The imaginary unit i and the square root of a negative number, standard form and the families of complex numbers, adding and subtracting, multiplying with FOIL, dividing by a complex conjugate, and plotting in the complex plane with absolute value.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 4 — Quadratic Functions and Factoring
Perform Operations with Complex Numbers
Objectives
Five outcomes. The first one is a definition; the rest are consequences of it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 275-281 — the lesson these objectives are drawn from
Warm-up
Lesson 4.5 hit one wall: a square equal to a negative number has no real solution.
Discussion prompt
Solve 2x squared plus 11 equals negative 37 as far as the methods of Lesson 4.5 will take you. Where exactly does it stop, and why?
Hint: Isolate the squared term first.
Answer:
\[ 2x^2 + 11 = -37 \;\Longrightarrow\; 2x^2 = -48 \;\Longrightarrow\; x^2 = -24 \]
The method stops here. Every real number squares to something non-negative, so no real x can work. This lesson does not find a trick to get around that — it enlarges the number system so that a square root of a negative number exists, and then the same method runs to completion.
Concept
Mathematicians defined a single new number, the imaginary unit i, whose square is negative one. Combining it with the real numbers gives the complex numbers, written a plus bi, and every rule of arithmetic you already know carries over unchanged.
imaginary unit i — The number defined by i equals the square root of negative one, so that i squared equals negative one. It is used to write the square root of any negative number.
\[ i = \sqrt{-1}, \qquad i^2 = -1 \]
This is not a trick or an approximation. It is an extension of the number system, in exactly the same spirit as the extension from whole numbers to negatives, or from fractions to irrationals: each was invented so that a class of previously unsolvable equations could be solved.
Figure (svg): The definition of the imaginary unit and the rule for the square root of a negative number
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 275-275
Section
Section 1
Concept
For a positive number r, the square root of negative r is i times the square root of r. Squaring that expression returns negative r, which is the check that the definition does what was wanted.
\[ \sqrt{-r} = i\sqrt{r}, \qquad (i\sqrt{r})^2 = i^2 r = -r \]
The simplification rules of Lesson 4.5 still apply inside the radical: pull out the largest perfect-square factor first, then attach the i.
Figure (svg): The definition of the imaginary unit and the rule for the square root of a negative number
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 275-275 — The Square Root of a Negative Number
Picture it
One equation, two properties that follow from it.
Figure (svg): The definition of the imaginary unit and the rule for the square root of a negative number
Property 2 is what makes the whole thing legitimate: the new number really does square to a negative, so it really does solve the equation that had no answer.
Worked example
Example 1. The same four steps as Lesson 4.5, with one extra.
\[ \text{Solve } 2x^2 + 11 = -37. \]
Isolate the squared term
Why: Subtracting 11 gives 2x squared equal to negative 48, and dividing by 2 gives x squared equal to negative 24.
\[ x ^{2} = -24 \]
Take square roots of both sides
Why: Both signs are written, exactly as before.
\[ x = +- \sqrt{-24} \]
Write in terms of i
Why: The square root of negative 24 is i times the square root of 24.
\[ x = +- i \sqrt{24} \]
Simplify the radical
Why: Twenty-four is 4 times 6, so the 4 comes out as 2.
\[ x = +- 2 i \sqrt{6} \]
Figure (svg): The solution to Worked example solve an equation with no real solutions shown as a ladder of expressions, one row per algebraic move
\[ x = \pm 2i\sqrt{6} \]
Verify: square the answer and substitute
Why: Two i root 6, squared, is 4 times i squared times 6, which is 24 times negative one, or negative 24. Then 2 times negative 24 is negative 48, plus 11 is negative 37 — the right side exactly. The negative solution behaves identically because squaring erases its sign.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 275-275
Matching
Pull out the square factor, then attach the i.
Match the pairs
Why: Three of the four radicands contain the square factor 4, which comes out as a 2 in front. The last has no square factor at all, so the i sits directly against the radical. The order of operations is always the same: strip the minus sign into an i, then simplify what is left exactly as in Lesson 4.5.
Worked example
Guided Practice 1 to 6.
\[ \text{Solve } x^2 = -13, \; x^2 = -38, \; x^2 + 11 = 3, \; x^2 - 8 = -36, \; 3x^2 - 7 = -31, \; 5x^2 + 33 = 3. \]
The two direct ones
Why: Thirteen and 38 have no square factors, so the radicals stay as they are.
\[ x = +- i \sqrt{13}; + - i \sqrt{38} \]
The two needing one move
Why: Subtracting 11 gives negative 8, and adding 8 gives negative 28.
\[ x ^{2} = -8; x ^{2} = -28 \]
Simplify those two
Why: Eight is 4 times 2 and 28 is 4 times 7.
\[ +- 2 i \sqrt{2}; + - 2 i \sqrt{7} \]
The two with coefficients
Why: Three x squared equal to negative 24 gives x squared equal to negative 8; five x squared equal to negative 30 gives negative 6.
\[ +- 2 i \sqrt{2}; + - i \sqrt{6} \]
Figure (svg): The solution to Worked example six more equations shown as a ladder of expressions, one row per algebraic move
\[ \pm i\sqrt{13}, \; \pm i\sqrt{38}, \; \pm 2i\sqrt{2}, \; \pm 2i\sqrt{7}, \; \pm 2i\sqrt{2}, \; \pm i\sqrt{6} \]
Verify: check the third and fifth agree
Why: Both reduce to x squared equal to negative 8 and so have the same solutions, even though one started with a constant to move and the other with a coefficient to divide by. Squaring 2i root 2 gives 4 times negative one times 2, which is negative 8, as required.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 275-275
Trap
\[ \sqrt{-4}\cdot\sqrt{-9} \]
Apply the product property from Lesson 4.5
Why: The two radicands are multiplied under a single radical.
\[ \sqrt{(-4)(-9)} = \sqrt{36} = 6 \quad \text{(wrong)} \]
Converting to i first gives 2i times 3i, which is 6i squared, or negative 6. The two answers differ in sign.
\[ \sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6 \]
Write every negative radicand in terms of i before multiplying
Why: The product property was stated for non-negative radicands only, and this is exactly where it fails.
\[ \sqrt{ab} = \sqrt{a}\sqrt{b} \quad \text{requires } a \ge 0 \text{ and } b \ge 0 \]
The condition on the property was not decoration. Convert first, then use ordinary multiplication, and the i squared takes care of the sign automatically.
Fill the middle
Example 1, at the radical.
Fill in the blanks
x = \pm\sqrt2 = \pm i\sqrt___ = \pm ___i\sqrt___
Why: Twenty-four is 4 times 6, so the root of 4 comes out as 2 and root 6 stays under the radical. Writing 2i root 6 rather than i root 24 matters for the same reason it mattered in Lesson 4.5: the radicand must have no perfect-square factor left for the expression to count as simplified.
Prediction
Commit before reasoning.
Predict first
Which earlier extension of the number system is i most like?
Correct: Extending the whole numbers to include negatives.
\[ \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C} \]
Why: Negative numbers were introduced so that 5 minus 8 would have an answer, fractions so that 2 divided by 3 would, and irrationals so that the square root of 2 would. Each time the rules of arithmetic were preserved and one class of unsolvable problems became solvable. The imaginary unit is the same move applied to the square root of a negative, and it is the last such extension you will need: with i in hand, every polynomial equation has solutions.
Sorting
Ask whether a square root of a negative appears.
Sort into buckets
Sort each solution set.
Every one of these has two solutions. What changes is which number system they live in, and that is decided entirely by the sign after the square is isolated.
Section
Section 2
Concept
A complex number in standard form is a plus bi, where a is the real part and bi the imaginary part. To add or subtract two of them, combine the real parts and the imaginary parts separately, exactly as you would collect like terms.
complex number — A number a plus bi with a and b real. If b is not zero it is an imaginary number; if in addition a is zero it is a pure imaginary number. Two complex numbers are equal exactly when their real parts agree and their imaginary parts agree.
\[ (a + bi) + (c + di) = (a + c) + (b + d)i \]
Because b can be zero, every real number is also a complex number. The real numbers did not go anywhere; they are the complex numbers with no imaginary part.
Figure (svg): Nested regions showing complex numbers containing the real numbers and the imaginary numbers, with pure imaginary numbers inside the latter
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 276-276 — Sums and Differences of Complex Numbers
Picture it
One system containing the reals and the imaginaries side by side.
Figure (svg): Nested regions showing complex numbers containing the real numbers and the imaginary numbers, with pure imaginary numbers inside the latter
Notice that real and imaginary do not overlap, and that pure imaginary is a special case of imaginary, not a separate family. The number 5 minus 5i is imaginary but not pure imaginary.
Worked example
Example 2, all three parts.
\[ \text{Simplify } (8 - i) + (5 + 4i), \; (7 - 6i) - (3 - 6i), \; 10 - (6 + 7i) + 4i. \]
First: add the parts separately
Why: Eight plus 5 is 13, and negative 1 plus 4 is 3.
\[ 13 + 3 i \]
Second: subtract the parts separately
Why: Seven minus 3 is 4, and negative 6 plus 6 is 0.
\[ 4 + 0 i = 4 \]
Note what the second result is
Why: The imaginary part vanished, so the answer is a real number — written as 4, not as 4 plus 0i.
Third: work left to right
Why: Ten minus 6 is 4 with negative 7i, then adding 4i gives negative 3i.
\[ 4 - 3 i \]
Figure (svg): The solution to Worked example add and subtract shown as a ladder of expressions, one row per algebraic move
\[ 13 + 3i, \quad 4, \quad 4 - 3i \]
Verify: reverse the second subtraction
Why: Adding the answer back: 4 plus the quantity 3 minus 6i is 7 minus 6i, which is the number subtracted from. Subtraction and addition undo each other here exactly as they do for real numbers, which is the point of defining the operations part by part.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 276-276
Sorting
Look at whether a is zero, b is zero, or neither.
Sort into buckets
Sort each complex number.
Every item here is a complex number. The three buckets are subdivisions of one system, which is what the nested diagram shows.
Worked example
Example 3. A real situation where complex addition is the natural language.
\[ \text{A series circuit has a } 5 \text{ ohm resistor, a } 3 \text{ ohm inductor and a } 4 \text{ ohm capacitor. Find its impedance.} \]
Convert each component to an impedance
Why: A resistance R contributes R, an inductive reactance L contributes L times i, and a capacitive reactance C contributes negative C times i.
\[ 5, 3 i, -4 i \]
Add the impedances
Why: Impedance in a series circuit is the sum of the individual impedances.
\[ 5 + 3 i + (-4 i) \]
Combine the imaginary parts
Why: Three i minus 4i is negative i.
\[ 5 - i \]
Interpret the answer
Why: The real part is the resistance and the imaginary part is the net reactance, which here is capacitive because it is negative.
\[ 5 - i\text{ ohms} \]
Figure (svg): The solution to Worked example impedance in a circuit shown as a ladder of expressions, one row per algebraic move
\[ 5 - i \text{ ohms} \]
Verify: change one component and see what happens
Why: Guided Practice 10 replaces the capacitor with a 7 ohm one, giving 5 plus 3i minus 7i, or 5 minus 4i. The real part is unchanged because only the reactances moved, which confirms that the two parts really do track separate physical quantities rather than being an arbitrary bookkeeping device.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 277-277
Error analysis
A student subtracts one complex number from another.
Annotate
On: \( (3 + 7i) - (8 - 2i) = 3 - 8 + 7i - 2i = -5 + 5i \)
The same slip as distributing a negative over a bracket in Lesson 1.2. Rewrite the subtraction as adding the opposite of the whole number and it cannot happen.
Fill the middle
Guided Practice 9.
Fill in the blanks
-4 - (1 + i) - (5 + 9i) = -10 - 10i
Why: The real parts give negative 4 minus 1 minus 5, which is negative 10, and the imaginary parts give negative 1 minus 9, which is negative 10. Both subtractions apply to the whole bracket, and the fact that both parts came out as negative 10 is a coincidence worth noticing so it does not become an assumption.
Two truths and a lie
All three are about the families of complex numbers.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and the textbook asks about it directly in Exercise 2. Imaginary requires b to be non-zero, and a real number such as negative 1 is complex with b equal to zero. The containment runs one way only: all imaginary numbers are complex, but not all complex numbers are imaginary.
Prediction
Commit before reasoning.
Predict first
If x plus yi equals 5 minus 3i, with x and y real, what are x and y?
Correct: x equals 5 and y equals negative 3.
\[ a + bi = c + di \;\Longleftrightarrow\; a = c \text{ and } b = d \]
Why: Two complex numbers are equal exactly when their real parts agree and their imaginary parts agree, so a single complex equation carries two real equations inside it. That is genuinely useful: it lets one equation determine two unknowns, and it is why complex numbers are convenient in physics and engineering rather than merely legal.
Section
Section 3
Concept
Complex numbers are multiplied by the distributive property or FOIL, exactly as binomials are. The one extra step is that any i squared produced along the way is replaced by negative one, which turns an imaginary term into a real one.
\[ (9 - 2i)(-4 + 7i) = -36 + 63i + 8i - 14i^2 = -22 + 71i \]
The i squared term always changes sign and moves into the real part, which is why the real part of a product is rarely just the product of the two real parts.
Figure (svg): A grid showing the four products of FOIL for two complex numbers, with the i squared term converted
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 277-277 — Multiply complex numbers
Picture it
Example 4b, laid out as a grid.
Figure (svg): A grid showing the four products of FOIL for two complex numbers, with the i squared term converted
Three of the four cells are ordinary algebra. The fourth, highlighted, is the only place the new number does anything, and it contributes to the real part rather than the imaginary one.
Worked example
Example 4, both parts.
\[ \text{Simplify } 4i(-6 + i) \text{ and } (9 - 2i)(-4 + 7i). \]
Distribute in the first
Why: Four i times negative 6 is negative 24i, and 4i times i is 4i squared.
\[ -24 i + 4 i ^{2} \]
Replace i squared
Why: Four times negative one is negative 4, which is a real term.
\[ -24 i - 4 \]
Write in standard form
Why: The real part comes first.
\[ -4 - 24 i \]
FOIL the second
Why: The four products are negative 36, 63i, 8i and negative 14i squared, and the last becomes positive 14.
\[ -36 + 14 + 71 i \]
Figure (svg): The solution to Worked example multiply complex numbers shown as a ladder of expressions, one row per algebraic move
\[ -4 - 24i \qquad \text{and} \qquad -22 + 71i \]
Verify: check the second product's real part
Why: The real contributions are negative 36 from the outer constants and positive 14 from the i squared term, giving negative 22. Had the i squared been left alone, the answer would have kept an i squared term and would not be in standard form at all — that substitution is what makes the product a complex number rather than an unfinished expression.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 277-278
Fill the middle
Example 4a, at the substitution.
Fill in the blanks
4i(-6 + i) = -24i + 4i^2 = -24i + 4(-1) = -4 - 24i
Why: Four times negative one is negative 4, a real number, so it becomes the real part of the answer. Note the reordering at the end: standard form always writes the real part first, even when the imaginary term was produced first.
Worked example
Guided Practice 11 and 12.
\[ \text{Simplify } i(9 - i) \text{ and } (3 + i)(5 - i). \]
Distribute the first
Why: I times 9 is 9i, and i times negative i is negative i squared.
\[ 9 i - i ^{2} \]
Replace and reorder
Why: Negative i squared is positive 1, so the answer is 1 plus 9i.
\[ 1 + 9 i \]
FOIL the second
Why: The four products are 15, negative 3i, 5i and negative i squared.
\[ 15 + 2 i - i ^{2} \]
Replace and combine
Why: Negative i squared is positive 1, so the real part is 16.
\[ 16 + 2 i \]
Figure (svg): The solution to Worked example two more products shown as a ladder of expressions, one row per algebraic move
\[ 1 + 9i \qquad \text{and} \qquad 16 + 2i \]
Verify: notice where the real part came from
Why: In both, the real part of the answer is larger than a naive guess, because negative i squared contributed a positive number. In the second, 3 times 5 is 15 but the answer's real part is 16 — the extra 1 came entirely from the imaginary terms multiplying each other.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 278-278
Trap
\[ (3 + i)(5 - i) = 15 - 3i + 5i - i^2 \]
Collect like terms and stop
Why: The i squared term is treated as if it were another kind of term to be collected.
\[ 15 + 2i - i^2 \quad \text{(wrong: not standard form)} \]
The expression is correct but unfinished. Standard form is a plus bi, and an i squared term belongs in neither slot.
\[ 15 + 2i - i^2 = 15 + 2i - (-1) = 16 + 2i \]
Substitute negative one for i squared before collecting
Why: The substitution is not optional tidying; it is what turns the expression into a complex number.
\[ i^2 = -1 \;\Longrightarrow\; -i^2 = +1 \]
Watch the sign carefully: the term is negative i squared, so replacing i squared by negative one makes it positive one. Sign errors here are the single most common source of wrong real parts.
Matching
FOIL, then substitute.
Match the pairs
Why: In every one of these the i squared term contributed to the real part. Two of them started with a purely imaginary factor, which is why their answers still have both parts: multiplying by i does not keep a number imaginary, it rotates it.
Prediction
Commit before reasoning.
Predict first
Multiply 1, then i, then i squared, then i cubed, by i each time. What happens?
Correct: The values cycle through i, negative 1, negative i, 1, and repeat.
\[ i^1 = i, \; i^2 = -1, \; i^3 = -i, \; i^4 = 1, \; i^5 = i, \; \dots \]
Why: Multiplying by i takes 1 to i, i to negative 1, negative 1 to negative i, and negative i back to 1, so the powers of i repeat with period four. In the complex plane of the last idea this is a quarter turn about the origin each time, which is why complex multiplication turns up wherever rotation does. It also gives a quick way to evaluate any power of i: divide the exponent by 4 and use the remainder.
Comparison
Fill the blanks. Only one row differs.
Comparison matrix
| Step | (3 + x)(5 - x) | (3 + i)(5 - i) |
|---|---|---|
| The four products | 15, -3x, 5x, -x^2 | 15, -3i, 5i, -i^2 |
| Combine the middle | +2x | +2i |
| The squared term | stays as -x^2 | becomes +1 |
| The result | -x^2 + 2x + 15 | 16 + 2i |
Everything is identical until the squared term. Because i squared is a number rather than a term, the complex product collapses to two terms while the polynomial keeps three.
Section
Section 4
Concept
A quotient is not in standard form while an i remains in the denominator. Multiplying the numerator and denominator by the complex conjugate of the denominator makes the denominator real, after which the quotient splits into a real part and an imaginary part.
complex conjugates — The numbers a plus bi and a minus bi. Their product, a squared plus b squared, is always a real number.
\[ (a + bi)(a - bi) = a^2 + b^2 \]
This is the same manoeuvre as rationalising a denominator in Lesson 4.5, with the same reason for working: the cross terms cancel and the squared term loses its radical, or here its i.
Figure (svg): Dividing by a complex conjugate, with the cancelling middle terms highlighted
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 278-278 — Complex conjugates
Picture it
Example 5: 7 plus 5i over 1 minus 4i.
Figure (svg): Dividing by a complex conjugate, with the cancelling middle terms highlighted
The denominator's two imaginary terms cancel and its i squared term flips sign, leaving 1 plus 16, or 17. From there the quotient is just a fraction with a whole-number bottom.
Worked example
Example 5. Multiply by the conjugate, then split.
\[ \text{Write } \frac{7 + 5i}{1 - 4i} \text{ in standard form.} \]
Identify the conjugate of the denominator
Why: The denominator is 1 minus 4i, so its conjugate is 1 plus 4i.
\[ \text{multiply by } \frac{1 + 4 i}{1 + 4 i} \]
Multiply out the numerator
Why: The four products are 7, 28i, 5i and 20i squared, and the last becomes negative 20.
\[ -13 + 33 i \]
Multiply out the denominator
Why: The four products are 1, 4i, negative 4i and negative 16i squared; the middle terms cancel and the last becomes positive 16.
\[ 17 \]
Split into standard form
Why: Dividing each part by 17 gives the real and imaginary parts separately.
\[ -\frac{13}{17} + (\frac{33}{17}) i \]
Figure (svg): The solution to Worked example divide complex numbers shown as a ladder of expressions, one row per algebraic move
\[ -\tfrac{13}{17} + \tfrac{33}{17}i \]
Verify: multiply the answer back by the denominator
Why: Multiplying negative 13 over 17 plus 33 over 17 i by 1 minus 4i gives, after clearing the 17, the quantity negative 13 plus 33i times 1 minus 4i, which is negative 13 plus 52i plus 33i plus 132, or 119 plus 85i; dividing by 17 gives 7 plus 5i, the original numerator. Division really is the inverse of multiplication here.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 278-278
Matching
Change only the sign between the parts.
Match the pairs
Why: Only the sign of the imaginary part changes; the real part is copied unchanged, including its own sign. The last one is the useful check on that: its conjugate is negative 2 minus 5i, not 2 minus 5i.
Worked example
Guided Practice 13 and 14.
\[ \text{Write } \frac{5}{1 + i} \text{ and } \frac{5 + 2i}{3 - 2i} \text{ in standard form.} \]
First: multiply by 1 minus i
Why: The denominator becomes 1 plus 1, which is 2, and the numerator becomes 5 minus 5i.
\[ \frac{5 - 5 i}{2} \]
Split the first
Why: Each part is divided by 2.
\[ \frac{5}{2} - (\frac{5}{2}) i \]
Second: multiply by 3 plus 2i
Why: The denominator becomes 9 plus 4, which is 13.
\[ \text{denominator } 13 \]
Multiply out the second numerator
Why: The four products are 15, 10i, 6i and 4i squared, and the last becomes negative 4.
\[ 11 + 16 i \]
Split the second
Why: Each part is divided by 13.
\[ \frac{11}{13} + (\frac{16}{13}) i \]
Figure (svg): The solution to Worked example two more quotients shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{5}{2} - \tfrac{5}{2}i \qquad \text{and} \qquad \tfrac{11}{13} + \tfrac{16}{13}i \]
Verify: check the denominators against the shortcut
Why: For 1 plus i the conjugate product is 1 squared plus 1 squared, or 2; for 3 minus 2i it is 3 squared plus 2 squared, or 13. Both match the long multiplication, so the formula a squared plus b squared can be used directly and only the numerator needs FOIL.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 278-278
Error analysis
A student divides by multiplying top and bottom by the numerator's conjugate.
Annotate
On: \( \frac{7 + 5i}{1 - 4i}\cdot\frac{7 - 5i}{7 - 5i} = \frac{49 + 25}{7 - 28i - 35i + 20i^2} \)
Read the denominator, change the sign between its two parts, and multiply by that. Whatever is happening in the numerator is irrelevant to the choice.
Fill the middle
Example 5, at the denominator.
Fill in the blanks
(1 - 4i)(1 + 4i) = 1 + 4i - 4i - 16i^2 = 1 + 16 = 17
Why: The imaginary terms cancel and negative 16i squared becomes positive 16, giving 17. In general the product is a squared plus b squared, which is a sum rather than a difference — the opposite of the radical conjugates in Lesson 4.5, and the sign flip comes from i squared being negative.
Ranking
Writing a quotient in standard form.
Put in order
Why: The last step is the one that produces standard form: until the numerator is split, the answer is a single fraction rather than a real part plus an imaginary part. The denominator can be written down directly from a squared plus b squared, so in practice only the numerator needs expanding.
Prediction
Commit before reasoning.
Predict first
Multiplying a plus bi by a minus bi gives a squared plus b squared. Where did the imaginary part go?
Correct: The two cross terms are exact opposites and cancel.
\[ (a+bi)(a-bi) = a^2 - abi + abi - b^2i^2 = a^2 + b^2 \]
Why: The outer term is negative abi and the inner term is positive abi, so they sum to zero for every a and b. What remains is a squared minus b squared times i squared, and since i squared is negative one that becomes a squared plus b squared. Both the cancellation and the sign flip are automatic, which is why the conjugate trick never fails.
Section
Section 5
Concept
Every complex number a plus bi corresponds to a point in the complex plane: a units along the horizontal real axis and b units along the vertical imaginary axis. The absolute value of a complex number is its distance from the origin.
absolute value of a complex number — For z equal to a plus bi, the non-negative real number given by the square root of a squared plus b squared. It is the distance between z and the origin in the complex plane.
\[ \lvert a + bi \rvert = \sqrt{a^2 + b^2} \]
The real numbers occupy the horizontal axis, so the familiar number line is still there as one line inside a whole plane. That is why complex numbers cannot be ordered: there is no sensible way to say which of two points in a plane is larger.
Figure (svg): Four complex numbers plotted in the complex plane, with the real axis horizontal and the imaginary axis vertical
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 278-279 — Absolute Value of a Complex Number
Picture it
Example 6: 3 minus 2i, negative 2 plus 4i, 3i, and negative 4 minus 3i.
Figure (svg): Four complex numbers plotted in the complex plane, with the real axis horizontal and the imaginary axis vertical
The pure imaginary number 3i lies on the vertical axis, and any real number would lie on the horizontal one. Position encodes both parts at once.
Worked example
Example 6, all four.
\[ \text{Plot } 3 - 2i, \; -2 + 4i, \; 3i, \; -4 - 3i. \]
First: read the two parts
Why: The real part 3 moves right, and the imaginary part negative 2 moves down.
\[ \text{right } 3,\text{ down } 2 \]
Second
Why: Real part negative 2 moves left; imaginary part 4 moves up.
\[ \text{left } 2,\text{ up } 4 \]
Third
Why: The real part is zero, so there is no horizontal movement at all.
\[ \text{up } 3,\text{ on the imaginary axis} \]
Fourth
Why: Real part negative 4 moves left; imaginary part negative 3 moves down.
\[ \text{left } 4,\text{ down } 3 \]
Figure (svg): The solution to Worked example plot complex numbers shown as a ladder of expressions, one row per algebraic move
\[ 3 - 2i, \; -2 + 4i, \; 3i, \; -4 - 3i \]
Verify: check which axis each point relates to
Why: The third point lies exactly on the imaginary axis because its real part is zero, and no point here lies on the real axis because none has imaginary part zero. Plotting a real number would put it on the horizontal axis, which is precisely the number line you have used since Lesson 1.1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 278-278
Matching
Square both parts, add, take the root.
Match the pairs
Why: Two of these come out as whole numbers because their parts form a Pythagorean triple or because one part is zero. The other two stay as radicals, which is normal — an absolute value is only whole when the two parts happen to be legs of a right triangle with a whole hypotenuse.
Worked example
Example 7 and Guided Practice 15 to 18.
\[ \text{Find } \lvert -4 + 3i \rvert, \; \lvert -3i \rvert, \; \lvert 4 - i \rvert, \; \lvert -3 - 4i \rvert, \; \lvert 2 + 5i \rvert, \; \lvert -4i \rvert. \]
First: square both parts and add
Why: Negative 4 squared is 16 and 3 squared is 9, giving 25.
\[ \sqrt{25} = 5 \]
Second: treat it as 0 plus negative 3i
Why: Zero squared plus negative 3 squared is 9.
\[ \sqrt{9} = 3 \]
The next two
Why: Sixteen plus 1 is 17; 9 plus 16 is 25.
\[ \sqrt{17}; 5 \]
The last two
Why: Four plus 25 is 29; 0 plus 16 is 16.
\[ \sqrt{29}; 4 \]
Figure (svg): The solution to Worked example absolute values shown as a ladder of expressions, one row per algebraic move
\[ 5, \; 3, \; \sqrt{17}, \; 5, \; \sqrt{29}, \; 4 \]
Verify: notice two numbers with the same absolute value
Why: Negative 4 plus 3i and negative 3 minus 4i both have absolute value 5, because both are 5 units from the origin — they lie on the same circle. Absolute value therefore does not determine a complex number, only its distance from zero, which is exactly the same situation as for real numbers where 5 and negative 5 share an absolute value.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 279-279
Error analysis
A student computes an absolute value.
Annotate
On: \( \lvert -4 + 3i \rvert = -4 + 3 = -1 \)
The squaring is what removes the signs, which is why the answer is always non-negative. It is the Pythagorean theorem, and a hypotenuse is never negative.
Fill the middle
Example 7a.
Fill in the blanks
\lvert -4 + 3i \rvert = \sqrt25 = \sqrt___} = 5
Why: Negative 4 squared is positive 16, and 3 squared is 9, giving 25 under the radical. The signs disappear at the squaring step, which is why negative 4 plus 3i and 4 plus 3i have the same absolute value: they are mirror images across the imaginary axis and equally far from the origin.
Sorting
Look at which part is zero, if either.
Sort into buckets
Sort each complex number by its position.
The two axes are exactly the two special families from the second idea, drawn rather than described.
Prediction
Commit before reasoning.
Predict first
Is 3 plus 4i greater than 5, or less, or neither?
Correct: Neither — complex numbers cannot be ordered.
\[ \lvert 3 + 4i \rvert = 5 = \lvert 5 \rvert, \quad \text{but } 3 + 4i \neq 5 \]
Why: There is no way to define less than on the complex numbers that keeps the rules real inequalities obey. Both numbers do have absolute value 5, so they are the same distance from the origin, but that is a statement about distances rather than about the numbers. This is the one thing genuinely lost in the extension: you gain solutions to every quadratic and give up the number line's ordering.
Comparison
Fill the blanks. The extension was designed to preserve as much as possible.
Comparison matrix
| Property | Real numbers | Complex numbers |
|---|---|---|
| Adding and multiplying | commutative, associative, distributive | all still hold |
| Every quadratic solvable | no, x^2 = -1 fails | yes, always |
| Pictured as | a line | a plane |
| Ordering, less than | yes | no |
| Absolute value | distance from 0 on the line | distance from 0 in the plane |
The last two rows are the honest trade: the plane is richer than the line, but a plane has no natural left-to-right order.
Pattern
One routine per operation, all resting on i squared equals negative one.
Every answer should end in the form a plus bi, with no i squared anywhere and no i in a denominator.
OpenStax Algebra and Trigonometry 2e, §2.4 Complex Numbers §2.4
Check
Solving with i. Simplify the radical too.
Check your understanding
Solve 2x^2 + 11 = -37.
Answer: A
Why: Isolating gives x^2 = -24, so x = +- i sqrt(24) = +- 2i sqrt(6) after simplifying.
Check
Multiplying. Watch the i squared term.
Check your understanding
Write (3 + i)(5 - i) in standard form.
Answer: A
Why: FOIL gives 15 + 2i - i^2, and -i^2 is +1, so the real part is 16.
Check
Absolute value. Square before adding.
Check your understanding
Find |-3 - 4i|.
Answer: A
Why: The value is sqrt((-3)^2 + (-4)^2) = sqrt(9 + 16) = sqrt(25) = 5.
Real world
An alternating-current circuit has a resistor of 4 ohms, an inductor of 9 ohms of reactance and a capacitor of 6 ohms of reactance, all in series. Ohm's law still holds, with voltage equal to current times impedance.
Discussion prompt
Find the impedance, its absolute value, and then the voltage if a current of 3 amps flows. What does the absolute value mean physically?
Hint: Impedance is 4 plus 9i minus 6i. Voltage is a complex product.
Answer:
\[ Z = 4 + 9i - 6i = 4 + 3i \text{ ohms}, \qquad \lvert Z \rvert = \sqrt{16 + 9} = 5 \]
\[ V = IZ = 3(4 + 3i) = 12 + 9i \text{ volts} \]
The absolute value, 5 ohms, is the total opposition the circuit presents to current — the quantity an ordinary meter reads. The real and imaginary parts say how that opposition splits between resistance, which dissipates energy, and reactance, which stores and returns it.
This is why engineers use complex numbers rather than tracking two separate real quantities: the single number 4 plus 3i carries both facts, and adding components in series becomes ordinary complex addition. The absolute value from the last idea turns out to be exactly the measurable magnitude.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the square root of negative 4 times the square root of negative 9 equal to 6?
Correct: No — it is negative 6.
\[ \sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6 \]
Why: Each radical becomes an imaginary number first: 2i times 3i is 6i squared, which is negative 6. The product property of Lesson 4.5 was stated only for non-negative radicands, and this is precisely the case it excludes. Applying it anyway is the single most common error with imaginary numbers, and it produces an answer with the wrong sign rather than an obviously absurd one, which is what makes it dangerous.
Explain it
They have just been told that a negative number has no square root and are suspicious of this whole lesson.
Discussion prompt
In four sentences or fewer, explain what i is, why inventing it is legitimate, and what it buys you.
Hint: Compare it with the invention of negative numbers.
Answer:
The claim that a negative has no square root was always a claim about real numbers specifically. Mathematicians defined a new number i whose square is negative one, and kept every rule of arithmetic otherwise unchanged, so nothing you already knew stops being true.
That is the same move that produced negative numbers, when people wanted 5 minus 8 to have an answer, and fractions, when they wanted 2 divided by 3 to have one. What it buys is that every quadratic equation now has solutions, with no exceptions — which is what the next two lessons rely on.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For radicals, strip the minus into an i before doing anything else, then simplify as in Lesson 4.5. For i squared, write the substitution as its own line rather than doing it in your head, keeping any minus sign in front of it. For conjugates, look only at the denominator and change only the sign between its parts. For absolute value, square both parts before adding anything. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Choose the two complex numbers 3 plus 2i and 1 minus 4i and put them through everything. Top left: draw a complex plane and plot both, marking the absolute value of each as a right triangle with its legs labelled. Top right: compute their sum, their difference and their product, showing where each i squared term went. Bottom left: divide the first by the second in full, writing the conjugate you used and the real denominator it produced. Bottom right: draw the nested diagram of complex, real, imaginary and pure imaginary numbers, and place each of your six results into the right region. In a margin, write the four values of i to the first, second, third and fourth powers, and say what pattern the powers follow.
If any of your six results is real, that is not a mistake — check whether it came from a subtraction that cancelled the imaginary parts, which is exactly what happened in Example 2b.
Recap
Five things, all consequences of one definition.
| If you see | Then |
|---|---|
| A negative under a radical | Write it as i times the positive root |
| An i squared in an expression | Replace it by -1, keeping any minus in front |
| An i in a denominator | Multiply by the complex conjugate |
| A number with b equal to 0 | It is real, and lies on the horizontal axis |
| A number with a equal to 0 | It is pure imaginary, on the vertical axis |
| Absolute value bars | Square both parts, add, take the root |
Every quadratic equation now has solutions, but you still have no method that finds them in every case. Lesson 4.7 supplies one by turning any quadratic into the square-equals-a-number shape of Lesson 4.5.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.6 Perform Operations with Complex Numbers §4.6, pp. 275-281 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.