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Mathematics LibreTexts

17.9: Use the Complex Number System

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Learning Objectives

By the end of this section, you will be able to:

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

Before you get started, take this readiness quiz.

  1. Given the numbers 4,7,0.¯5,73,3,81, list the
    1. rational numbers
    2. irrational numbers
    3. real numbers
      If you missed this problem, review Example 1.42.
  2. Multiply: (x3)(2x+5).
    If you missed this problem, review Example 5.28.
  3. Rationalize the denominator: 553
    If you missed this problem, review Example 5.32.

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, we are looking for a real number x so that x2=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. 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 i as the number whose square is 1.

Definition 17.9.1

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

i2=1 or i=1

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

Definition 17.9.2

Square Root of a Negative Number

If b is a positive real number, then

b=bi

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 to emphasize the i is not under the radical. But the b=bi is considered standard form.

Example 17.9.1

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

  1. 25
  2. 7
  3. 12

Solution:

a.

25

Use the definition of the square root of negative numbers.

25i

Simplify.

5i

b.

7

Use the definition of the square root of negative numbers.

7i

Simplify.

Be careful that it is clear that i is not under the radical sign.

c.

12

Use the definition of the square root of negative numbers.

12i

Simplify 12.

23i

Exercise 17.9.1

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

  1. 81
  2. 5
  3. 18
Answer
  1. 9i
  2. 5i
  3. 32i
Exercise 17.9.2

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

  1. 36
  2. 3
  3. 27
Answer
  1. 6i
  2. 3i
  3. 33i

Now that we are familiar with the imaginary number i, we can expand the real numbers 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.

Definition 17.9.3

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

The image shows the expression a plus b i. The number a is labeled “real part” and the number b i is labeled “imaginary part”.
Figure 8.8.1

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

If b=0, then a+bi becomes a+0i=a, and is a real number.

If b0, then a+bi is an imaginary number.

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

We summarize this here.

  a+bi  
b=0

a+0i

a

Real number
b0 a+bi Imaginary number
a=0R

0+bi

bi

Pure imaginary numbe4
Table 8.8.1

The standard form of a complex number is a+bi, so this explains why the preferred form is b=bi when b>0.

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

The table has four rows and three columns. The first row is a header and the second column entry a plus b i. In the second row is b equals zero, a plus 0 i, and “Real number”. The third row contains b is not equal to 0, a plus b i, and “Imaginary number”. The fourth row contains a = 0, 0 plus b i, and “Pure imaginary number”.
Figure 8.8.2

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.

Example 17.9.2

Add: 12+27.

Solution:

12+27

Use the definition of the square root of negative numbers.

12i+27i

Simplify the square roots.

23i+33i

Add.

53i

Exercise 17.9.3

Add: 8+32.

Answer

62i

Exercise 17.9.4

Add: 27+48

Answer

73i

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

Example 17.9.3

Simplify:

  1. (43i)+(5+6i)
  2. (25i)(52i)

Solution:

a.

(43i)+(5+6i)

Use the Associative Property to put the real parts and the imaginary parts together.

(4+5)+(3i+6i)

Simplify.

9+3i

b.

(25i)(52i)

Distribute.

25i5+2i

Use the Associative Property to put the real parts and the imaginary parts together.

255i+2i

Simplify.

33i

Exercise 17.9.5

Simplify:

  1. (2+7i)+(42i)
  2. (84i)(2i)
Answer
  1. 6+5i
  2. 63i
Exercise 17.9.6

Simplify:

  1. (32i)+(54i)
  2. (4+3i)(26i)
Answer
  1. 26i
  2. 2+9i

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.

Example 17.9.4

Multiply: 2i(75i)

Solution:

2i(75i)

Distribute.

14i10i2

Simplify i2.

14i10(1)

Multiply.

14i+10

Write in standard form.

10+14i

Exercise 17.9.7

Multiply: 4i(53i).

Answer

12+20i

Exercise 17.9.8

Multiply: 3i(2+4i).

Answer

126i

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

Example 17.9.5

Multiply: (3+2i)(43i).

Solution:

(3+2i)(43i)

Use FOIL.

129i+8i6i2

Simplify i2 and combine like terms.

12i6(1)

Multiply.

12i+6

Combine the real parts.

18i

Exercise 17.9.9

Multiple: (53i)(12i).

Answer

117i

Exercise 17.9.10

Multiple: (43i)(2+i).

Answer

510i

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

Example 17.9.6

Multiply: (3+2i)2

Solution:

  .
Use the Product of Binomial Squares Pattern, (a+b)2=a2+2ab+b2. .
Simplify. .
Simplify i2. .
Simplify. .
Table 8.8.2
Exercise 17.9.11

Multiply using the Binomial Squares pattern: (25i)2.

Answer

21+20i

Exercise 17.9.12

Multiply using the Binomial Squares pattern: (5+4i)2.

Answer

940i

Since the square root of a negative number is not a real number, 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.This is one place students tend to make errors, so be careful when you see multiplying with a negative square root.

Example 17.9.7

Multiply: 364.

Solution:

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

364

Write as complex numbers using b=bi.

36i4i

Simplify.

6i2i

Multiply.

12i2

Simplify i2 and multiply.

12

Exercise 17.9.13

Multiply: 494.

Answer

14

Exercise 17.9.14

Multiply: 3681.

Answer

54

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.

Example 17.9.8

Multiply: (312)(5+27).

Solution:

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

(312)(5+27)

Write as complex numbers using b=bi.

(323i)(5+33i)

Use FOIL.

15+93i103i63i2

Combine like terms and simplify i2.

153i6(3)

Multiply and combine like terms.

333i

Exercise 17.9.15

Multiply: (412)(348).

Answer

12223i

Exercise 17.9.16

Multiply: (2+8)(318).

Answer

6+122i

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 (ab),(a+b).

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

Definition 17.9.4

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

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

Example 17.9.9

Multiply: (32i)(3+2i).

Solution:

(32i)(3+2i)

Use FOIL

9+6i6i4i2

Combine like terms and simplify i2.

94(1)

Multiply and combine like terms.

13

Exercise 17.9.17

Multiply: (43i)(4+3i).

Answer

25

Exercise 17.9.18

Multiply: (2+5i)(25i).

Answer

29

From our study of polynomials, we know the product of conjugates is always of the form (ab)(a+b)=a2b2.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.

The quantity a minus b in parentheses times the quantity a plus b in parentheses is written above the expression showing the product of 3 minus 2 i in parentheses and 3 plus 2 i in parentheses. In the next line a squared minus b squared is written above the expression 3 squared minus the quantity 2 i in parentheses squared. Simplifying we get 9 minus 4 i squared. This is equal to 9 minus 4 times negative 1. The final result is 13.
Figure 8.8.8

Notice this is the same result we found in Example 8.8.9.

When we multiply complex conjugates, the product of the last terms will always have an i2 which simplifies to 1.

(abi)(a+bi)a2(bi)2a2b2i2a2b2(1)a2+b2

This leads us to the Product of Complex Conjugates Pattern: (abi)(a+bi)=a2+b2

Definition 17.9.5

Product of Complex Conjugates

If a and b are real numbers, then

(abi)(a+bi)=a2+b2

Example 17.9.10

Multiply using the Product of Complex Conjugates Pattern: (82i)(8+2i).

Solution:

  .
Use the Product of Complex Conjugates Pattern, (abi)(a+bi)=a2+b2. .
Simplify the squares. .
Add. .
Table 8.8.3
Exercise 17.9.19

Multiply using the Product of Complex Conjugates Pattern: (310i)(3+10i).

Answer

109

Exercise 17.9.20

Multiply using the Product of Complex Conjugates Pattern: (5+4i)(54i).

Answer

41

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.

Example 17.9.11 how to divide complex numbers

Divide: 4+3i34i.

Solution:

Step 1: Write both the numerator and denominator in standard form. They are both in standard form. 4+3i34i
Step 2: Multiply the numerator and denominator by the complex conjugate of the denominator. The complex conjugate of 34i is 3+4i. (4+3i)(3+4i)(34i)(3+4i)
Step 3: Simplify and write the result in standard form.

Use the pattern (abi)(a+bi)=a2+b2 in the denominator.

Combine like terms.

Simplify.

Write the result in standard form.

12+16i+9i+12i29+1612+25i122525i25i
Table 8.8.4
Exercise 17.9.21

Divide: 2+5i52i.

Answer

i

Exercise 17.9.22

Divide: 1+6i6i.

Answer

i

We summarize the steps here.

How to Divide Complex Numbers

  1. Write both the numerator and denominator in standard form.
  2. Multiply both the numerator and denominator by the complex conjugate of the denominator.
  3. Simplify and write the result in standard form.
Example 17.9.12

Divide, writing the answers in standard form: 35+2i.

Solution:

35+2i

Multiply the numerator and denominator by the complex conjugate of the denominator.

3(52i)(5+2i)(52i)

Multiply in the numerator and use the Product of Complex Conjugates Pattern in the denominator.

15+6i52+22

Simplify.

15+6i29

Write in standard form.

1529+629i

Exercise 17.9.23

Divide, writing the answer in standard form: 414i.

Answer

417+1617i

Exercise 17.9.24

Divide, writing the answer in standard form: 21+2i.

Answer

25+45i

Be careful as you find the conjugate of the denominator.

Example 17.9.13

Divide: 5+3i4i.

Solution:

5+3i4i

Write the denominator in standard form.

5+3i0+4i

Multiply the numerator and denominator by the complex conjugate of the denominator.

(5+3i)(04i)(0+4i)(04i)

Simplify.

(5+3i)(4i)(4i)(4i)

Multiply.

20i12i216i2.

Simplify the i2.

20i+1216

Rewrite in standard form.

12162016i

Simplify the fractions.

3454i

Exercise 17.9.25

Divide: 3+3i2i.

Answer

3232i

Exercise 17.9.26

Divide: 2+4i5i.

Answer

4525i

Simplify Powers of i

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

i1i2i3i4i1i2ii2i21i(1)(1)i1

i5i6i7i8i4ii4i2i4i3i4i41i1i21i311ii2i311i

We summarize this now.

i1=ii5=ii2=1i6=1i3=ii7=ii4=1i8=1

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=1, we rewrite each power, in, as a product using i4 to a power and another power of i.

We rewrite it in the form in=(i4)qir, 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, we divide 57 by 4 and we get 14 with a remainder of 1. In other words, 57=414+1. So we write i57=(14)14i1 and then simplify from there.

.
Figure 8.8.13
Example 17.9.14

Simplify: i86.

Solution:

i86

Divide 86 by 4 and rewrite i86 in the in=(i4)qir form.

(14)21i2

\require{enclose} \begin{array}{rll}     21 && \hbox{(divide the exponent by the number of repeats in the imaginary number i)} \\[-3pt]    4\enclose{longdiv}{86}\kern-.2ex \\[-3pt]       \underline{8\phantom{0}}  && \hbox{(four goes into eight twice with zero remainders)} \\[-3pt]       06\phantom{}   \\[-3pt]       \underline{\phantom{0}4\phantom{}} && \hbox{(four goes into 6 one time with 2 remainders)}  \\[-3pt]       \phantom{0}2  && \hbox{(remainder is 2 so calculate the imaginary number with an exponent of two)}  \\[-3pt]   \end{array} Simplify.

 

(1)^{21} \cdot(-1)

Simplify.

-1

Exercise \PageIndex{27}

Simplify: i^{74}.

Answer

-1

Exercise \PageIndex{28}

Simplify: i^{92}.

Answer

1

Access these online resources for additional instruction and practice with the complex number system.

  • Expressing Square Roots of Negative Numbers with i
  • Subtract and Multiply Complex Numbers
  • Dividing Complex Numbers
  • Rewriting Powers of i

Key Concepts

  • Square Root of a Negative Number
    • If b is a positive real number, then \(\sqrt{-b}=\sqrt{b} i\
  a+bi  
b=0

a+0\cdot i

a

Real number
b\neq 0 a+bi Imaginary number
a=0

0+bi

bi

Pure imaginary number
Table 8.8.1
    • A complex number is in standard form when written as a + bi, where a, b are real numbers.
      The diagram has a rectangle with the labels “Complex Numbers” and a plus b i. A second rectangle has the labels “Real Numbers”, a plus b i, b = 0. A third rectangle has the labels “Imaginary Numbers”, a plus b i, b not equal to 0. Arrows go from the Real Numbers rectangle and Imaginary Numbers rectangle and point toward the Complex Numbers rectangle.
      Figure 8.8.2
  • Product of Complex Conjugates
    • If a, b are real numbers, then
      (a−bi)(a+bi)=a^{2}+b^{2}
  • How to Divide Complex Numbers
    1. Write both the numerator and denominator in standard form.
    2. Multiply the numerator and denominator by the complex conjugate of the denominator.
    3. Simplify and write the result in standard form.

Glossary

complex conjugate pair
A complex conjugate pair is of the form a+bi, a-bi.
complex number
A complex number is of the form a+bi, where a and b are real numbers. We call a the real part and b the imaginary part.
complex number system
The complex number system is made up of both the real numbers and the imaginary numbers.
imaginary unit
The imaginary unit i is the number whose square is –1. i^{2}=-1 or i=\sqrt{−1}.
standard form
A complex number is in standard form when written as a+bi, where a, b are real numbers.

17.9: Use the Complex Number System is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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