MGU Mathematics Series — Algebra II
Part II: Quadratics, Complex Numbers, and Polynomial Functions
Imaginary Unit · Operations · Conjugates · Quadratic Solutions · Complex Plane
Chapter 4 of 14
MGU Mathematics Series  ·  Algebra II  ·  Library Textbook Edition

Chapter 4:
Complex Numbers

The imaginary unit, complex arithmetic, conjugates, quadratic solutions with complex roots, and the complex plane — expanding the number system beyond the real line

■ Math Level: Algebra II ■

■ 14 Sections  •  Part II of 5 ■

Follows Chapter 3: Quadratic Functions in Depth  ·  Prepares students for Chapter 5: Polynomial Functions

Chapter 4 — Front Matter

Purpose and Overview

Every number system in mathematics was created to solve a problem that the previous system could not solve. Whole numbers could not handle subtraction in all cases, so negative numbers were introduced. Integers could not handle division in all cases, so rational numbers were introduced. Rational numbers could not express every measurement, so irrational numbers were introduced. Together, these sets form the real number system — and for a long time, that felt like enough.

Then mathematicians encountered a simple question that no real number could answer: what number, when squared, gives a negative result? No positive number works. No negative number works. Zero does not work. The real number system has no solution to equations like \(x^2 = -1\).

Rather than declare such equations unsolvable, mathematicians did what they had always done. They extended the number system. Chapter 4 introduces complex numbers — a system that includes all real numbers and adds new numbers built from the square root of negative one. Complex numbers appear in electrical engineering, quantum physics, signal processing, fluid dynamics, computer graphics, and many branches of modern science and mathematics. Understanding them begins here, in Algebra II.

Opening Question

A student solving \(x^2 + 9 = 0\) subtracts 9 from both sides and gets \(x^2 = -9\). She knows that squaring any real number gives a result that is zero or positive. She concludes the equation has no solution and moves on. Is she correct? What would it take for this equation to have a solution? What kind of number would satisfy \(x^2 = -9\)? By the end of this chapter, you will be able to answer these questions completely.

14Sections
12Objectives
10Key Terms
12+Worked Examples

Learning Objectives

  1. 1Explain why the real number system needs to be extended.
  2. 2Define the imaginary unit \(i\) and use it correctly.
  3. 3Simplify square roots of negative numbers using \(i\).
  4. 4Write complex numbers in standard form \(a + bi\).
  5. 5Add, subtract, and multiply complex numbers.
  6. 6Compute powers of \(i\) using the cyclic pattern.
  7. 7Identify and use complex conjugates.
  8. 8Divide complex numbers by multiplying by the conjugate.
  9. 9Solve quadratic equations with complex solutions.
  10. 10Plot complex numbers on the complex plane.
  11. 11Compute the magnitude of a complex number.
  12. 12Recognize and avoid common errors with complex numbers.
Chapter 4 — Contents

Sections at a Glance

Sections 4.1 – 4.4
Extending the Number System
Why Real Numbers Fall Short · Defining \(i\) · Simplifying Negative Radicands · Standard Form
Section 4.1

Why the Real Numbers Are Not Enough

Consider the equation \(x^2 = -1\). Squaring any positive number gives a positive result. Squaring any negative number also gives a positive result, because a negative times a negative is positive. Squaring zero gives zero. No real number, when squared, ever gives a negative result.

This is not a deficiency in technique. It is a genuine limitation of the real number system. The real numbers are built on the number line, and the number line contains no solution to \(x^2 = -1\).

Historically, mathematicians working on cubic equations found that even when equations had real solutions, the path to finding those solutions sometimes passed through square roots of negative numbers. They could not ignore the problem any longer. The response was not to give up — it was to define a new number. Mathematicians agreed to let the symbol \(\sqrt{-1}\) represent a new kind of number: one that, when squared, gives \(-1\). This number was named the imaginary unit and given the symbol \(i\).

Why "Imaginary" Is Misleading

The word imaginary comes from historical skepticism. Early mathematicians were uncertain whether numbers built from \(\sqrt{-1}\) were legitimate. The word stuck, but the skepticism did not. Complex numbers are now as legitimate and useful as any other numbers in mathematics. When you work with them, you are not doing something approximate or fictional — you are working inside a well-defined, logically consistent number system that includes the real numbers as a special case.

MGU Library Connection
  • The history of complex numbers involves Cardano, Bombelli, Euler, and Gauss — see the MGU History of Mathematics reference pages for the full story
  • The Complex Number Reference — Appendix E of this textbook
Section 4.2

The Imaginary Unit

Definition of the Imaginary Unit

\(i = \sqrt{-1}\)

\(i^2 = -1\)

Both forms are important. The first says \(i\) is the square root of \(-1\). The second says squaring \(i\) gives \(-1\). You will use both forms depending on the situation.

The imaginary unit \(i\) is not a variable. It is a specific, defined number — just as \(\pi\) is a specific number, not a variable. When you write \(3i\), you are writing a number, not an expression in terms of an unknown. Any real multiple of \(i\) is called a pure imaginary number. Examples include \(2i\), \(-5i\), \(\frac{3}{4}i\), and \(\sqrt{7}\,i\).

Practice 4.2Concept Check
  1. Explain in your own words why no real number satisfies \(x^2 = -4\).
  2. A student writes \(i = \sqrt{-1}\) and concludes that \(i\) is "just like \(\sqrt{2}\) or \(\sqrt{3}\)." In what sense is this true? In what sense is it different?
  3. Evaluate \(i^2\), \((2i)^2\), and \((-3i)^2\). What pattern do you observe?
Section 4.3

Simplifying Square Roots of Negative Numbers

Key Principle

For any positive real number \(n\):

\(\sqrt{-n} = i\sqrt{n}\)

This follows because \((i\sqrt{n})^2 = i^2 \cdot n = (-1)(n) = -n\,\checkmark\)

Example 4.3.1
Simplifying \(\sqrt{-25}\)
  • ProblemSimplify \(\sqrt{-25}\).
  • ThinkSeparate the negative sign from the radicand. Write \(\sqrt{-25} = \sqrt{-1 \cdot 25}\), then use \(\sqrt{-1} = i\).
  • Solve\[\sqrt{-25} = \sqrt{-1 \cdot 25} = \sqrt{-1} \cdot \sqrt{25} = i \cdot 5 = 5i\]
  • CheckVerify: \((5i)^2 = 25i^2 = 25(-1) = -25\). ✓
Answer: \(\sqrt{-25} = 5i\)
Example 4.3.2
Simplifying \(\sqrt{-18}\)
  • ProblemSimplify \(\sqrt{-18}\).
  • ThinkSeparate the negative, then simplify \(\sqrt{18}\) by pulling out perfect squares: \(\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\).
  • Solve\[\sqrt{-18} = i\sqrt{18} = i\sqrt{9 \cdot 2} = i \cdot 3\sqrt{2} = 3i\sqrt{2}\]
  • Check\((3i\sqrt{2})^2 = 9i^2 \cdot 2 = 9(-1)(2) = -18\). ✓
Answer: \(\sqrt{-18} = 3i\sqrt{2}\)
Critical Warning — The Product Rule Does Not Apply to Negative Radicands

The product rule \(\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}\) applies only when both \(a\) and \(b\) are non-negative. Do not write \(\sqrt{-4} \cdot \sqrt{-9} = \sqrt{36} = 6\). This is incorrect. The correct approach: convert to imaginary form first, then multiply. \(\sqrt{-4} \cdot \sqrt{-9} = (2i)(3i) = 6i^2 = 6(-1) = -6\). Note the answer is \(-6\), not \(+6\).

Practice 4.3Simplifying Negative Radicands
  1. \(\sqrt{-49}\)
  2. \(\sqrt{-100}\)
  3. \(\sqrt{-72}\)
  4. \(\sqrt{-50}\)
  5. \(\sqrt{-200}\)
  6. \(-\sqrt{-36}\)
  7. \(\sqrt{-7}\)
  8. \(\sqrt{-12}\)

For problems 7 and 8, leave the simplified radical in place.

Section 4.4

Complex Numbers in Standard Form

Standard Form of a Complex Number

A complex number is a number of the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i = \sqrt{-1}\).

\(a\) is the real part  ·  \(b\) is the imaginary part (the coefficient of \(i\), not the term \(bi\))

Complex NumberReal PartImaginary PartClassification
\(3 + 4i\)34Complex (general)
\(-2 - 7i\)\(-2\)\(-7\)Complex (general)
\(5\)50Real number
\(6i\)06Pure imaginary
\(0\)00Zero

Notice that every real number is a complex number — one where the imaginary part is zero. The set of complex numbers, denoted \(\mathbb{C}\), contains all real numbers as a subset, along with pure imaginary numbers and all combinations where both \(a \neq 0\) and \(b \neq 0\).

Diagram Note — The Number System Hierarchy

Draw a nested set diagram: Natural Numbers ⊂ Whole Numbers ⊂ Integers ⊂ Rational Numbers ⊂ Real Numbers ⊂ Complex Numbers. Pure imaginary numbers like \(3i\) sit inside \(\mathbb{C}\) but outside the Real Numbers region. Every number system you have studied is a subset of \(\mathbb{C}\).

Example 4.4.1
Identifying Parts of a Complex Number
  • ProblemFor each complex number, identify the real part and the imaginary part: (a) \(7 - 3i\)   (b) \(-\sqrt{2} + i\)   (c) \(-9i\)   (d) \(4\)
  • Solve (a) \(7 - 3i\): Real part = 7, Imaginary part = \(-3\)
    (b) \(-\sqrt{2} + i\): Real part = \(-\sqrt{2}\), Imaginary part = 1
    (c) \(-9i = 0 + (-9)i\): Real part = 0, Imaginary part = \(-9\)
    (d) \(4 = 4 + 0i\): Real part = 4, Imaginary part = 0
Answer: Real and imaginary parts as identified above.
Equality of Complex Numbers

Two complex numbers \(a + bi\) and \(c + di\) are equal if and only if \(a = c\) and \(b = d\). Both the real parts and the imaginary parts must match.

Practice 4.4Standard Form
  1. Write each in standard form \(a + bi\) and identify the real and imaginary parts: (a) \(-3 + \sqrt{-64}\)   (b) \(2\sqrt{-9}\)   (c) \(\sqrt{-50} - 1\)   (d) \(\sqrt{25} + \sqrt{-16}\)
  2. For each pair, determine whether the two complex numbers are equal. Explain your reasoning. (a) \(4 + 2i\) and \(2 + 4i\)   (b) \(3 + 0i\) and \(3\)   (c) \(0 + 5i\) and \(5\)
  3. A student says: "The imaginary part of \(7 + 3i\) is \(3i\)." Explain what is wrong with this statement.
Sections 4.5 – 4.10
Arithmetic of Complex Numbers
Addition and Subtraction · Multiplication · Powers of i · Conjugates · Division · Quadratic Solutions
Section 4.5

Adding and Subtracting Complex Numbers

Adding and subtracting complex numbers follows the same logic as combining like terms. Real parts combine with real parts, and imaginary parts combine with imaginary parts.

Addition and Subtraction Rules

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

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

Example 4.5.1
Adding Complex Numbers
  • ProblemAdd \((3 + 4i) + (1 - 7i)\).
  • ThinkCollect real parts together and imaginary parts together.
  • Solve\[(3 + 4i) + (1 - 7i) = (3 + 1) + (4 - 7)i = 4 + (-3)i = 4 - 3i\]
  • CheckReal parts: \(3 + 1 = 4\). Imaginary parts: \(4 - 7 = -3\). ✓
Answer: \(4 - 3i\)
Example 4.5.2
Subtracting Complex Numbers
  • ProblemSubtract \((5 - 2i) - (-3 + 6i)\).
  • ThinkDistribute the subtraction sign across the second complex number, then combine like parts.
  • Solve\[(5 - 2i) - (-3 + 6i) = 5 - 2i + 3 - 6i = (5 + 3) + (-2 - 6)i = 8 - 8i\]
  • CheckReal parts: \(5 - (-3) = 8\). Imaginary parts: \(-2 - 6 = -8\). ✓
Answer: \(8 - 8i\)
Think of Complex Numbers as Binomials

Think of \(a + bi\) as a binomial where \(i\) behaves like a variable. Addition and subtraction work exactly like polynomial addition and subtraction — just combine like terms. The key difference from polynomials comes during multiplication, when \(i^2 = -1\) produces something genuinely new.

Practice 4.5Addition and Subtraction
  1. \((2 + 5i) + (3 - 2i)\)
  2. \((-4 + i) + (6 - 3i)\)
  3. \((7 - 4i) - (2 + 5i)\)
  4. \((-1 + 3i) - (-4 - 7i)\)
  5. \((3i) + (2 - 3i)\)
  6. \((5 - i) - (5 + i)\)
Section 4.6

Multiplying Complex Numbers

When multiplying two complex numbers, use the distributive property exactly as you would when multiplying two binomials. Then apply the rule \(i^2 = -1\) to simplify. You do not need to memorize a formula — if you distribute carefully and replace \(i^2\) with \(-1\), you will always reach the correct result.

Example 4.6.1
Multiplying Two Complex Numbers
  • ProblemMultiply \((2 + 3i)(4 - i)\).
  • ThinkDistribute just like multiplying binomials (FOIL). Then replace \(i^2\) with \(-1\).
  • Solve \[\begin{aligned} (2 + 3i)(4 - i) &= 2(4) + 2(-i) + 3i(4) + 3i(-i)\\ &= 8 - 2i + 12i - 3i^2\\ &= 8 + 10i - 3(-1)\\ &= 8 + 10i + 3\\ &= 11 + 10i \end{aligned}\]
  • CheckReal: \(8 + 3 = 11\). Imaginary: \(-2 + 12 = 10\). ✓
Answer: \(11 + 10i\)
Example 4.6.2
Squaring a Complex Number
  • ProblemCompute \((1 + 5i)^2\).
  • ThinkApply \((a + b)^2 = a^2 + 2ab + b^2\). The middle term is essential — do not drop it.
  • Solve \[\begin{aligned} (1 + 5i)^2 &= 1^2 + 2(1)(5i) + (5i)^2\\ &= 1 + 10i + 25i^2\\ &= 1 + 10i + 25(-1)\\ &= 1 + 10i - 25\\ &= -24 + 10i \end{aligned}\]
  • CheckExpand directly: \((1 + 5i)(1 + 5i) = 1 + 5i + 5i + 25i^2 = 1 + 10i - 25 = -24 + 10i\). ✓
Answer: \(-24 + 10i\)
Student Warning — Never Drop the Middle Term

When squaring a complex number \((a + bi)^2\), do not write just \(a^2 + b^2i^2\). You must include the middle term \(2abi\). For example, \((2 + 5i)^2 \neq 4 + 25i^2\). The correct expansion is \(4 + 20i + 25i^2 = 4 + 20i - 25 = -21 + 20i\). Forgetting the middle term is one of the most common errors in this chapter.

Practice 4.6Multiplication
  1. \((2 + i)(3 - 4i)\)
  2. \((1 + 5i)(1 - 5i)\)
  3. \((3 - 2i)^2\)
  4. \((4i)(2 - 3i)\)
  5. \((-1 + i)(-1 - i)\)
  6. \((2 + 3i)(2 + 3i)\)
Section 4.7

Powers of \(i\)

When you raise \(i\) to successive powers, the values follow a repeating cycle of length 4:

PowerComputationValue
\(i^1\)\(i\)\(i\)
\(i^2\)defined\(-1\)
\(i^3\)\(i^2 \cdot i = (-1)(i)\)\(-i\)
\(i^4\)\(i^2 \cdot i^2 = (-1)(-1)\)\(1\)
\(i^5\)\(i^4 \cdot i = 1 \cdot i\)\(i\)
Finding Any Power of \(i\)

Divide the exponent by 4 and use the remainder:

Remainder 0 → \(1\)    Remainder 1 → \(i\)    Remainder 2 → \(-1\)    Remainder 3 → \(-i\)

Example 4.7.1
Computing \(i^{23}\)
  • ProblemSimplify \(i^{23}\).
  • ThinkDivide 23 by 4 and find the remainder.
  • Solve\[23 = 4 \cdot 5 + 3 \quad \Rightarrow \quad \text{remainder} = 3 \quad \Rightarrow \quad i^{23} = i^3 = -i\]
  • Check\(i^{20} \cdot i^3 = (i^4)^5 \cdot i^3 = 1^5 \cdot (-i) = -i\). ✓
Answer: \(i^{23} = -i\)
Example 4.7.2
Computing \(i^{100}\)
  • ProblemSimplify \(i^{100}\).
  • Solve\[100 = 4 \cdot 25 + 0 \quad \Rightarrow \quad \text{remainder} = 0 \quad \Rightarrow \quad i^{100} = 1\]
Answer: \(i^{100} = 1\)
Practice 4.7Powers of i
  1. \(i^9\)
  2. \(i^{30}\)
  3. \(i^{47}\)
  4. \(i^{80}\)
  5. \(i^{13}\)
  6. \(i^{-1}\) (Hint: \(i^{-1} = i^3\) because \(i^4 = 1\))
Section 4.8

Complex Conjugates

The complex conjugate of \(a + bi\) is \(a - bi\). To form the conjugate, keep the real part the same and change the sign of the imaginary part.

Complex NumberIts Conjugate
\(3 + 4i\)\(3 - 4i\)
\(-2 - 5i\)\(-2 + 5i\)
\(7i\)\(-7i\)
\(6\)\(6\)
The Product of Conjugates Is Always Real

\((a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2\)

All imaginary parts disappear. The result is always the non-negative real number \(a^2 + b^2\).

Example 4.8.1
Multiplying a Complex Number by Its Conjugate
  • ProblemMultiply \((3 + 4i)(3 - 4i)\).
  • ThinkThese are conjugates. The product must be \(3^2 + 4^2\).
  • Solve\[(3 + 4i)(3 - 4i) = 3^2 - (4i)^2 = 9 - 16i^2 = 9 - 16(-1) = 9 + 16 = 25\]
  • CheckExpand: \(9 - 12i + 12i - 16i^2 = 9 - 16(-1) = 25\). ✓
Answer: 25
Practice 4.8Conjugates
  1. Write the conjugate of each: (a) \(6 + 5i\)   (b) \(-3 - 8i\)   (c) \(4i\)   (d) \(7\)   (e) \(\sqrt{2} - i\)
  2. Find the product of each complex number and its conjugate: (a) \(2 + 5i\)   (b) \(-3 + i\)   (c) \(6i\)   (d) \(5 - 12i\)
  3. A student claims: "The product of any complex number and its conjugate equals the square of the complex number." Is this true? Give an example to support or refute the claim.
Section 4.9

Dividing Complex Numbers

The goal of dividing complex numbers is to write the result in standard form \(a + bi\) with no \(i\) in the denominator. The method: multiply both numerator and denominator by the conjugate of the denominator. This eliminates \(i\) from the denominator without changing the value of the fraction.

Example 4.9.1
Dividing by a Pure Imaginary Number
  • ProblemSimplify \(\dfrac{6}{3i}\).
  • ThinkMultiply numerator and denominator by \(i\) to eliminate \(i\) from the denominator.
  • Solve\[\frac{6}{3i} \cdot \frac{i}{i} = \frac{6i}{3i^2} = \frac{6i}{3(-1)} = \frac{6i}{-3} = -2i\]
  • CheckDoes \(3i \cdot (-2i) = -6i^2 = -6(-1) = 6\)? ✓
Answer: \(-2i\)
Example 4.9.2
Dividing Two Complex Numbers
  • ProblemSimplify \(\dfrac{2 + 3i}{1 - 4i}\).
  • ThinkMultiply numerator and denominator by the conjugate of the denominator: \(1 + 4i\).
  • Solve Numerator: \((2 + 3i)(1 + 4i) = 2 + 8i + 3i + 12i^2 = 2 + 11i - 12 = -10 + 11i\)

    Denominator: \((1 - 4i)(1 + 4i) = 1^2 + 4^2 = 1 + 16 = 17\)

    \[\frac{2 + 3i}{1 - 4i} = \frac{-10 + 11i}{17} = -\frac{10}{17} + \frac{11}{17}i\]
  • CheckMultiply the result by \((1 - 4i)\) and verify you get \(2 + 3i\). This makes a good student exercise.
Answer: \(-\dfrac{10}{17} + \dfrac{11}{17}i\)
Example 4.9.3
Dividing When the Denominator Is Already Real
  • ProblemSimplify \(\dfrac{4 - 6i}{2}\).
  • ThinkThe denominator is already real — just divide each part.
  • Solve\[\frac{4 - 6i}{2} = \frac{4}{2} - \frac{6i}{2} = 2 - 3i\]
Answer: \(2 - 3i\)
Always Multiply by the Conjugate of the Denominator

Multiply by the conjugate of the denominator, not the conjugate of the numerator. The denominator is what needs to become real. Multiplying by the wrong conjugate produces an incorrect result.

Practice 4.9Division
  1. \(\dfrac{10}{2i}\)
  2. \(\dfrac{3 + i}{1 + 2i}\)
  3. \(\dfrac{4 - 3i}{2 + i}\)
  4. \(\dfrac{6 - 4i}{3i}\)
  5. \(\dfrac{5}{3 - 4i}\)
  6. \(\dfrac{1 + i}{1 - i}\)
Section 4.10

Complex Solutions of Quadratic Equations

In Chapter 3, you learned that the discriminant \(b^2 - 4ac\) determines the nature of the solutions to a quadratic equation. With complex numbers, a negative discriminant is no longer a dead end.

Discriminant \(b^2 - 4ac\)Nature of Solutions
PositiveTwo distinct real solutions
ZeroOne real solution (repeated root)
NegativeTwo complex conjugate solutions (non-real)
Example 4.10.1
Solving \(x^2 + 9 = 0\)
  • ProblemSolve \(x^2 + 9 = 0\).
  • ThinkIsolate \(x^2\), then take the square root. Expect an imaginary result.
  • Solve\[x^2 = -9 \quad \Rightarrow \quad x = \pm\sqrt{-9} = \pm 3i\]
  • Check\((3i)^2 = 9i^2 = -9\). ✓   \((-3i)^2 = 9i^2 = -9\). ✓
Answer: \(x = 3i\) or \(x = -3i\)   (complex conjugate pair: \(0 + 3i\) and \(0 - 3i\))
Example 4.10.2
Solving with the Quadratic Formula — Complex Solutions
  • ProblemSolve \(x^2 - 4x + 13 = 0\).
  • ThinkUse the quadratic formula with \(a = 1\), \(b = -4\), \(c = 13\). Check the discriminant first.
  • Solve Discriminant: \(b^2 - 4ac = 16 - 52 = -36\)   (negative → complex solutions) \[x = \frac{-(-4) \pm \sqrt{-36}}{2(1)} = \frac{4 \pm 6i}{2} = 2 \pm 3i\]
  • CheckSubstitute \(x = 2 + 3i\): \[(2+3i)^2 - 4(2+3i) + 13 = (4 + 12i - 9) - 8 - 12i + 13 = -5 + 12i - 8 - 12i + 13 = 0\,\checkmark\]
Answer: \(x = 2 + 3i\) or \(x = 2 - 3i\)
Example 4.10.3
Solving by Completing the Square
  • ProblemSolve \(x^2 + 2x + 5 = 0\) by completing the square.
  • Solve \[\begin{aligned} x^2 + 2x &= -5\\ x^2 + 2x + 1 &= -5 + 1 = -4\\ (x + 1)^2 &= -4\\ x + 1 &= \pm\sqrt{-4} = \pm 2i\\ x &= -1 \pm 2i \end{aligned}\]
  • CheckQuadratic formula gives: \(\dfrac{-2 \pm \sqrt{4-20}}{2} = \dfrac{-2 \pm \sqrt{-16}}{2} = \dfrac{-2 \pm 4i}{2} = -1 \pm 2i\). ✓
Answer: \(x = -1 + 2i\) or \(x = -1 - 2i\)
Complex Solutions Come in Conjugate Pairs

The two complex solutions of a quadratic equation with real coefficients are always complex conjugates. If you find one solution \(a + bi\), the other is always \(a - bi\). This is not a coincidence — it is a theorem that holds for all polynomials with real coefficients, and you will encounter it again in Chapter 6.

Practice 4.10Quadratic Equations — Complex Solutions
  1. \(x^2 + 4 = 0\)
  2. \(x^2 + 49 = 0\)
  3. \(x^2 - 4x + 8 = 0\)
  4. \(x^2 - 6x + 13 = 0\)
  5. \(2x^2 + x + 1 = 0\)
  6. \(3x^2 - 2x + 2 = 0\)
  7. \(x^2 + 10 = 4x\)
  8. \(2x^2 + 3x + 5 = 0\)
Sections 4.11 – 4.12
Geometry of Complex Numbers
The Complex Plane · Plotting · Magnitude · Direction as an Introduction
Section 4.11

The Complex Plane

A complex number \(a + bi\) has two components: a real part \(a\) and an imaginary part \(b\). A single number line cannot represent both. The appropriate representation is a two-dimensional coordinate system called the complex plane, also called the Argand plane.

In the complex plane, the horizontal axis is the real axis, representing the real part \(a\). The vertical axis is the imaginary axis, representing the imaginary part \(b\). Each complex number \(a + bi\) corresponds to the point \((a, b)\).

Diagram Instruction — The Complex Plane

Draw a coordinate plane. Label the horizontal axis "Real axis" and the vertical axis "Imaginary axis." Plot and label: \(3 + 4i\) at \((3, 4)\),  \(-2 + i\) at \((-2, 1)\),  \(-3 - 2i\) at \((-3, -2)\),  \(4\) at \((4, 0)\) on the real axis,  \(2i\) at \((0, 2)\) on the imaginary axis. Note that real numbers lie on the horizontal axis, and pure imaginary numbers lie on the vertical axis.

Example 4.11.1
Plotting on the Complex Plane
  • ProblemPlot and label each complex number: (a) \(2 + 3i\)   (b) \(-4 + i\)   (c) \(3 - 2i\)   (d) \(-5i\)
  • ThinkIdentify each real and imaginary part, then plot as a point \((a, b)\) in the complex plane.
  • Answer (a) \(2 + 3i\): point \((2, 3)\)
    (b) \(-4 + i\): point \((-4, 1)\)
    (c) \(3 - 2i\): point \((3, -2)\)
    (d) \(-5i = 0 - 5i\): point \((0, -5)\), on the imaginary axis

The complex plane shows that complex numbers exist in a two-dimensional world. This is why they cannot be ordered the way real numbers can — ordering is a one-dimensional concept, and the complex plane is two-dimensional. You cannot say one complex number is "greater than" another in any consistent way.

Practice 4.11The Complex Plane
  1. Plot and label each on the complex plane: \(3 + 2i\), \(-4 + i\), \(2 - 5i\), \(-3i\), \(5\).
  2. What complex number corresponds to the point \((-2, 4)\) on the complex plane?
  3. A complex number and its conjugate are plotted on the complex plane. Describe the geometric relationship between their positions. What kind of symmetry do they have?
  4. Four complex numbers are: \(1 + i\), \(-1 + i\), \(-1 - i\), \(1 - i\). Plot them. What geometric shape do they form?
Section 4.12

Magnitude and Direction as an Introduction

Magnitude of a Complex Number

The magnitude (also called absolute value or modulus) of \(a + bi\) is its distance from the origin in the complex plane:

\[|a + bi| = \sqrt{a^2 + b^2}\]

The magnitude is always a non-negative real number. This follows directly from the Pythagorean theorem.

Example 4.12.1
Computing the Magnitude of Complex Numbers
  • ProblemFind the magnitude of (a) \(3 + 4i\),   (b) \(-5 + 12i\),   (c) \(-7i\).
  • Solve (a) \(|3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

    (b) \(|-5 + 12i| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)

    (c) \(|-7i| = \sqrt{0^2 + (-7)^2} = \sqrt{49} = 7\)
Answer: 5, 13, and 7, respectively.

The direction of a complex number in the complex plane is the angle it makes with the positive real axis, measured counterclockwise. This angle is called the argument of the complex number. At the Algebra II level, direction is introduced as a concept but not computed in full generality, since the calculation involves inverse trigonometric functions that belong to Precalculus and Trigonometry. You will return to this in Chapter 13.

Notice that a complex number and its conjugate always have the same magnitude, since \(|a + bi| = |a - bi| = \sqrt{a^2 + b^2}\). Geometrically, conjugates are reflections of each other across the real axis — equidistant from the origin.

MGU Library Connection
  • Polar form, magnitude, and argument are developed in the MGU Precalculus chapter on Complex Numbers in Polar Form
  • De Moivre's Theorem for computing powers and roots geometrically — see MGU Trigonometry reference pages
  • Applications in electrical engineering phasors and quantum mechanics — MGU Physics and Engineering connections
Practice 4.12Magnitude
  1. \(|5 + 12i|\)
  2. \(|-3 + 4i|\)
  3. \(|1 + i|\)
  4. \(|-6i|\)
  5. \(|8 - 15i|\)
  6. \(|7 + 24i|\)

After computing, verify that each conjugate has the same magnitude as the original.

Sections 4.13 – 4.14
Common Errors and the Road Ahead
Seven Common Mistakes · Preparing for Polynomial Roots
Section 4.13

Common Complex Number Mistakes

Mistake 1 — Applying the radical product rule to negative radicands
\(\sqrt{-4} \cdot \sqrt{-9} = \sqrt{36} = 6\)

Error: The product rule \(\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}\) does not apply when both \(a\) and \(b\) are negative. Always convert to imaginary form first: \(\sqrt{-4} \cdot \sqrt{-9} = 2i \cdot 3i = 6i^2 = -6\). The correct answer is \(-6\).

Mistake 2 — Forgetting that \(i^2 = -1\), not \(+1\)
\((3i)^2 = 9\)

Error: \((3i)^2 = 9i^2 = 9(-1) = -9\). Squaring \(i\) produces \(-1\), not \(+1\). The result of squaring a pure imaginary number is always a negative real number.

Mistake 3 — Dropping the middle term when squaring
\((2 + 5i)^2 = 4 + 25i^2 = -21\)

Error: The middle term \(2 \cdot 2 \cdot 5i = 20i\) was omitted. Correct: \((2 + 5i)^2 = 4 + 20i + 25i^2 = 4 + 20i - 25 = -21 + 20i\).

Mistake 4 — Calling \(bi\) the imaginary part
Imaginary part of \(3 + 4i\) is \(4i\)

Error: The imaginary part is \(4\), not \(4i\). The imaginary part is the real coefficient of \(i\). The term \(bi\) is the imaginary term, not the imaginary part. The imaginary part is always a real number.

Mistake 5 — Multiplying by the wrong conjugate
To simplify \(\dfrac{2 + i}{3 - 2i}\), multiply by \(\dfrac{3 - 2i}{3 - 2i}\)

Error: Always multiply by the conjugate of the denominator, which is \(3 + 2i\) here. Multiplying by \(\dfrac{3 - 2i}{3 - 2i}\) just squares the denominator, which does not eliminate \(i\).

Mistake 6 — Leaving \(i\) in the denominator
Reporting \(\dfrac{5}{2 + i}\) as the final answer

Error: Standard form requires \(a + bi\) with no \(i\) in the denominator. Multiply by the conjugate: \(\dfrac{5(2 - i)}{(2 + i)(2 - i)} = \dfrac{10 - 5i}{5} = 2 - i\).

Mistake 7 — Computing powers of \(i\) incorrectly
\(i^{15}\) computed by listing \(i^1\) through \(i^{15}\) one at a time (error-prone)

Better method: \(15 \div 4 = 3\) remainder \(3\), so \(i^{15} = i^3 = -i\). Using the remainder rule is fast, reliable, and works for any positive integer exponent.

Practice 4.13Error Analysis
  1. A student simplifies \(\sqrt{-16} \cdot \sqrt{-4}\) and gets 8. Identify and correct the error.
  2. A student computes \((3 + 2i)^2\) and gets \(9 + 4i^2 = 5\). What was omitted? What is the correct answer?
  3. Solving \(x^2 + 25 = 0\), a student writes \(x = \pm 5\). What is wrong? What are the correct solutions?
  4. To divide \(\dfrac{2 + i}{3 - i}\), a student multiplied by \(\dfrac{3 - i}{3 - i}\). Explain the error and complete the division correctly.
  5. A student states: "The imaginary part of \(7 - 3i\) is \(-3i\)." Correct the statement.
Section 4.14

Preparing for Polynomial Roots

Chapter 4 established that numbers can be complex and that quadratic equations with negative discriminants have two complex conjugate solutions. This was not just a completion exercise — it was a preview of one of the most important theorems in algebra.

In Chapter 5 and Chapter 6, you will study polynomial functions of degree greater than two. You will ask: how many roots does a polynomial of degree \(n\) have? Can all those roots be found? Are some of them complex? The answer comes from the Fundamental Theorem of Algebra: every polynomial of degree \(n\) with complex coefficients has exactly \(n\) roots in the complex number system, counted with multiplicity.

This theorem would be meaningless without the complex numbers you have just studied. The real number system is not large enough to guarantee that every polynomial has roots. The complex number system is. Everything in Chapter 4 — especially complex conjugates, the product of conjugates, and complex solutions of quadratic equations — connects directly to that work.

Practice 4.14Readiness Check
  1. Solve \(x^2 + 16 = 0\). Write solutions in standard form \(a + bi\).
  2. Compute the discriminant of \(x^2 - 2x + 10 = 0\). What does it tell you about the number and type of solutions? Then solve the equation.
  3. If \(2 + 5i\) is a solution to a quadratic equation with real coefficients, what is the other solution? Why?
  4. A polynomial of degree 3 has one real root. How many complex roots does it have? Why must they come in pairs?
  5. Verify by substitution that \(x = -1 + 2i\) satisfies \(x^2 + 2x + 5 = 0\).
Chapter 4 — Back Matter

Chapter Summary

Key Terms Review

imaginary unit
The number \(i\), defined by \(i^2 = -1\). It is a specific number, not a variable.
pure imaginary number
A complex number \(0 + bi = bi\) where \(b \neq 0\). The real part is zero.
complex number
A number of the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i = \sqrt{-1}\).
real part
The value \(a\) in the complex number \(a + bi\). Always a real number.
imaginary part
The value \(b\) in \(a + bi\) — the coefficient of \(i\). Not the term \(bi\) itself.
standard form
The form \(a + bi\) for a complex number, with \(i\) not in the denominator.
complex conjugate
The conjugate of \(a + bi\) is \(a - bi\). Conjugates differ only in the sign of the imaginary part.
magnitude
The distance from the origin: \(|a + bi| = \sqrt{a^2 + b^2}\). Also called modulus or absolute value.
complex plane
A coordinate plane with a real axis (horizontal) and imaginary axis (vertical). Each \(a + bi\) plots as \((a, b)\).
argument
The angle a complex number makes with the positive real axis in the complex plane. Developed fully in Precalculus.

Concept Review Questions

  1. Why can no real number satisfy the equation \(x^2 = -4\)?
  2. What does the definition \(i^2 = -1\) mean? Why is it not enough to just say \(i = \sqrt{-1}\)?
  3. Explain the difference between a pure imaginary number and a complex number. Give an example of each.
  4. Why is every real number also a complex number?
  5. How are operations on complex numbers similar to operations on polynomials? How are they different?
  6. Explain why \(\sqrt{-4} \cdot \sqrt{-9} \neq \sqrt{36}\). What is the correct value?
  7. Describe the cyclic pattern for powers of \(i\). Why does the pattern repeat every four steps?
  8. What is the complex conjugate of \(a + bi\)? Why is the product of conjugates always a real number?
  9. Why do we multiply by the conjugate of the denominator when dividing complex numbers?
  10. What does the discriminant tell you about the nature of solutions to a quadratic equation?
  11. Describe the complex plane. How does it differ from the standard \(xy\)-coordinate plane?
  12. What is the magnitude of a complex number, and how is it calculated geometrically?
  13. Two complex numbers are equal. What must be true about their real parts and imaginary parts?
  14. A student says: "The imaginary part of \(4 + 7i\) is \(7i\)." What is the error?

Skill Practice

Part AImaginary Number Practice

Simplify each expression completely.

  1. \(\sqrt{-49}\)
  2. \(\sqrt{-72}\)
  3. \(\sqrt{-200}\)
  4. \(-\sqrt{-100}\)
  5. \((5i)^2\)
  6. \((-3i)^2\)
  7. \(i^9\)
  8. \(i^{30}\)
  9. \(i^{47}\)
  10. \(i^{100}\)
  11. \(i^{37}\)
  12. \(i^{80}\)
Part BAddition and Subtraction
  1. \((2 + 5i) + (3 - 2i)\)
  2. \((-4 + i) + (6 - 3i)\)
  3. \((7 - 4i) - (2 + 5i)\)
  4. \((-1 + 3i) - (-4 - 7i)\)
  5. \((3i) + (2 - 3i)\)
  6. \((5 - i) - (5 + i)\)
Part CMultiplication
  1. \((2 + i)(3 - 4i)\)
  2. \((1 + 5i)(1 - 5i)\)
  3. \((3 - 2i)^2\)
  4. \((4i)(2 - 3i)\)
  5. \((-1 + i)(-1 - i)\)
  6. \((2 + 3i)^2\)
Part DConjugates and Division
  1. \(\dfrac{10}{2i}\)
  2. \(\dfrac{3 + i}{1 + 2i}\)
  3. \(\dfrac{4 - 3i}{2 + i}\)
  4. \(\dfrac{6 - 4i}{3i}\)
  5. \(\dfrac{5}{3 - 4i}\)
  6. \(\dfrac{1 + i}{1 - i}\)
Part EQuadratic Equations with Complex Solutions

Solve each equation. Write solutions in standard form. Identify whether solutions are real or complex.

  1. \(x^2 + 4 = 0\)
  2. \(x^2 + 2x + 5 = 0\)
  3. \(x^2 - 6x + 13 = 0\)
  4. \(2x^2 + x + 1 = 0\)
  5. \(x^2 + 49 = 0\)
  6. \(x^2 - 4x + 8 = 0\)
  7. \(3x^2 - 2x + 2 = 0\)
  8. \(x^2 + 10 = 4x\)
Part FComplex Plane and Magnitude
  1. Plot and label: \(3 + 2i\), \(-4 + i\), \(2 - 5i\), \(-3i\), \(5\) on the complex plane.
  2. Find the magnitude of each: (a) \(5 + 12i\)   (b) \(-3 + 4i\)   (c) \(1 + i\)   (d) \(-6i\)
  3. Which complex number has greater magnitude: \(3 + 4i\) or \(4 + 3i\)? Explain.
  4. A complex number has magnitude 10 and imaginary part 6. Find all possible complex numbers satisfying these conditions.

Applications

  1. Electrical Engineering: An electrical circuit's impedance is \(Z = 6 + 8i\) ohms, where the real part is resistance and the imaginary part is reactance. Find the magnitude of the impedance. (This represents the total opposition to current flow.)
  2. Signal Processing: Two signals are represented as \(A = 3 + 4i\) and \(B = 1 - 2i\). Compute \(A + B\) and \(A - B\). Interpret what each result means in terms of real and imaginary components.
  3. Geometry: Four complex numbers — \(1 + i\), \(-1 + i\), \(-1 - i\), \(1 - i\) — are plotted on the complex plane. What geometric shape do they form? What is the magnitude of each?
  4. Number Theory: A complex number \(z = a + bi\) has magnitude 5. Find at least four different values of \(z\) that satisfy this. What equation do \(a\) and \(b\) satisfy?
  5. Conjugate Products: A computer graphics algorithm uses the property that \((a + bi)(a - bi)\) is always real to simplify a calculation. If \(a + bi = 5 + 7i\), compute the real result of the product with its conjugate. Explain why this property is useful.
  6. Quadratic Modeling: A physics problem leads to the equation \(x^2 - 6x + 25 = 0\). Solve for \(x\). Write solutions in standard form and interpret what complex solutions mean in the physical context.

Error Analysis

Each problem shows a student's work. Identify the error and provide the correct solution.

Problem 1
\(\sqrt{-16} \cdot \sqrt{-4} = \sqrt{64} = 8\)

What is the error? What is the correct answer?

Problem 2
\((3 + 2i)^2 = 9 + 4i^2 = 9 - 4 = 5\)

What was omitted? What is the correct expansion?

Problem 3
Solving \(x^2 + 25 = 0\): \(x^2 = -25\), so \(x = \pm\sqrt{25} = \pm 5\)

Identify the error. What are the correct solutions?

Problem 4
To divide \(\dfrac{2 + i}{3 - i}\), the student multiplied by \(\dfrac{3 - i}{3 - i}\)

Explain why this is wrong. Complete the division correctly.

Problem 5
Imaginary part of \(7 - 3i\) is \(-3i\)

Correct the statement and explain the distinction.

Problem 6
\((4 + 3i)(4 - 3i) = 16 - 9i^2 = 25i\)

Two errors appear here. Identify both. What is the correct product?

Chapter 4 Checkpoint

Chapter 4 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 4
Section 1 — Imaginary Numbers
  1. Simplify \(\sqrt{-121}\).
  2. Simplify \(\sqrt{-50}\).
  3. Compute \((6i)^2\).
  4. Simplify \((-i)^3\).
Section 2 — Standard Form
  1. Write \(-3 + \sqrt{-64}\) in standard form \(a + bi\).
  2. Identify the real part and imaginary part of \(9 - 2i\).
Section 3 — Operations
  1. Add: \((2 - 7i) + (-5 + 4i)\)
  2. Subtract: \((3 + 2i) - (7 - i)\)
  3. Multiply: \((2 + 3i)(4 - i)\)
  4. Multiply: \((1 - 6i)^2\)
Section 4 — Powers of i
  1. Simplify \(i^{37}\).
  2. Simplify \(i^{80}\).
Section 5 — Conjugates and Division
  1. Write the conjugate of \(-5 + 2i\) and compute their product.
  2. Divide: \(\dfrac{3 - i}{2 + 3i}\)
  3. Divide: \(\dfrac{10}{1 - i}\)
Section 6 — Quadratic Solutions
  1. Solve \(x^2 + 36 = 0\).
  2. Solve \(x^2 - 4x + 29 = 0\).
  3. Solve \(2x^2 + 3x + 5 = 0\).
Section 7 — Complex Plane and Magnitude
  1. Plot \(4 - 3i\) and \(-4 + 3i\) on the complex plane. Describe their geometric relationship.
  2. Find the magnitude of \(8 - 15i\).
  3. A complex number has magnitude 10 and imaginary part 6. Find all possible complex numbers satisfying these conditions.
Challenge Problems
  1. Show that if \(a + bi\) is a solution to \(x^2 - 2x + 10 = 0\), then its conjugate \(a - bi\) is also a solution. Use the quadratic formula, then verify by substitution.
  2. Compute \(\left(\dfrac{1 + i}{\sqrt{2}}\right)^8\). Show all steps using the power of \(i\) rules.
Looking Ahead
Chapter 5: Polynomial Functions  ·  Chapter 6: Polynomial Equations and Theorems

Complex numbers are not a detour. They are the foundation of the next two chapters.

In Chapter 5, you will study polynomial functions of degree 3 and higher — analyzing their graphs, zeros, turning points, and end behavior. Some of those zeros will be real. Others will be complex. The techniques of this chapter — especially complex conjugates and quadratic solutions — will be tools you reach for repeatedly.

In Chapter 6, the theory of polynomial roots becomes central. You will apply the Fundamental Theorem of Algebra, which guarantees that a polynomial of degree \(n\) has exactly \(n\) roots in \(\mathbb{C}\). You will discover that complex roots of polynomials with real coefficients always come in conjugate pairs — exactly the conjugate structure you studied in Sections 4.8 through 4.10. The result is this: no polynomial equation is truly unsolvable. Every polynomial has all of its roots within the complex number system.

Looking further ahead: in Chapter 13, you will return to the complex plane with trigonometric tools, computing complex numbers in polar form and encountering De Moivre's Theorem. In Precalculus and beyond, complex numbers appear in vectors, rotations, oscillations, and Fourier transforms — connecting algebra to the physical world in ways that are impossible with only real numbers.

MGU Library Connection
  • Complex Number Reference — Appendix E of this textbook
  • Fundamental Theorem of Algebra — stated and applied in Chapter 6
  • Polar form of complex numbers and De Moivre's Theorem — MGU Precalculus, Chapter on Complex Numbers in Polar Form
  • Trigonometric form and applications — MGU Trigonometry reference pages
  • MGU Dictionary entries: imaginary unit, complex number, conjugate, magnitude, complex plane, argument