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Mathematics LibreTexts

5.7: Complex Numbers and Their Operations

  • Anonymous
  • LibreTexts

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Learning Objectives

  • Define the imaginary unit and complex numbers.
  • Add and subtract complex numbers.
  • Multiply and divide complex numbers.

Introduction to Complex Numbers

Up to this point the square root of a negative number has been left undefined. For example, we know that 9 is not a real number.

9=? or (?)2=9

There is no real number that when squared results in a negative number. We begin to resolve this issue by defining the imaginary unit26, i, as the square root of 1.

i=1 and i2=1

To express a square root of a negative number in terms of the imaginary unit i, we use the following property where a represents any non-negative real number:

a=1a=1a=ia

With this we can write.

9=19=19=i3=3i

If 9=3i, then we would expect that 3i squared will equal 9:

(3i)2=9i2=9(1)=9

In this way any square root of a negative real number can be written in terms of the imaginary unit. Such a number is often called an imaginary number27.

Example 5.7.1

Rewrite in terms of the imaginary unit i.

  1. 7
  2. 25
  3. 72

Solution

  1. 7=17=17=i7
  2. 25=125=125=i5=5i
  3. 72=1362=1362=i62=6i2

When an imaginary number involves a radical, we place i in front of the radical. Consider the following:

6i2=62i

Since multiplication is commutative, these numbers are equivalent. However, in the form 62i, the imaginary unit i is often misinterpreted to be part of the radicand. To avoid this confusion, it is a best practice to place i in front of the radical and use 6i2

A complex number28 is any number of the form,

a+bi

where a and b are real numbers. Here, a is called the real part29 and b is called the imaginary part30. For example, 34i is a complex number with a real part of 3 and an imaginary part of 4. It is important to note that any real number is also a complex number. For example, 5 is a real number; it can be written as 5+0i with a real part of 5 and an imaginary part of 0. Hence, the set of real numbers, denoted , is a subset of the set of complex numbers, denoted .

C={a+bi|a,b}

de7ef4a536b8b16aa99d881cde47715f.png
Figure 5.7.1

Complex numbers are used in many fields including electronics, engineering, physics, and mathematics. In this textbook we will use them to better understand solutions to equations such as x2+4=0. For this reason, we next explore algebraic operations with them.

Adding and Subtracting Complex Numbers

Adding or subtracting complex numbers is similar to adding and subtracting polynomials with like terms. We add or subtract the real parts and then the imaginary parts.

Example 5.7.2

Add (52i)+(7+3i).

Solution

Add the real parts and then add the imaginary parts.

(52i)+(7+3i)=52i+7+3i=5+72i+3i=12+i

Answer

12+i

To subtract complex numbers, we subtract the real parts and subtract the imaginary parts. This is consistent with the use of the distributive property.

Example 5.7.3:

Subtract (107i)(9+5i).

Solution

Distribute the negative sign and then combine like terms.

(107i)(9+5i)=107i95i=1097i5i=112i

Answer:

112i

In general, given real numbers a, b, c, and d:

(a+bi)+(c+di)=(a+c)+(b+d)i(a+bi)(c+di)=(ac)+(bd)i

Example 5.7.4:

Simplify (5+i)+(23i)(47i).

Solution

(5+i)+(23i)(47i)=5+i+23i4+7i=3+5i

Answer:

3+5i

In summary, adding and subtracting complex numbers results in a complex number.

Multiplying and Dividing Complex Numbers

Multiplying complex numbers is similar to multiplying polynomials. The distributive property applies. In addition, we make use of the fact that i2=1 to simplify the result into standard form a+bi.

Example 5.7.5:

Multiply 6i(23i).

Solution

We begin by applying the distributive property.

6i(23i)=(6i)2(6i)3iDistribute.=12i+18i2Substitutei2=1.=12i+18(1)Simplify.=12i18=1812i

Answer:

1812i

Example 5.7.6:

Multiply (34i)(4+5i).

Solution

(34i)(4+5i)=34+35i4i44i5iDistribute.=12+15i16i20i2Substitutei2=1.=12+15i16i20(1)=12i+20=32i

Answer:

32i

In general, given real numbers a, b, c, and d:

(a+bi)(c+di)=ac+adi+bci+bdi2=ac+adi+bci+bd(1)=ac+(ad+bc)ibd=(acbd)+(ad+bc)i

Exercise 5.7.1

Simplify: (32i)2.

Answer

512i

www.youtube.com/v/OVtqcB2z9v8

Given a complex number a+bi, its complex conjugate31 is abi.We next explore the product of complex conjugates.

Example 5.7.7:

Multiply (5+2i)(52i).

Solution

(5+2i)(52i)=5552i+2i52i2i=2510i+10i4i2=254(1)=25+4=29

Answer:

29

In general, the product of complex conjugates32 follows:

(a+bi)(abi)=a2abi+biab2i2=a2abi+abib2(1)=a2+b2

Note that the result does not involve the imaginary unit; hence, it is real. This leads us to the very useful property

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

To divide complex numbers, we apply the technique used to rationalize the denominator. Multiply the numerator and denominator by the conjugate of the denominator. The result can then be simplified into standard form a+bi.

Example 5.7.8:

Divide 123i.

Solution

In this example, the conjugate of the denominator is 2+3i. Therefore, we will multiply by 1 in the form (2+3i)(2+3i).

123i=1(23i)(2+3i)(2+3i)=(2+3i)22+32=2+3i4+9=2+3i13

To write this complex number in standard form, we make use of the fact that 13 is a common denominator.

2+3i13=213+3i13=213+313i

Answer

213+313i

Example 5.7.9:

Divide: 15i4+i.

Solution

15i4+i=(15i)(4+i)(4i)(4i)=4i20i+5i242+12=421i+5(1)16+1=421i516+1=121i17=1172117i

Answer

1172117i

In general, given real numbers a, b, c and d where c and d are not both 0:

(a+bi)(c+di)=(a+bi)(c+di)(cdi)(cdi)=acadi+bcibdi2c2+d2=(ac+bd)+(bcad)ic2+d2=(ac+bdc2+d2)+(bcadc2+d2)i

Example 5.7.10:

Divide: 83i2i.

Solution

Here we can think of 2i=0+2i and thus we can see that its conjugate is 2i=02i.

83i2i=(83i)(2i)(2i)(2i)=16i+6i24i2=16i+6(1)4(1)=16i64=616i4=6416i4=324i

Because the denominator is a monomial, we could multiply numerator and denominator by 1 in the form of ii and save some steps reducing in the end.

83i2i=(83i)(2i)ii=8i3i22i2=8i3(1)2(1)=8i+32=8i2+32=4i32

Answer

324i

Exercise 5.7.2

Divide 3+2i1i.

Answer

12+52i

www.youtube.com/v/4r2kolq-_T0

When multiplying and dividing complex numbers we must take care to understand that the product and quotient rules for radicals require that both a and b are positive. In other words, if na and nb are both real numbers then we have the following rules.

Productruleforradicals:nab=nanbQuotientruleforradicals:nab=nanb

For example, we can demonstrate that the product rule is true when a and b are both positive as follows:

49=3623=66=6

However, when a and b are both negative the property is not true.

49?=362i3i=66i2=66=6

Here 4 and 9 both are not real numbers and the product rule for radicals fails to produce a true statement. Therefore, to avoid some common errors associated with this technicality, ensure that any complex number is written in terms of the imaginary unit i before performing any operations.

Example 5.7.11:

Multiply 615.

Solution

Begin by writing the radicals in terms of the imaginary unit i.

615=i6i15

Now the radicands are both positive and the product rule for radicals applies.

615=i6i15=i615=(1)90=(1)910=(1)310=310

Answer

310

Example 5.7.12:

Multiply: 10(610).

Solution

Begin by writing the radicals in terms of the imaginary unit i and then distribute.

10(610)=i10(i610)=i260i100=(1)415i100=(1)215i10=21510i

Answer:

21510i

In summary, multiplying and dividing complex numbers results in a complex number.

Exercise 5.7.3

Simplify: (2i2)2(3i5)2.

Answer

12+6i5

www.youtube.com/v/TONiI5oqTTg

Key Takeaways

  • The imaginary unit i is defined to be the square root of negative one. In other words,i=1and i2=1.
  • Complex numbers have the form a+bi where a and b are real numbers.
  • The set of real numbers is a subset of the complex numbers.
  • The result of adding, subtracting, multiplying, and dividing complex numbers is a complex number.
  • The product of complex conjugates, a+bi and abi, is a real number. Use this fact to divide complex numbers. Multiply the numerator and denominator of a fraction by the complex conjugate of the denominator and then simplify.
  • Ensure that any complex number is written in terms of the imaginary unit i before performing any operations.

Exercise 5.7.4

Rewrite in terms of imaginary unit i.

  1. 81
  2. 64
  3. 4
  4. 36
  5. 20
  6. 18
  7. 50
  8. 48
  9. 45
  10. 8
  11. 116
  12. 29
  13. 0.25
  14. 1.44
Answer

1. 9i

3. 2i

5. 2i5

7. 5i2

9. 3i5

11. i4

13. 0.5i

Exercise 5.7.5

Write the complex number in standard form a+bi.

  1. 524
  2. 359
  3. 2+38
  4. 4218
  5. 3246
  6. 2+7510
  7. 63512
  8. 72+824
Answer

1. 54i

3. 2+6i2

5. 1263i

7. 51274i

Exercise 5.7.6

Given that i12=1 compute the following powers of i.

  1. i3
  2. i4
  3. i5
  4. i6
  5. i15
  6. i24
Answer

1. i

3. i

5. i

Exercise 5.7.7

Perform the operations.

  1. (3+5i)+(74i)
  2. (67i)+(52i)
  3. (83i)+(5+2i)
  4. (10+15i)+(1520i)
  5. (12+34i)+(1618i)
  6. (2516i)+(11032i)
  7. (5+2i)(83i)
  8. (7i)(69i)
  9. (95i)(8+12i)
  10. (11+2i)(137i)
  11. (114+32i)(4734i)
  12. (3813i)(1212i)
  13. (2i)+(3+4i)(65i)
  14. (7+2i)(6i)(34i)
  15. (13i)(112i)(16+16i)
  16. (134i)+(52+i)(1458i)
  17. (53i)(2+7i)(110i)
  18. (611i)+(2+3i)(84i)
  19. 16(31)
  20. 100+(9+7)
  21. (1+1)(11)
  22. (381)(539)
  23. (5225)(3+41)
  24. (121)(349)
Answer

1. 10+i

3. 3i

5. 23+58i

7. 3+5i

9. 1717i

11. 12+94i

13. 1+8i

15. 5623i

17. 2

19. 3+5i

21. 2i

23. 814i

Exercise 5.7.8

Perform the operations.

  1. i(1i)
  2. i(1+i)
  3. 2i(74i)
  4. 6i(12i)
  5. 2i(34i)
  6. 5i(2i)
  7. (2+i)(23i)
  8. (35i)(12i)
  9. (1i)(89i)
  10. (1+5i)(5+2i)
  11. (4+3i)2
  12. (1+2i)2
  13. (25i)2
  14. (5i)2
  15. (1+i)(1i)
  16. (2i)(2+i)
  17. (42i)(4+2i)
  18. (6+5i)(65i)
  19. (12+23i)(1312i)
  20. (2313i)(1232i)
  21. (2i)3
  22. (13i)3
  23. 2(26)
  24. 1(1+8)
  25. 6(106)
  26. 15(310)
  27. (232)(2+32)
  28. (1+5)(15)
  29. (134)(2+9)
  30. ( 2 - 3 \sqrt { - 1 } ) ( 1 + 2 \sqrt { - 16 } )
  31. ( 2 - 3 i \sqrt { 2 } ) ( 3 + i \sqrt { 2 } )
  32. ( - 1 + i \sqrt { 3 } ) ( 2 - 2 i \sqrt { 3 } )
  33. \frac { - 3 } { i }
  34. \frac { 5 } { i }
  35. \frac { 1 } { 5 + 4 i }
  36. \frac { 1 } { 3 - 4 i }
  37. \frac { 15 } { 1 - 2 i }
  38. \frac { 29 } { 5 + 2 i }
  39. \frac { 20 i } { 1 - 3 i }
  40. \frac { 10 i } { 1 + 2 i }
  41. \frac { 10 - 5 i } { 3 - i }
  42. \frac { 5 - 2 i } { 1 - 2 i }
  43. \frac { 5 + 10 i } { 3 + 4 i }
  44. \frac { 2 - 4 i } { 5 + 3 i }
  45. \frac { 26 + 13 i } { 2 - 3 i }
  46. \frac { \overline { 4 } + 2 i } { 1 + i }
  47. \frac { 3 - i } { 2 i }
  48. \frac { - 5 + 2 i } { 4 i }
  49. \frac { 1 } { a - b i }
  50. \frac { 1 } { a + b i }
  51. \frac { 1 - \sqrt { - 1 } } { 1 + \sqrt { - 1 } }
  52. \frac { 1 + \sqrt { - 9 } } { 1 - \sqrt { - 9 } }
  53. \frac { - \sqrt { - 6 } } { \sqrt { 18 } + \sqrt { - 4 } }
  54. \frac { \sqrt { - 12 } } { \sqrt { 2 } - \sqrt { - 27 } }
Answer

1. 1 + i

3. 8 + 14 i

5. - 8 - 6 i

7. 7 - 4 i

9. - 1 - 17 i

11. 7 + 24 i

13. - 21 - 20 i

15. 2

17. 20

19. \frac { 1 } { 2 } - \frac { 1 } { 36 } i

21. 2-11i

23. - 2 - 2 i \sqrt { 3 }

25. 6 + 2 i \sqrt { 15 }

27. 22

29. 20-9i

31. 12 - 7 i \sqrt { 2 }

33. 3i

35. \frac { 5 } { 41 } - \frac { 4 } { 41 } i

37. 3+6i

39. -6+2i

41. \frac { 7 } { 2 } - \frac { 1 } { 2 } i

43. \frac { 11 } { 5 } - \frac { 2 } { 5 } i

45. 1 + 8 i

47. - \frac { 1 } { 2 } - \frac { 3 } { 2 } i

49. \frac { a } { a ^ { 2 } + b ^ { 2 } } + \frac { b } { a ^ { 2 } + b ^ { 2 } } i

51. - i

53. - \frac { \sqrt { 6 } } { 11 } - \frac { 3 \sqrt { 3 } } { 11 } i

Exercise \PageIndex{9}

Given that i ^ { - n } = \frac { 1 } { i ^ { n } } compute the following powers of i.

  1. i ^ { - 1 }
  2. i ^ { - 2 }
  3. i ^ { - 3 }
  4. i ^ { - 4 }
Answer

1. -i

3. i

Exercise \PageIndex{10}

Perform the operations and simplify.

  1. 2 i ( 2 - i ) - i ( 3 - 4 i )
  2. i ( 5 - i ) - 3 i ( 1 - 6 i )
  3. 5 - 3 ( 1 - i ) ^ { 2 }
  4. 2 ( 1 - 2 i ) ^ { 2 } + 3 i
  5. ( 1 - i ) ^ { 2 } - 2 ( 1 - i ) + 2
  6. ( 1 + i ) ^ { 2 } - 2 ( 1 + i ) + 2
  7. ( 2 i \sqrt { 2 } ) ^ { 2 } + 5
  8. ( 3 i \sqrt { 5 } ) ^ { 2 } - ( i \sqrt { 3 } ) ^ { 2 }
  9. ( \sqrt { 2 } - i ) ^ { 2 } - ( \sqrt { 2 } + i ) ^ { 2 }
  10. ( i \sqrt { 3 } + 1 ) ^ { 2 } - ( 4 i \sqrt { 2 } ) ^ { 2 }
  11. \left( \frac { 1 } { 1 + i } \right) ^ { 2 }
  12. \left( \frac { 1 } { 1 + i } \right) ^ { 3 }
  13. ( a - b i ) ^ { 2 } - ( a + b i ) ^ { 2 }
  14. \left( a ^ { 2 } + a i + 1 \right) \left( a ^ { 2 } - a i + 1 \right)
  15. Show that both -2i and 2i satisfy x ^ { 2 } + 4 = 0.
  16. Show that both -i and i satisfy x^{2}+1=0.
  17. Show that both 3-2i and 3+2i satisfy x^{2}-6x+13=0.
  18. Show that both 5-i and 5+i satisfy x^{2}-10x+26=0.
  19. Show that 3 , - 2 i, and 2i are all solutions to x ^ { 3 } - 3 x ^ { 2 } + 4 x - 12 = 0.
  20. Show that -2, 1, -i, and 1+i are all solutions to x ^ { 3 } - 2 x + 4 = 0.
Answer

1. -2+i

3. 5+6i

5. 0

7. -3

9. - 4 i \sqrt { 2 }

11. - \frac {i } { 2 }

13. -4abi

15. Proof

17. Proof

19. Proof

Exercise \PageIndex{11}

  1. Research and discuss the history of the imaginary unit and complex numbers.
  2. How would you define i^{0} and why?
  3. Research what it means to calculate the absolute value of a complex number | a + b i |. Illustrate your finding with an example.
  4. Explore the powers of i. Look for a pattern and share your findings.
Answer

1. Answer may vary

3. Answer may vary

Footnotes

26Defined as i = \sqrt { - 1 } where i^{2} = −1.

27A square root of any negative real number.

28A number of the form a + bi, where a and b are real numbers.

29The real number a of a complex number a + bi.

30The real number b of a complex number a + bi.

31Two complex numbers whose real parts are the same and imaginary parts are opposite. If given a + bi, then its complex conjugate is a − bi.

32The real number that results from multiplying complex conjugates: (a + bi) (a − bi) = a^{2} + b^{2}.


This page titled 5.7: Complex Numbers and Their Operations is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous via source content that was edited to the style and standards of the LibreTexts platform.

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