations with Complex Solutions

Solve each equation.

  1. x2+4=0

  2. x2+2x+2=0

Solution

  1. Use the square root method.

    x2+4=0Original equationx2=−4Add−4 to both sides.x=±√−4=±(√4)i=±2iSolve for x.

    The solution set is {−2i,2i}. You should check these solutions.

  2. x2+2x+2=1⋅x2+2x+2=0

    x=−2±√22−4(1)(2)2(1)Use the quadratic formulawith a=1,b=2, and c =2.=−2±√−42Simplify.=−2±2i2=2(−1±i)2=−1±i

    The solution set is {−1−i, −1+i}. You should check these solutions.

Practice Problem 8

  1. Solve.

    1. 4x2+9=0

    2. x2=4x−13

The Discriminant

Recall that in the quadratic formula

x=−b±√b2−4ac2a,

the quantity b2−4ac under the radical sign is called the discriminant. If a, b, and c are real numbers (not necessarily integers) and a≠0, the discriminant reveals the type of solutions of the equation ax2+bx+c=0, as shown in the table below.

Discriminant Description of Solutions
b2−4ac>0 There are two unequal real solutions.
b2−4ac=0 There is one real solution.
b2−4ac<0 There are two nonreal complex solutions, and they are conjugates.

Example 9 Using the Discriminant

Use the discriminant to determine the number and type of solutions of each quadratic equation.

  1. x2−4x+2=0

  2. 2t2+2t+19=0

  3. 4x2+4x+1=0

Solution

Equation b2−4ac Conclusion

  1. x2−4x+2=0

 (−4)2−4(1)(2)=8>0 Two unequal real solutions

  1. 2t2+2t+19=0

 (2)2−4(2)(19)=−148<0 Two nonreal complex solutions

  1. 4x2+4x+1=0

 (4)2−4(4)(1)=0 Exactly one real solution

Practice Problem 9

  1. Determine the number and type of solutions.

    1. 9x2−6x+1=0

    2. x2−5x+3=0

    3. 2x2−3x+4=0.

Section 1.4 Exercises

Concepts and Vocabulary

  1. We define i=_ so that i2=_.

  2. A complex number in the form a+bi is said to be in                   .

  3. For b>0, √−b=_.

  4. The conjugate of a+bi is                   , and the conjugate of a−bi is                   .

  5. True or False. The product of a complex number and its conjugate is a real number.

  6. True or False. Division by a nonzero complex number z is accomplished by multiplying the numerator and denominator by ˉz.

  7. True or False. A pure imaginary number does not have a conjugate.

  8. True or False. The standard form of 1i is a+bi where a=1 and b=−1.

Building Skills

In Exercises 9–32, perform each operation and write the result in the standard form a+bi.

  1. (5+2i)+(3+i)

  2. (6+i)+(1+2i)

  3. (4−3i)−(5+3i)

  4. (3−5i)−(3+2i)

  5. (−2−3i)+(−3−2i)

  6. (−5−3i)+(2−i)

  7. 3(5+2i)

  8. 4(3+5i)

  9. −4(2−3i)

  10. −7(3−4i)

  11. 3i(5+i)

  12. 2i(4+3i)

  13. 4i(2−5i)

  14. −3i(5−2i)

  15. (3+i)(2+3i)

  16. (4+3i)(2+5i)

  17. (2−3i)(2+3i)

  18. (4−3i)(4+3i)

  19. (3+4i)(4−3i)

  20. (−2+3i)(−3+10i)

  21. (√3−12i)2

  22. (−√5−13i)2

  23. (2−√−16)(3+5i)

  24. (5−2i)(3+√−25)

In Exercises 33–38, write the conjugate ˉz of each complex number z Then find zˉz.

  1. z=2−3i

  2. z=4+5i

  3. z=12−2i

  4. z=23+12i

  5. z=√2−3i

  6. z=√5+√3 i

In Exercises 39–52, write each quotient in the standard form a+bi.

  1. 5−i

  2. 2−3i

  3. −11+i

  4. 12−i

  5. 5i2+i

  6. 3i2−i

  7. 2+3i1+i

  8. 3+5i4+i

  9. 2−5i4−7i

  10. 3+5i1−3i

  11. 2+√−41+i

  12. 5−√−93+2i

  13. −2+√−252−3i

  14. −5−√−45−√−9

In Exercises 53–62, solve each equation.

  1. x2+5=1

  2. 4x2+9=0

  3. z2−2z+2=0

  4. x2−6x+11=0

  5. 2x2−20x+49=−7

  6. 4y2+4y+5=0

  7. 8(x2−x)=x2−3

  8. t(t+1)=3t2+1

  9. 9k2+25=0

  10. 3k2+4=0

Applying the Concepts

  1. Series circuits. Assuming that the impedance of a resistor in a circuit is Z1=4+3i ohms and the impedance of a second resistor is Z2=5−2i ohms, find the total impedance of the two resistors when placed in series (sum of the two impedances).

  2. Parallel circuits. If the two resistors in Exercise 63 are connected in parallel, the total impedance is given by

    Z1Z2Z1+Z2.

    Find the total impedance, assuming that the resistors in Exercise 63 are connected in parallel.

    As with impedance, the current I and voltage V in a circuit can be represented by complex numbers. The three quantities (voltage, V; impedance, Z; and current, I) are related by the equation Z=VI. Thus, if two of these values are given, the value of the third can be found from the equation.

In Exercises 65–70, use the equation V=ZI to find the value that is not specified.

  1. Finding impedance: I=7+5iV=35+70i

  2. Finding impedance: I=7+4iV=45+88i

  3. Finding voltage: Z=5−7iI=2+5i

  4. Finding voltage: Z=7−8iI=13+16i

  5. Finding current: V=12+10iZ=12+6i

  6. Finding current: V=29+18iZ=25+6i

Beyond the Basics

In Exercises 71–80, find each power of i and simplify the expression.

  1. i17

  2. i125

  3. i−7

  4. i−24

  5. i10+7

  6. 9+i3

  7. 3i5−2i3

  8. 5i6−3i4

  9. 2i3(1+i4)

  10. 5i5(i3−i)

  11. Prove that the reciprocal of a+bi, where both a and b are nonzero, is aa2+b2−ba2+b2i.

In Exercises 82–84, let z=a+bi. Prove each statement.

  1. a=z+ˉz2

  2. b=z−ˉz2i

  3. The product zˉz=0 if and only if z=0.

  4. Show that 1−2i5−5i=2−3i9−7i.

  5. Write in standard form:

    1+i1−i÷2+i1+2i

  6. Solve for z:

    (1+3i)z+(2+4i)=7−3i

  7. Let z=2−3i and w=1+2i.

    1. Show that ¯(zw)=(ˉz)(ˉw).

    2. Show that (zw)=ˉzˉw.

Critical Thinking/Writing/Discussion

  1. State whether each of the following is true or false. Explain your reasoning.

    1. Every real number is a complex number.

    2. Every complex number is a real number.

    3. Every complex number is an imaginary number.

    4. A real number is a complex number whose imaginary part is 0.

    5. The product of a complex number and its conjugate is a real number.

    6. The equality z=ˉz holds if and only if z is a real number.

  2. Find the least positive n for which (1+i1−i)n=1.

  3. Show that the set of complex numbers does not have the ordering properties of the set of real numbers. [Hint: Assume that ordering properties hold. Then by the law of trichotomy, i=0 or i<0 or i>0. Show that each leads to a contradiction.]

  4. Suppose that multiplication of complex numbers had been defined as

    (a+bi)(c+di)=ac+(bd)i.

    Find two complex numbers z and w, z≠0,w≠0 but zw=0.

Getting Ready for the Next Section

In Exercises 93 and 94, factor each expression by grouping.

  1. x3−x2−9x+9

  2. −x3−2x2+25x+50

In Exercises 95–100, find the LCD for each group of rational expressions.

  1. 1x−1, 1x

  2. 32−x, xx−2

  3. 15x+3, 2x+1x−3

  4. 3xx2−25, 95−x

  5. x+1(x−3)(2−x), 12(x−2)(x+1)

  6. x4x2−1, 72x−1, 1x

In Exercises 101–108, simplify each expression.

  1. 272/3

  2. 84/3

  3. 45/2

  4. 253/2

  5. (√3+x2)2

  6. (1+√x+1)2

  7. ((3x2+7)1/3)3

  8. ((5x−2)3/5)5/3

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.
Reset
3.147.72.15