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Lesson 3 of 3

Complex Numbers and Quadratic Equations · Lesson 3 of 3

Chapter Summary and Practice

“Connect complex arithmetic, powers, conjugation and geometry through worked reasoning and a full chapter practice set.”

Learning Objectives

• Connect the imaginary unit and standard form to the purpose of extending real numbers. • Choose efficient arithmetic, inverse and power-cycle methods for mixed calculations. • Use conjugate and modulus relationships to prove statements while checking denominators. • Relate algebraic conditions to distance and reflection in the Argand plane. • Derive component identities by expanding products and powers step by step. • Solve the complete chapter’s mixed practice and explain why each method applies.

One number system, several connected views

The chapter starts with an equation that real numbers cannot solve. It introduces i, builds numbers a + ib, and develops their arithmetic. The Argand plane then turns their two components into coordinates, making modulus and conjugation visible as distance and reflection.

These are connected tools rather than separate procedures. Standard form reveals the components needed for equality; the conjugate removes an imaginary denominator; modulus can shorten a proof involving a product or quotient. When reviewing, ask which representation reveals the structure of the problem before beginning a long calculation.

Chapter conceptCentral ideaWhat to check
The imaginary uniti² = −1 extends the real number systemThe number system in which an equation is being solved
Standard form and equalitya + ib keeps both real components visibleIm z is b; both components must match
Addition and subtractionCollect like componentsDistribute a subtraction sign over both terms
Multiplication and algebraic lawsDistribute and replace i² by −1The imaginary-term product contributes a real term
Inverse and divisionA nonzero number has a reciprocala and b must not both be zero
Integer powers of iThe values repeat every four exponentsA valid remainder or reciprocal for negative powers
Negative square roots√(−t) = i√t for t > 0The selected radical versus the two equation roots
Polynomial identitiesDistribution works for complex inputsSigns in squared and cubed expressions
Conjugate and modulusa − ib and √(a² + b²)Conjugate is complex; modulus is real and nonnegative
Product and quotient propertieszz̄ = |z|² and |z₁z₂| = |z₁||z₂|Quotients require a nonzero denominator
Argand representationa + ib ↔ (a, b)Real axis, imaginary axis and coordinate order
GeometryModulus is origin distance; conjugate is reflectionEqual modulus does not imply equal number
Development of the subjectNegative roots led to a larger algebraic systemConnect the historical questions with the mathematics

Standard form makes comparison possible

A long expression may contain several powers of i and both real and imaginary terms. Simplifying to a + ib is therefore a useful first task. Once standard form is reached, its parts can be compared, conjugated or used in a modulus calculation without guessing.

Keep i² = −1 visible during expansion. A term containing an even power of i becomes real; one containing an odd power is a real multiple of i. The four-power cycle reduces both types. If an equality contains unknown real components, comparing parts produces ordinary real equations.

Core arithmetic relationshipsLaTeX
The components a, b, c and d are real numbers. Addition and multiplication are associative and commutative; multiplication distributes over addition.

The same algebraic laws give the familiar square, cube and factorisation identities. A plus or minus inside the binomial affects the odd-power cross terms. When an input includes i, expand with the identity first and then reduce each power of i.

Polynomial identities in one placeLaTeX
Use the upper signs throughout for a sum, or the lower signs throughout for a difference. These identities hold for every complex z₁ and z₂.

For integer powers of i, divide the exponent into groups of four and a remainder. A whole group contributes 1, while the remainder determines one of the four possible outputs. Negative exponents follow the same rule because i⁻¹ = −i.

Integer power cycleLaTeX
k is any integer, including negative integers and zero.
Example — A mixed power expression

Problem
Evaluate [i¹⁸ + (1/i)²⁵]³ in standard form.

  1. 1.Because 18 has remainder 2 when divided by 4, i¹⁸ = −1.
  2. 2.The inverse 1/i is −i. Thus (1/i)²⁵ = i⁻²⁵. Write −25 = 4(−7) + 3 to obtain i⁻²⁵ = −i.
  3. 3.The bracket is therefore −1 − i. Square it: (−1 − i)² = 1 + 2i + i² = 2i.
  4. 4.Multiply by the remaining factor: 2i(−1 − i) = −2i − 2i² = 2 − 2i.
  5. 5.Separating exponent reduction from the final cube kept the calculation short while showing every sign change.

For negative real square roots, convert the selected radical before multiplying. The symbol √(−t) selects i√t for positive t, whereas solving a squared equation supplies both signs. Remember that √(−1)√(−1) = −1 but √[(-1)(-1)] = 1, so the radical product rule cannot be applied to two negative real factors.

Radical value and equation rootsLaTeX
The ± sign belongs to the equation’s solution set, not to the selected radical value.
Example — Comparing components after conjugating

Problem
Find real x and y if (x − iy)(3 + 5i) is the conjugate of −6 − 24i.

  1. 1.The conjugate on the right is −6 + 24i. First perform this sign change before comparing components.
  2. 2.Expand the left side: 3x + 5xi − 3yi − 5yi² = (3x + 5y) + i(5x − 3y).
  3. 3.The real equations are 3x + 5y = −6 and 5x − 3y = 24.
  4. 4.Multiply the first by 3 and the second by 5: 9x + 15y = −18 and 25x − 15y = 120. Add to get 34x = 102, hence x = 3.
  5. 5.Substitute in 3x + 5y = −6: 9 + 5y = −6, so y = −3. Checking 5x − 3y gives 15 + 9 = 24 as required.
The imaginary component must be read after simplification

An unsimplified term involving i² may look imaginary even though it contributes to the real part. Expand and reduce powers first. Also apply a conjugate to its whole argument before equating components.

Choose the right meaning of each bar

A conjugate and a modulus use different symbols and give different kinds of output. Conjugation produces another complex number; modulus produces a nonnegative real number. Their relationship zz̄ = |z|² often provides a bridge from a complex expression to a real calculation.

For z = a + ib, the conjugate preserves a and reverses b. The negative of z reverses both. The reciprocal is the conjugate divided by a² + b², provided that this sum is positive. These operations may agree in special cases, but their definitions remain different.

Conjugate, modulus and reciprocalLaTeX
The nonzero condition is a² + b² > 0, so either a or b may be zero individually.
ReImO-5-4-3-2-112345-4-3-2-11234z = 3 + 2iz̄ = 3 − 2i−z = −3 − 2iAll three have modulus √13, but they are different points.
Three operations and their geometric consequences— Compare the coordinates of z, its conjugate and its negative. Equal modulus records a shared distance, not equality of the numbers.
Example — A modulus after simplifying two fractions

Problem
Find the modulus of (1 + i)/(1 − i) − (1 − i)/(1 + i).

  1. 1.For the first fraction, multiply by (1 + i)/(1 + i): its denominator is 2 and its numerator (1 + i)² = 2i, so its value is i.
  2. 2.For the second, multiply by (1 − i)/(1 − i): the denominator is 2 and numerator (1 − i)² = −2i, so its value is −i.
  3. 3.The difference is i − (−i) = 2i.
  4. 4.Its modulus is √(0² + 2²) = 2. The point is (0, 2), two units from the origin.

The modulus product and quotient properties often avoid expanding the final expression. Conjugation preserves modulus, so a nonzero number divided by its conjugate has modulus 1. Powers can be treated as repeated products: for a nonnegative integer n, |zⁿ| = |z|ⁿ.

Properties for mixed proofsLaTeX
Example — A squared numerator in a modulus proof

Problem
If a + ib = (x + i)²/(2x² + 1), with real x, prove a² + b² = (x² + 1)²/(2x² + 1)².

  1. 1.For real x, 2x² + 1 is positive. Hence the displayed quotient is defined and its denominator’s modulus is 2x² + 1.
  2. 2.Apply the quotient and product rules: |a + ib| = |x + i|²/(2x² + 1).
  3. 3.Since |x + i| = √(x² + 1), its square is x² + 1. Thus |a + ib| = (x² + 1)/(2x² + 1).
  4. 4.Square both sides. The left side is a² + b²; the right side is (x² + 1)²/(2x² + 1)².
  5. 5.A component check is also possible: a = (x² − 1)/(2x² + 1) and b = 2x/(2x² + 1). Their squared numerators add to (x² − 1)² + 4x² = (x² + 1)².
Example — A modulus statement about a square root

Problem
Suppose real x and y satisfy (x − iy)² = (a − ib)/(c − id), where a, b, c, d are real and c, d are not both zero. Prove (x² + y²)² = (a² + b²)/(c² + d²).

  1. 1.Write w = x − iy. This form of the given condition also covers either square root w of the quotient, without selecting a sign.
  2. 2.Take moduli: |w²| = |a − ib|/|c − id|. The denominator modulus is positive by the given condition.
  3. 3.Use |w²| = |w|² = x² + y². The right side is √(a² + b²)/√(c² + d²).
  4. 4.Square both sides to obtain (x² + y²)² = (a² + b²)/(c² + d²).
  5. 5.The argument concerns the magnitude of any square root. It does not need an additional formula for square roots of arbitrary complex numbers.

Products and cubes reveal component identities

Some mixed problems ask for a statement about real and imaginary parts rather than the entire complex number. Expand just far enough to identify those parts. The earlier product and cube identities supply everything needed, so the proof should explain how the requested components arise.

Example — Real part of a product

Problem
For any complex z₁ and z₂, prove Re(z₁z₂) = Re z₁ Re z₂ − Im z₁ Im z₂.

  1. 1.Write z₁ = a + ib and z₂ = c + id with real components. Then Re z₁ = a, Im z₁ = b, Re z₂ = c and Im z₂ = d.
  2. 2.The product formula gives z₁z₂ = (ac − bd) + i(ad + bc).
  3. 3.Its real part is ac − bd. Replace a, b, c and d by the corresponding part names to obtain the required equality.
  4. 4.The minus sign comes from (ib)(id) = −bd. This directly connects the component identity to multiplication, rather than treating it as a new unrelated rule.
Example — A component identity from a cube

Problem
If (x + iy)³ = u + iv for real x, y, u and v, with x ≠ 0 and y ≠ 0, show u/x + v/y = 4(x² − y²).

  1. 1.Use the cube-of-a-sum identity: x³ + 3x²iy + 3x(iy)² + (iy)³.
  2. 2.The squared term is −3xy² and the cubed imaginary term is −iy³. Thus (x + iy)³ = (x³ − 3xy²) + i(3x²y − y³).
  3. 3.Compare parts: u = x³ − 3xy² and v = 3x²y − y³.
  4. 4.Divide using the nonzero assumptions: u/x = x² − 3y² and v/y = 3x² − y².
  5. 5.Adding yields 4x² − 4y² = 4(x² − y²). If x or y were zero, the original requested quotient would not be defined.
Example — A product of four complex numbers

Problem
If (a + ib)(c + id)(e + if)(g + ih) = A + iB, prove (a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B².

  1. 1.Take the modulus of each side. Repeatedly apply the product-modulus property to the four factors.
  2. 2.The left side becomes √(a² + b²)√(c² + d²)√(e² + f²)√(g² + h²). The right side is √(A² + B²).
  3. 3.All these quantities are nonnegative real numbers. Square the equality to obtain the stated product of squared moduli.
  4. 4.No division was used, so the conclusion also holds when one or more factors is zero. A lengthy expansion of the four-factor product was unnecessary.
Retain every denominator condition in a proof

A true polynomial identity can be used at any complex input, but a quotient has a smaller domain. Check nonzero denominators before dividing or cancelling. In particular, u/x + v/y requires x and y to be nonzero even though the preceding cube expansion does not.

Recognise unit modulus and repeating powers

An expression involving conjugates and a number of modulus one often contains a hidden equal-distance relationship. A power equation may instead reduce to the four-value cycle of i. Recognising the pattern helps you choose a short proof, but each simplification still needs a reason.

Example — A quotient involving a point on the unit circle

Problem
If α and β are different complex numbers and |β| = 1, find |(β − α)/(1 − conjugate(α)β)|.

  1. 1.Because |β| = 1, β·conjugate(β) = 1. Expand β·conjugate(β − α) to obtain β[conjugate(β) − conjugate(α)] = 1 − conjugate(α)β.
  2. 2.Take moduli: |1 − conjugate(α)β| = |β|·|conjugate(β − α)| = |β − α|.
  3. 3.The assumption α ≠ β makes β − α nonzero, so its modulus is positive. The equal denominator modulus is therefore positive too.
  4. 4.Apply the quotient-modulus rule: the required value is |β − α|/|β − α| = 1.
  5. 5.The proof used conjugation and unit modulus to identify equal lengths; it did not require expanding the unknown coordinates of α and β.
Example — An integer equation using a modulus

Problem
Find the number of nonzero integer solutions of |1 − i|ˣ = 2ˣ.

  1. 1.The modulus is |1 − i| = √(1² + (−1)²) = √2. Thus the equation is (√2)ˣ = 2ˣ.
  2. 2.For x > 0, raising the smaller positive base √2 to the same positive integer power gives a smaller value, so equality is impossible.
  3. 3.For x < 0, the powers are reciprocals of the positive powers. The inequality reverses, so equality is again impossible.
  4. 4.At x = 0, both sides equal 1. It is the only integer solution and is excluded by the word nonzero.
  5. 5.Therefore the number of nonzero integer solutions is 0. The exponent applies to the modulus, not just to i inside its argument.
Example — The least positive power returning to one

Problem
Find the least positive integer m for which [(1 + i)/(1 − i)]ᵐ = 1.

  1. 1.Rationalise the fraction: (1 + i)/(1 − i) = (1 + i)²/[(1 − i)(1 + i)] = 2i/2 = i.
  2. 2.The condition becomes iᵐ = 1. The first positive powers are i, −1, −i and 1.
  3. 3.The first return to 1 occurs at m = 4. More generally, all positive multiples of 4 satisfy the equation.
  4. 4.The word least requires identifying the first occurrence, rather than giving any exponent that works.

How the larger number system developed

The difficulty of negative square roots was recognised long before a consistent complex-number system was established. The chapter’s historical account connects that difficulty to concrete algebraic questions. It also shows how symbolic calculation and a two-component definition gradually supported one another.

Earlier writers, including Mahavira in Ganitasara Sangraha and Bhaskara in Bijaganita, described the difficulty of taking square roots of negative quantities within real arithmetic. In the sixteenth century Cardan considered two quantities whose sum is 10 and whose product is 40. His calculation led to expressions involving √(−15), which can now be interpreted directly using i.

Example — A historical question in modern notation

Problem
Find x and y if x + y = 10 and xy = 40.

  1. 1.From the sum, y = 10 − x. Substitute into the product to get x(10 − x) = 40.
  2. 2.Expand and rearrange: 10x − x² = 40, so x² − 10x + 40 = 0.
  3. 3.Complete the square by separating the constant: x² − 10x + 25 = −15, hence (x − 5)² = −15.
  4. 4.Thus x − 5 = ±i√15. The pair of values is x = 5 + i√15 and y = 5 − i√15, or the same two numbers in the reverse order.
  5. 5.Their sum is 10 and their product is 25 − (i√15)² = 25 + 15 = 40. The conjugate pair gives a real sum and a real product.
From negative roots to ordered pairs

The historical account traces the difficulty of negative roots through ancient Greek mathematics and Indian mathematical writings. It describes Albert Girard connecting their acceptance with polynomial root counts, Euler introducing the notation i, and W. R. Hamilton treating a complex number as an ordered pair (a, b). The pair viewpoint agrees with the Argand representation: two real components specify one complex number. These developments helped turn difficult-looking expressions into a coherent number system.

The historical calculation returns us to the chapter’s starting idea: an equation can lack real solutions and still have meaningful solutions in a larger number system. The arithmetic is checked through familiar relationships such as sum, product and conjugation. This chapter establishes those relationships rather than requiring a separate rule for each new-looking expression.

Check understanding across the chapter

The following questions move between arithmetic, notation, domains and geometry. Before choosing an answer, identify which definition or property applies. A plausible-looking result should still be checked for its sign, components or denominator.

Quiz

Quick check

Which expression is the inverse of 2i?

Quick check

What is √(−16) under the selected radical convention?

Quick check

What is the real part of (1 + 2i)(3 − i)?

Quick check

For any integer k, what is i^(4k + 2)?

Quick check

What is the modulus of (3 + 4i)/(1 − i)?

Quick check

A number is represented by (−2, −3). Which point represents its conjugate?

Quick check

When using the identity u/x + v/y = 4(x² − y²) derived from (x + iy)³ = u + iv, which extra condition is needed?

Quick check

If |β| = 1 and α ≠ β, what is |(β − α)/(1 − conjugate(α)β)|?

Quick check

Which method most directly proves a product of squared-modulus factors equals A² + B² when their complex product is A + iB?

Practice Problems

Practice Problems
  1. Evaluate [i¹⁸ + (1/i)²⁵]³ in standard form, separating the power reductions from the cube.
  2. For any complex z₁ and z₂, prove Re(z₁z₂) = Re z₁ Re z₂ − Im z₁ Im z₂ using real components.
  3. Reduce [1/(1 − 4i) − 2/(1 + i)]·[(3 − 4i)/(5 + i)] to standard form. Show the rationalisation of each denominator.
  4. If x − iy is a square root of (a − ib)/(c − id), with all components real and c, d not both zero, prove (x² + y²)² = (a² + b²)/(c² + d²).
  5. If z₁ = 2 − i and z₂ = 1 + i, find |(z₁ + z₂ + 1)/(z₁ − z₂ + 1)|. Check the denominator first.
  6. If a + ib = (x + i)²/(2x² + 1) with real x, prove a² + b² = (x² + 1)²/(2x² + 1)².
  7. Let z₁ = 2 − i and z₂ = −2 + i. Find Re[(z₁z₂)/(conjugate of z₁)] and Im[1/(z₁·conjugate of z₁)].
  8. Find real x and y if (x − iy)(3 + 5i) is the conjugate of −6 − 24i. Verify both component equations.
  9. Find the modulus of (1 + i)/(1 − i) − (1 − i)/(1 + i), and represent the simplified number in the Argand plane.
  10. If (x + iy)³ = u + iv with real x, y, u, v and xy ≠ 0, prove u/x + v/y = 4(x² − y²). Explain why the nonzero assumptions are needed.
  11. If α ≠ β and |β| = 1, find |(β − α)/(1 − conjugate(α)β)| and prove that the denominator cannot be zero.
  12. Find the number of nonzero integer solutions of |1 − i|ˣ = 2ˣ. Consider positive, negative and zero x separately.
  13. If (a + ib)(c + id)(e + if)(g + ih) = A + iB, prove (a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B².
  14. Find the least positive integer m for which [(1 + i)/(1 − i)]ᵐ = 1. Give all other positive integer solutions too.
  15. Solve u² + 9 = 0 in the complex numbers and distinguish the solution set from the selected value √(−9).
  16. For z = −4 + 3i, find Re z, Im z, −z, z̄, |z| and 1/z. Check the inverse by multiplication.
  17. Plot 3 + 2i, −3 + 2i, −3 − 2i and 3 − 2i. Identify conjugate pairs and compare their moduli.
  18. Derive (z₁ − z₂)² and z₁² − z₂² from distribution. Apply one of these identities to simplify a complex expression of your choice.
  19. Find all real x and y satisfying 2x + i(3x − y) = −4 + 7i.
  20. Explain, using a numerical example, why modulus cannot be distributed over addition as though it were multiplication.
  21. Show that √(−2)√(−8) differs from √[(-2)(-8)] under the selected radical convention.
  22. Use modern complex notation to solve x + y = 10 and xy = 40, and explain why both the sum and product of the resulting conjugate pair are real.

Key Takeaways

Key Takeaways

• Complex numbers extend real numbers through i² = −1 while retaining familiar addition and multiplication laws. • Standard form makes both components visible for equality, arithmetic and interpretation. • The power cycle and careful treatment of negative radicals control many sign-sensitive calculations. • Conjugation, modulus and inverse are distinct but connected by zz̄ = |z|². • The Argand plane turns components into coordinates, modulus into distance and conjugation into reflection. • Products and powers can often be handled efficiently by modulus properties or polynomial identities. • A complete solution explains each transformation and preserves every condition needed for division.

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Complex Numbers on the Argand Plane

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