Intermediate Algebra 2e · free preview

8.8 Use the Complex Number System

Learning objectives: Evaluate the square root of a negative number; Add and subtract complex numbers; Multiply complex numbers; Divide complex numbers; Simplify powers of $i$

1. Learning objectives

Learning objectives

Learning objectives

  • Evaluate the square root of a negative number
  • Add and subtract complex numbers
  • Multiply complex numbers
  • Divide complex numbers
  • Simplify powers of ii

2. Evaluate the Square Root of a Negative Number

Evaluate the Square Root of a Negative Number

Whenever we have a situation where we have a square root of a negative number we say there is no real number that equals that square root. For example, to simplify −1,\sqrt{-1}, we are looking for a real number x so that x2{}^{2} = –1. Since all real numbers squared are positive numbers, there is no real number that equals –1 when squared.

Mathematicians have often expanded their numbers systems as needed. They added 0 to the counting numbers to get the whole numbers. When they needed negative balances, they added negative numbers to get the integers. When they needed the idea of parts of a whole they added fractions and got the rational numbers. Adding the irrational numbers allowed numbers like 5.\sqrt{5}. All of these together gave us the real numbers and so far in your study of mathematics, that has been sufficient.

But now we will expand the real numbers to include the square roots of negative numbers. We start by defining the imaginary unit ii as the number whose square is –1.

The imaginary unit i is the number whose square is –1.

i2=−1 or i=−1{i}^{2}=-1\,\mathrm{or}\,i=\sqrt{-1}

We will use the imaginary unit to simplify the square roots of negative numbers.

If b is a positive real number, then

−b=b i\sqrt{-b}=\sqrt{b}\,i

We will use this definition in the next example. Be careful that it is clear that the i is not under the radical. Sometimes you will see this written as −b=ib\sqrt{-b}=i\sqrt{b} to emphasize the i is not under the radical. But the −b=b i\sqrt{-b}=\sqrt{b}\,i is considered standard form.

3. Example 8.76

Example 8.76

Example 8.76

Write each expression in terms of i and simplify if possible:

ⓐ −25\sqrt{-25} ⓑ −7\sqrt{-7} ⓒ −12.\sqrt{-12}.

Solution

ⓐ

−25\sqrt{-25}

Use the definition of the square root of negative numbers. · 25 i\sqrt{25}\,i Simplify. · 5i5i

ⓑ

−7\sqrt{-7}

Use the definition of the square root of negative numbers. · 7i\sqrt{7}i Simplify. · Be careful that it is clear that ii is not under the radical sign.

ⓒ

−12\sqrt{-12}

Use the definition of the square root of negative numbers. · 12 i\sqrt{12}\,i Simplify 12.\sqrt{12}. · 23 i2\sqrt{3}\,i

Now that we are familiar with the imaginary number i, we can expand our concept of the number system to include imaginary numbers. The complex number system includes the real numbers and the imaginary numbers. A complex number is of the form a + bi, where a, b are real numbers. We call a the real part and b the imaginary part.

A complex number is of the form a + bi, where a and b are real numbers.

A complex number is in standard form when written as a+bi,a+bi, where a and b are real numbers.

If b=0,b=0, then a+bia+bi becomes a+0⋅i=a,a+0\cdot{}i=a, and is a real number.

If b≠0,b\ne{}0, then a+bia+bi is an imaginary number.

If a=0,a=0, then a+bia+bi becomes 0+bi=bi,0+bi=bi, and is called a pure imaginary number.

We summarize this here.

a+bia+bi

b=0b=0 · a+0⋅i a\begin{array}{c} a+0\cdot{}i \\ \,a \\ \end{array} · Real number b≠0b\ne{}0 · a+bia+bi · Imaginary number a=0a=0 · 0+bi bi\begin{array}{c} 0+bi \\ \,bi \\ \end{array} · Pure imaginary number

The standard form of a complex number is a+bi,a+bi, so this explains why the preferred form is −b=bi\sqrt{-b}=\sqrt{b}i when b>0.b>0.

The diagram helps us visualize the complex number system. It is made up of both the real numbers and the imaginary numbers.

4. Add or Subtract Complex Numbers

Add or Subtract Complex Numbers

We are now ready to perform the operations of addition, subtraction, multiplication and division on the complex numbers—just as we did with the real numbers.

Adding and subtracting complex numbers is much like adding or subtracting like terms. We add or subtract the real parts and then add or subtract the imaginary parts. Our final result should be in standard form.

5. Example 8.77

Example 8.77

Example 8.77

Add: −12+−27.\sqrt{-12}+\sqrt{-27}.

Solution

−12+−27\sqrt{-12}+\sqrt{-27}

Use the definition of the square root of negative numbers. · 12 i+27 i\sqrt{12}\,i+\sqrt{27}\,i Simplify the square roots. · 23 i+33 i2\sqrt{3}\,i+3\sqrt{3}\,i Add. · 53 i5\sqrt{3}\,i

Remember to add both the real parts and the imaginary parts in this next example.

6. Example 8.78

Example 8.78

Example 8.78

Simplify: ⓐ (4−3i)+(5+6i)(4-3i)+(5+6i) ⓑ (2−5i)−(5−2i).(2-5i)-(5-2i).

Solution

ⓐ

(4−3i)+(5+6i)(4-3i)+(5+6i)

Use the Associative Property to put the realparts and the imaginary parts together. · (4+5)+(−3i+6i)(4+5)+(-3i+6i) Simplify. · 9+3i9+3i

ⓑ

(2−5i)−(5−2i)(2-5i)-(5-2i)

Distribute. · 2−5i−5+2i2-5i-5+2i Use the Associative Property to put the realparts and the imaginary parts together. · 2−5−5i+2i2-5-5i+2i Simplify. · −3−3i-3-3i

7. Multiply Complex Numbers

Multiply Complex Numbers

Multiplying complex numbers is also much like multiplying expressions with coefficients and variables. There is only one special case we need to consider. We will look at that after we practice in the next two examples.

8. Example 8.79

Example 8.79

Example 8.79

Multiply: 2i(7−5i).2i(7-5i).

Solution

2i(7−5i)2i(7-5i)

Distribute. · 14i−10i214i-10{i}^{2} Simplify i2.{i}^{2}. · 14i−10(−1)14i-10(-1) Multiply. · 14i+1014i+10 Write in standard form. · 10+14i10+14i

In the next example, we multiply the binomials using the Distributive Property or FOIL.

9. Example 8.80

Example 8.80

Example 8.80

Multiply: (3+2i)(4−3i).(3+2i)(4-3i).

Solution

(3+2i)(4−3i)(3+2i)(4-3i)

Use FOIL. · 12−9i+8i−6i212-9i+8i-6{i}^{2} Simplify i2{i}^{2} and combine like terms. · 12−i−6(−1)12-i-6(-1) Multiply. · 12−i+612-i+6 Combine the real parts. · 18−i18-i

In the next example, we could use FOIL or the Product of Binomial Squares Pattern.

10. Example 8.81

Example 8.81

Example 8.81

Multiply: (3+2i)2{(3+2i)}^{2}

Solution

Since the square root of a negative number is not a real number, when we have the square roots of two negative numbers, we cannot use the Product Property for Radicals. In order to multiply square roots of negative numbers we should first write them as complex numbers, using −b=bi.\sqrt{-b}=\sqrt{b}i. This is one place students tend to make errors, so be careful when you see multiplying with a negative square root.

11. Example 8.82

Example 8.82

Example 8.82

Multiply: −36⋅−4.\sqrt{-36}\cdot\sqrt{-4}.

Solution

To multiply square roots of negative numbers, we first write them as complex numbers.

−36⋅−4\sqrt{-36}\cdot\sqrt{-4}

Write as complex numbers using −b=bi.\sqrt{-b}=\sqrt{b}i. · 36 i⋅4 i\sqrt{36}\,i\cdot\sqrt{4}\,i Simplify. · 6i⋅2i6i\cdot{}2i Multiply. · 12i212{i}^{2} Simplify i2{i}^{2} and multiply. · −12-12

In the next example, each binomial has a square root of a negative number. Before multiplying, each square root of a negative number must be written as a complex number.

12. Example 8.83

Example 8.83

Example 8.83

Multiply: (3−−12)(5+−27).(3-\sqrt{-12})(5+\sqrt{-27}).

Solution

To multiply square roots of negative numbers, we first write them as complex numbers.

(3−−12)(5+−27)(3-\sqrt{-12})(5+\sqrt{-27})

Write as complex numbers using −b=bi.\sqrt{-b}=\sqrt{b}i. · (3−23 i)(5+33 i)(3-2\sqrt{3}\,i)(5+3\sqrt{3}\,i) Use FOIL. · 15+93 i−103 i−6⋅3i215+9\sqrt{3}\,i-10\sqrt{3}\,i-6\cdot{}3{i}^{2} Combine like terms and simplify i2.{i}^{2}. · 15−3 i−6⋅(−3)15-\sqrt{3}\,i-6\cdot(-3) Multiply and combine like terms. · 33−3 i33-\sqrt{3}\,i

We first looked at conjugate pairs when we studied polynomials. We said that a pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference is called a conjugate pair and is of the form (a−b),(a+b).(a-b),(a+b).

A complex conjugate pair is very similar. For a complex number of the form a+bi,a+bi, its conjugate is a−bi.a-bi. Notice they have the same first term and the same last term, but one is a sum and one is a difference.

A complex conjugate pair is of the form a+bi,a+bi,a−bi.a-bi.

We will multiply a complex conjugate pair in the next example.

13. Example 8.84

Example 8.84

Example 8.84

Multiply: (3−2i)(3+2i).(3-2i)(3+2i).

Solution

(3−2i)(3+2i)(3-2i)(3+2i)

Use FOIL. · 9+6i−6i−4i29+6i-6i-4{i}^{2} Combine like terms and simplify i2.{i}^{2}. · 9−4(−1)9-4(-1) Multiply and combine like terms. · 13

From our study of polynomials, we know the product of conjugates is always of the form (a−b)(a+b)=a2−b2.(a-b)(a+b)={a}^{2}-{b}^{2}. The result is called a difference of squares. We can multiply a complex conjugate pair using this pattern.

The last example we used FOIL. Now we will use the Product of Conjugates Pattern.

Notice this is the same result we found in Example 8.84.

When we multiply complex conjugates, the product of the last terms will always have an i2{i}^{2} which simplifies to −1.-1.

(a−bi)(a+bi)a2−(bi)2a2−b2i2a2−b2(−1)a2+b2\begin{array}{c} (a-bi)(a+bi) \\ {a}^{2}-{(bi)}^{2} \\ {a}^{2}-{b}^{2}{i}^{2} \\ {a}^{2}-{b}^{2}(-1) \\ {a}^{2}+{b}^{2} \\ \end{array}

This leads us to the Product of Complex Conjugates Pattern: (a−bi)(a+bi)=a2+b2(a-bi)(a+bi)={a}^{2}+{b}^{2}

If a and b are real numbers, then

(a−bi)(a+bi)=a2+b2(a-bi)(a+bi)={a}^{2}+{b}^{2}

14. Example 8.85

Example 8.85

Example 8.85

Multiply using the Product of Complex Conjugates Pattern: (8−2i)(8+2i).(8-2i)(8+2i).

Solution

15. Divide Complex Numbers

Divide Complex Numbers

Dividing complex numbers is much like rationalizing a denominator. We want our result to be in standard form with no imaginary numbers in the denominator.

16. Example 8.86

Example 8.86

Example 8.86

How to Divide Complex Numbers

Divide: 4+3i3−4i.\frac{4+3i}{3-4i}.

We summarize the steps here.

How to divide complex numbers.

  1. Step 1. Write both the numerator and denominator in standard form.
  2. Step 2. Multiply the numerator and denominator by the complex conjugate of the denominator.
  3. Step 3. Simplify and write the result in standard form.

17. Example 8.87

Example 8.87

Example 8.87

Divide, writing the answer in standard form: −35+2i.\frac{-3}{5+2i}.

Solution

−35+2i\frac{-3}{5+2i}

Multiply the numerator and denominator by thecomplex conjugate of the denominator. · −3(5−2i)(5+2i)(5−2i)\frac{-3(5-2i)}{(5+2i)(5-2i)} Multiply in the numerator and use the Product ofComplex Conjugates Pattern in the denominator. · −15+6i52+22\frac{-15+6i}{{5}^{2}+{2}^{2}} Simplify. · −15+6i29\frac{-15+6i}{29} Write in standard form. · −1529+629i-\frac{15}{29}+\frac{6}{29}i

Be careful as you find the conjugate of the denominator.

18. Example 8.88

Example 8.88

Example 8.88

Divide: 5+3i4i.\frac{5+3i}{4i}.

Solution

5+3i4i\frac{5+3i}{4i}

Write the denominator in standard form. · 5+3i0+4i\frac{5+3i}{0+4i} Multiply the numerator and denominator bythe complex conjugate of the denominator. · (5+3i)(0−4i)(0+4i)(0−4i)\frac{(5+3i)(0-4i)}{(0+4i)(0-4i)} Simplify. · (5+3i)(−4i)(4i)(−4i)\frac{(5+3i)(-4i)}{(4i)(-4i)} Multiply. · −20i−12i2−16i2\frac{-20i-12{i}^{2}}{-16{i}^{2}} Simplify the i2.{i}^{2}. · −20i+1216\frac{-20i+12}{16} Rewrite in standard form. ·  1216−2016i\,\frac{12}{16}-\frac{20}{16}i Simplify the fractions. ·  34−54i\,\frac{3}{4}-\frac{5}{4}i

19. Simplify Powers of *i*

Simplify Powers of i

The powers of ii make an interesting pattern that will help us simplify higher powers of i. Let’s evaluate the powers of ii to see the pattern.

i1 i2 i3 i4i −1 i2⋅i i2⋅i2 −1⋅i (−1)(−1) −i 1i5 i6 i7 i8i4⋅i i4⋅i2 i4⋅i3 i4⋅i41⋅i 1⋅i2 1⋅i3 1⋅1i i2 i3 1 −1 −i\begin{array}{cccccccccc} {i}^{1} & & & \,{i}^{2} & & & \,{i}^{3} & & & \,{i}^{4} \\ i & & & \,-1 & & & \,{i}^{2}\cdot{}i & & & \,{i}^{2}\cdot{i}^{2} \\ & & & & & & \,-1\cdot{}i & & & \,(-1)(-1) \\ & & & & & & \,-i & & & \,1 \\ {i}^{5} & & & \,{i}^{6} & & & \,{i}^{7} & & & \,{i}^{8} \\ {i}^{4}\cdot{}i & & & \,{i}^{4}\cdot{i}^{2} & & & \,{i}^{4}\cdot{i}^{3} & & & \,{i}^{4}\cdot{i}^{4} \\ 1\cdot{}i & & & \,1\cdot{i}^{2} & & & \,1\cdot{i}^{3} & & & \,1\cdot{}1 \\ i & & & \,{i}^{2} & & & \,{i}^{3} & & & \,1 \\ & & & \,-1 & & & \,-i \\ \end{array}

We summarize this now.

i1=i i5=ii2=−1 i6=−1i3=−i i7=−ii4=1 i8=1\begin{array}{rllccrll} {i}^{1} & = & i & & & \,{i}^{5} & = & i \\ {i}^{2} & = & -1 & & & \,{i}^{6} & = & -1 \\ {i}^{3} & = & -i & & & \,{i}^{7} & = & -i \\ {i}^{4} & = & 1 & & & \,{i}^{8} & = & 1 \\ \end{array}

If we continued, the pattern would keep repeating in blocks of four. We can use this pattern to help us simplify powers of i. Since i4{}^{4} = 1, we rewrite each power, ini^{n}, as a product using i4{}^{4} to a power and another power of i.

We rewrite it in the form in=(i4)q⋅ir,{i}^{n}={({i}^{4})}^{q}\cdot{i}^{r}, where the exponent, q, is the quotient of n divided by 4 and the exponent, r, is the remainder from this division. For example, to simplify i57{}^{57}, we divide 57 by 4 and we get 14 with a remainder of 1. In other words, 57=4⋅14+1.57=4\cdot{}14+1. So we write i57=(14)14⋅i1{i}^{57}={({1}^{4})}^{14}\cdot{i}^{1} and then simplify from there.

20. Example 8.89

Example 8.89

Example 8.89

Simplify: i86.{i}^{86}.

Solution

21. Section 8.8 Exercises

Section 8.8 Exercises

Section 8.8 Exercises

Evaluate the Square Root of a Negative Number

In the following exercises, write each expression in terms of i and simplify if possible.

Add or Subtract Complex Numbers In the following exercises, add or subtract.

Multiply Complex Numbers

In the following exercises, multiply.

In the following exercises, multiply using the Product of Binomial Squares Pattern.

In the following exercises, multiply.

In the following exercises, multiply using the Product of Complex Conjugates Pattern.

Divide Complex Numbers

In the following exercises, divide.

Simplify Powers of i

In the following exercises, simplify.

Writing Exercises

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1−10,1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

22. Practice 1

  1. Write each expression in terms of i and simplify if possible:

ⓐ −81\sqrt{-81} ⓑ −5\sqrt{-5} ⓒ −18.\sqrt{-18}.

  1. Write each expression in terms of i and simplify if possible:

ⓐ −36\sqrt{-36} ⓑ −3\sqrt{-3} ⓒ −27.\sqrt{-27}.

  1. Add: −8+−32.\sqrt{-8}+\sqrt{-32}.
  1. Add: −27+−48.\sqrt{-27}+\sqrt{-48}.

23. Practice 2

  1. Simplify: ⓐ (2+7i)+(4−2i)(2+7i)+(4-2i) ⓑ (8−4i)−(2−i).(8-4i)-(2-i).
  1. Simplify: ⓐ (3−2i)+(−5−4i)(3-2i)+(-5-4i) ⓑ (4+3i)−(2−6i).(4+3i)-(2-6i).
  1. Multiply: 4i(5−3i).4i(5-3i).
  1. Multiply: −3i(2+4i).-3i(2+4i).

24. Practice 3

  1. Multiply: (5−3i)(−1−2i).(5-3i)(-1-2i).
  1. Multiply: (−4−3i)(2+i).(-4-3i)(2+i).
  1. Multiply using the Binomial Squares pattern: (−2−5i)2.{(-2-5i)}^{2}.
  1. Multiply using the Binomial Squares pattern: (−5+4i)2.{(-5+4i)}^{2}.

25. Practice 4

  1. Multiply: −49⋅−4.\sqrt{-49}\cdot\sqrt{-4}.
  1. Multiply: −36⋅−81.\sqrt{-36}\cdot\sqrt{-81}.
  1. Multiply: (4−−12)(3−−48).(4-\sqrt{-12})(3-\sqrt{-48}).
  1. Multiply: (−2+−8)(3−−18).(-2+\sqrt{-8})(3-\sqrt{-18}).

26. Practice 5

  1. Divide: 2+5i5−2i.\frac{2+5i}{5-2i}.
  1. Divide: 1+6i6−i.\frac{1+6i}{6-i}.
  1. Divide, writing the answer in standard form: 41−4i.\frac{4}{1-4i}.
  1. Divide, writing the answer in standard form: −2−1+2i.\frac{-2}{-1+2i}.

27. Practice 6

  1. Divide: 3+3i2i.\frac{3+3i}{2i}.
  1. Divide: 2+4i5i.\frac{2+4i}{5i}.
  1. Simplify: i75.{i}^{75}.
  1. Simplify: i92.{i}^{92}.

28. Practice 7

  1. ⓐ −16\sqrt{-16} ⓑ −11\sqrt{-11}ⓒ −8\sqrt{-8}
  1. ⓐ −100\sqrt{-100} ⓑ −13\sqrt{-13} ⓒ −45\sqrt{-45}
  1. −75+−48\sqrt{-75}+\sqrt{-48}
  1. −50+−18\sqrt{-50}+\sqrt{-18}

29. Practice 8

  1. (1+3i)+(7+4i)(1+3i)+(7+4i)
  1. (8−i)+(6+3i)(8-i)+(6+3i)
  1. (1−4i)−(3−6i)(1-4i)-(3-6i)
  1. (6+i)−(−2−4i)(6+i)-(-2-4i)

30. Practice 9

  1. (5−−36)+(2−−49)(5-\sqrt{-36})+(2-\sqrt{-49})
  1. (−7−−50)−(−32−−18)(-7-\sqrt{-50})-(-32-\sqrt{-18})
  1. 4i(5−3i)4i(5-3i)
  1. −6i(−3−2i)-6i(-3-2i)

31. Practice 10

  1. (4+3i)(−5+6i)(4+3i)(-5+6i)
  1. (−3+3i)(−2−7i)(-3+3i)(-2-7i)
  1. (3+4i)2{(3+4i)}^{2}
  1. (−2−3i)2{(-2-3i)}^{2}

32. Practice 11

  1. −25⋅−36\sqrt{-25}\cdot\sqrt{-36}
  1. −9⋅−100\sqrt{-9}\cdot\sqrt{-100}
  1. (−2−−27)(4−−48)(-2-\sqrt{-27})(4-\sqrt{-48})
  1. (2+−8)(−4+−18)(2+\sqrt{-8})(-4+\sqrt{-18})

33. Practice 12

  1. 3+4i4−3i\frac{3+4i}{4-3i}
  1. 2+i3−4i\frac{2+i}{3-4i}
  1. 32−3i\frac{3}{2-3i}
  1. −43−2i\frac{-4}{3-2i}

34. Practice 13

  1. 1+4i3i\frac{1+4i}{3i}
  1. 4+3i7i\frac{4+3i}{7i}
  1. −2−3i4i\frac{-2-3i}{4i}
  1. i41{i}^{41}

35. Practice 14

  1. i66{i}^{66}
  1. i137{i}^{137}
  1. Explain the relationship between real numbers and complex numbers.
  1. Why is −64=8i\sqrt{-64}=8i but −643=−4.\sqrt[3]{-64}=-4.

Practice this interactively

Free account · instant grading · spaced review that schedules itself.

Start this course for free