Learn on PengiOpenStax Algebra and TrigonometryChapter 2: Equations and Inequalities

Lesson 2.4: Complex Numbers

In this Grade 7 math lesson from OpenStax Algebra and Trigonometry, students learn to work with complex numbers by expressing square roots of negative numbers as multiples of the imaginary unit i and writing results in standard form a + bi. The lesson covers adding, subtracting, multiplying, and dividing complex numbers, simplifying powers of i, and plotting complex numbers on the complex plane. Students build on their knowledge of real numbers to understand how imaginary and complex numbers extend the number system.

Section 1

πŸ“˜ Complex Numbers

New Concept

Complex numbers extend the real number system by introducing the imaginary unit ii, where i2=βˆ’1i^2 = -1. This allows us to solve new equations and perform arithmeticβ€”addition, subtraction, multiplication, and divisionβ€”on numbers in the form a+bia+bi.

What’s next

Next, you'll master these operations through a series of interactive examples and practice cards, starting with how to plot complex numbers visually.

Section 2

Imaginary and Complex Numbers

Property

A complex number is a number of the form a+bia + bi where aa is the real part of the complex number, and bb is the imaginary part of the complex number.
If b=0b = 0, then a+bia + bi is a real number. If a=0a = 0 and bb is not equal to 00, the complex number is called a pure imaginary number.
An imaginary number is an even root of a negative number. The imaginary number ii is defined as the square root of βˆ’1-1, so βˆ’1=i\sqrt{-1} = i and i2=βˆ’1i^2 = -1.
To express an imaginary number in standard form, write βˆ’a\sqrt{-a} as aβˆ’1\sqrt{a}\sqrt{-1}, which simplifies to iai\sqrt{a}.

Examples

  • Express βˆ’16\sqrt{-16} in standard form. This is written as 16βˆ’1\sqrt{16}\sqrt{-1}, which simplifies to 4i4i. In standard form, this is 0+4i0 + 4i.
  • Express βˆ’75\sqrt{-75} in standard form. This becomes 75βˆ’1=25β‹…3βˆ’1=53i\sqrt{75}\sqrt{-1} = \sqrt{25 \cdot 3}\sqrt{-1} = 5\sqrt{3}i, or 0+5i30 + 5i\sqrt{3}.

Section 3

Plotting on the Complex Plane

Property

In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis.
A complex number a+bia+bi is represented by the ordered pair (a,b)(a, b).
To plot the point, move aa units along the horizontal (real) axis and bb units along the vertical (imaginary) axis.

Examples

  • Plot the complex number 5+3i5 + 3i on the complex plane. The real part is 55 and the imaginary part is 33, so we plot the ordered pair (5,3)(5, 3).
  • Plot the complex number βˆ’4βˆ’2i-4 - 2i on the complex plane. The real part is βˆ’4-4 and the imaginary part is βˆ’2-2, so we plot the point (βˆ’4,βˆ’2)(-4, -2).

Section 4

Adding and Subtracting Complex Numbers

Property

To add or subtract complex numbers, combine the real parts and combine the imaginary parts.
Adding complex numbers:

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

Subtracting complex numbers:

(a+bi)βˆ’(c+di)=(aβˆ’c)+(bβˆ’d)i(a + bi) - (c + di) = (a - c) + (b - d)i

Examples

  • Add (4βˆ’3i)+(1+6i)(4 - 3i) + (1 + 6i). We combine the real parts (4+1)(4+1) and the imaginary parts (βˆ’3+6)i(-3+6)i, which results in 5+3i5 + 3i.
  • Subtract (9+2i)βˆ’(5βˆ’4i)(9 + 2i) - (5 - 4i). We subtract the real parts (9βˆ’5)(9-5) and the imaginary parts (2βˆ’(βˆ’4))i(2 - (-4))i, which results in 4+6i4 + 6i.

Section 5

Multiplying Complex Numbers

Property

To multiply two complex numbers, use the distributive property or the FOIL method, just as with binomials. The key difference is to replace i2i^2 with βˆ’1-1. The general form is:

(a+bi)(c+di)=(acβˆ’bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i

Examples

  • Multiply 3(4+7i)3(4 + 7i). Distribute the real number to get (3β‹…4)+(3β‹…7i)(3 \cdot 4) + (3 \cdot 7i), which simplifies to 12+21i12 + 21i.
  • Multiply (3+2i)(5βˆ’4i)(3 + 2i)(5 - 4i). Using FOIL, we get 15βˆ’12i+10iβˆ’8i215 - 12i + 10i - 8i^2. Since i2=βˆ’1i^2 = -1, this becomes 15βˆ’2iβˆ’8(βˆ’1)=15βˆ’2i+8=23βˆ’2i15 - 2i - 8(-1) = 15 - 2i + 8 = 23 - 2i.

Section 6

Dividing Complex Numbers

Property

The complex conjugate of a complex number a+bia + bi is aβˆ’bia - bi.
To divide complex numbers, write the problem as a fraction.
Then, multiply the numerator and denominator by the complex conjugate of the denominator to eliminate the imaginary number in the denominator.
Simplify the result into standard form a+bia+bi.

Examples

  • Find the complex conjugate of βˆ’5+2i-5 + 2i. The real part stays the same and the sign of the imaginary part is changed, so the conjugate is βˆ’5βˆ’2i-5 - 2i.
  • Divide (3+4i)(3 + 4i) by (2βˆ’i)(2 - i). Multiply the numerator and denominator by the conjugate 2+i2+i: 3+4i2βˆ’iβ‹…2+i2+i=6+3i+8i+4i24βˆ’i2=2+11i5=25+115i\frac{3+4i}{2-i} \cdot \frac{2+i}{2+i} = \frac{6+3i+8i+4i^2}{4-i^2} = \frac{2+11i}{5} = \frac{2}{5} + \frac{11}{5}i.

Section 7

Simplifying Powers of i

Property

The powers of ii are cyclic, repeating every four powers:
i1=ii^1 = i
i2=βˆ’1i^2 = -1
i3=βˆ’ii^3 = -i
i4=1i^4 = 1
To simplify ini^n, divide the exponent nn by 4 and find the remainder rr. The simplified form is iri^r. If the remainder is 0, the result is i4=1i^4=1.

Examples

  • Evaluate i22i^{22}. Since 22Γ·4=522 \div 4 = 5 with a remainder of 22, we have i22=i2=βˆ’1i^{22} = i^2 = -1.
  • Evaluate i51i^{51}. Since 51Γ·4=1251 \div 4 = 12 with a remainder of 33, we have i51=i3=βˆ’ii^{51} = i^3 = -i.

Book overview

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Chapter 2: Equations and Inequalities

  1. Lesson 1

    Lesson 2.1: The Rectangular Coordinate Systems and Graphs

  2. Lesson 2

    Lesson 2.2: Linear Equations in One Variable

  3. Lesson 3

    Lesson 2.3: Models and Applications

  4. Lesson 4Current

    Lesson 2.4: Complex Numbers

  5. Lesson 5

    Lesson 2.5: Quadratic Equations

  6. Lesson 6

    Lesson 2.6: Other Types of Equations

  7. Lesson 7

    Lesson 2.7: Linear Inequalities and Absolute Value Inequalities

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Complex Numbers

New Concept

Complex numbers extend the real number system by introducing the imaginary unit ii, where i2=βˆ’1i^2 = -1. This allows us to solve new equations and perform arithmeticβ€”addition, subtraction, multiplication, and divisionβ€”on numbers in the form a+bia+bi.

What’s next

Next, you'll master these operations through a series of interactive examples and practice cards, starting with how to plot complex numbers visually.

Section 2

Imaginary and Complex Numbers

Property

A complex number is a number of the form a+bia + bi where aa is the real part of the complex number, and bb is the imaginary part of the complex number.
If b=0b = 0, then a+bia + bi is a real number. If a=0a = 0 and bb is not equal to 00, the complex number is called a pure imaginary number.
An imaginary number is an even root of a negative number. The imaginary number ii is defined as the square root of βˆ’1-1, so βˆ’1=i\sqrt{-1} = i and i2=βˆ’1i^2 = -1.
To express an imaginary number in standard form, write βˆ’a\sqrt{-a} as aβˆ’1\sqrt{a}\sqrt{-1}, which simplifies to iai\sqrt{a}.

Examples

  • Express βˆ’16\sqrt{-16} in standard form. This is written as 16βˆ’1\sqrt{16}\sqrt{-1}, which simplifies to 4i4i. In standard form, this is 0+4i0 + 4i.
  • Express βˆ’75\sqrt{-75} in standard form. This becomes 75βˆ’1=25β‹…3βˆ’1=53i\sqrt{75}\sqrt{-1} = \sqrt{25 \cdot 3}\sqrt{-1} = 5\sqrt{3}i, or 0+5i30 + 5i\sqrt{3}.

Section 3

Plotting on the Complex Plane

Property

In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis.
A complex number a+bia+bi is represented by the ordered pair (a,b)(a, b).
To plot the point, move aa units along the horizontal (real) axis and bb units along the vertical (imaginary) axis.

Examples

  • Plot the complex number 5+3i5 + 3i on the complex plane. The real part is 55 and the imaginary part is 33, so we plot the ordered pair (5,3)(5, 3).
  • Plot the complex number βˆ’4βˆ’2i-4 - 2i on the complex plane. The real part is βˆ’4-4 and the imaginary part is βˆ’2-2, so we plot the point (βˆ’4,βˆ’2)(-4, -2).

Section 4

Adding and Subtracting Complex Numbers

Property

To add or subtract complex numbers, combine the real parts and combine the imaginary parts.
Adding complex numbers:

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

Subtracting complex numbers:

(a+bi)βˆ’(c+di)=(aβˆ’c)+(bβˆ’d)i(a + bi) - (c + di) = (a - c) + (b - d)i

Examples

  • Add (4βˆ’3i)+(1+6i)(4 - 3i) + (1 + 6i). We combine the real parts (4+1)(4+1) and the imaginary parts (βˆ’3+6)i(-3+6)i, which results in 5+3i5 + 3i.
  • Subtract (9+2i)βˆ’(5βˆ’4i)(9 + 2i) - (5 - 4i). We subtract the real parts (9βˆ’5)(9-5) and the imaginary parts (2βˆ’(βˆ’4))i(2 - (-4))i, which results in 4+6i4 + 6i.

Section 5

Multiplying Complex Numbers

Property

To multiply two complex numbers, use the distributive property or the FOIL method, just as with binomials. The key difference is to replace i2i^2 with βˆ’1-1. The general form is:

(a+bi)(c+di)=(acβˆ’bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i

Examples

  • Multiply 3(4+7i)3(4 + 7i). Distribute the real number to get (3β‹…4)+(3β‹…7i)(3 \cdot 4) + (3 \cdot 7i), which simplifies to 12+21i12 + 21i.
  • Multiply (3+2i)(5βˆ’4i)(3 + 2i)(5 - 4i). Using FOIL, we get 15βˆ’12i+10iβˆ’8i215 - 12i + 10i - 8i^2. Since i2=βˆ’1i^2 = -1, this becomes 15βˆ’2iβˆ’8(βˆ’1)=15βˆ’2i+8=23βˆ’2i15 - 2i - 8(-1) = 15 - 2i + 8 = 23 - 2i.

Section 6

Dividing Complex Numbers

Property

The complex conjugate of a complex number a+bia + bi is aβˆ’bia - bi.
To divide complex numbers, write the problem as a fraction.
Then, multiply the numerator and denominator by the complex conjugate of the denominator to eliminate the imaginary number in the denominator.
Simplify the result into standard form a+bia+bi.

Examples

  • Find the complex conjugate of βˆ’5+2i-5 + 2i. The real part stays the same and the sign of the imaginary part is changed, so the conjugate is βˆ’5βˆ’2i-5 - 2i.
  • Divide (3+4i)(3 + 4i) by (2βˆ’i)(2 - i). Multiply the numerator and denominator by the conjugate 2+i2+i: 3+4i2βˆ’iβ‹…2+i2+i=6+3i+8i+4i24βˆ’i2=2+11i5=25+115i\frac{3+4i}{2-i} \cdot \frac{2+i}{2+i} = \frac{6+3i+8i+4i^2}{4-i^2} = \frac{2+11i}{5} = \frac{2}{5} + \frac{11}{5}i.

Section 7

Simplifying Powers of i

Property

The powers of ii are cyclic, repeating every four powers:
i1=ii^1 = i
i2=βˆ’1i^2 = -1
i3=βˆ’ii^3 = -i
i4=1i^4 = 1
To simplify ini^n, divide the exponent nn by 4 and find the remainder rr. The simplified form is iri^r. If the remainder is 0, the result is i4=1i^4=1.

Examples

  • Evaluate i22i^{22}. Since 22Γ·4=522 \div 4 = 5 with a remainder of 22, we have i22=i2=βˆ’1i^{22} = i^2 = -1.
  • Evaluate i51i^{51}. Since 51Γ·4=1251 \div 4 = 12 with a remainder of 33, we have i51=i3=βˆ’ii^{51} = i^3 = -i.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 2: Equations and Inequalities

  1. Lesson 1

    Lesson 2.1: The Rectangular Coordinate Systems and Graphs

  2. Lesson 2

    Lesson 2.2: Linear Equations in One Variable

  3. Lesson 3

    Lesson 2.3: Models and Applications

  4. Lesson 4Current

    Lesson 2.4: Complex Numbers

  5. Lesson 5

    Lesson 2.5: Quadratic Equations

  6. Lesson 6

    Lesson 2.6: Other Types of Equations

  7. Lesson 7

    Lesson 2.7: Linear Inequalities and Absolute Value Inequalities