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Lesson 1 of 3

Complex Numbers and Quadratic Equations · Lesson 1 of 3

Complex Numbers and Their Algebra

“Build a larger number system and learn how its arithmetic, powers, conjugates and modulus fit together.”

Learning Objectives

• Explain why negative squares require an extension of the real number system and use i² = −1. • Identify real and imaginary parts and solve equalities by comparing them. • Perform complex-number arithmetic and explain the algebraic laws and inverse conditions. • Reduce positive and negative powers of i and handle square roots of negative real numbers correctly. • Use familiar polynomial identities with complex numbers while showing intermediate steps. • Calculate modulus and conjugate and use their properties to simplify inverses and quotients.

When real numbers are not enough

Every real number has a nonnegative square. Therefore an equation such as x² + 1 = 0 cannot be solved by choosing a real value of x. We extend the number system so that algebra can describe such solutions without abandoning the arithmetic relationships we already know.

Rearranging the equation gives x² = −1. Introduce a new number i whose square is −1. This is its defining relationship, not an approximation. Whenever two factors of i appear together in a product, their product can be replaced by −1.

Definition
Imaginary unit

The number i is defined by i² = −1. Its negative −i also has square −1, because (−i)² = (−1)²i² = −1.

The defining relationshipLaTeX
The symbol i denotes a number in the extended system. It does not denote an unknown real number.
Example — Solving a simple equation with a negative square

Problem
Find all solutions of x² + 1 = 0 in the extended number system.

  1. 1.Rearrange to x² = −1. By definition i² = −1, so x = i is a solution.
  2. 2.Also (−i)² = −1, so x = −i is a solution.
  3. 3.Factor x² + 1 as (x − i)(x + i), because (x − i)(x + i) = x² − i² = x² + 1.
  4. 4.The two solutions are i and −i. The equation has no real solution, but it does have these two complex solutions.

This need also arises for a real-coefficient quadratic ax² + bx + c = 0, with a ≠ 0, when its discriminant D = b² − 4ac is negative. The discriminant is the expression under the square root in the familiar quadratic formula. A negative D produces a negative square root, which cannot be evaluated within the real numbers. Developing complex numbers gives that expression a meaning.

No real solution is not the same as no solution

The number system matters. The equation x² = −1 has no solution in the real numbers, while it has solutions in the complex numbers. The word imaginary is a mathematical name; it does not make calculations involving i arbitrary.

Two real components make one complex number

The extended system consists of numbers formed by combining a real component with a multiple of i. Keeping these two components visible gives us a standard form for calculation and comparison. It also lets the old real number system sit naturally inside the new one.

Definition
Complex number

A complex number is a number z = a + ib, where a and b are real numbers. The expressions a + ib and a + bi mean the same thing.

Real and imaginary partsLaTeX
Re means real part and Im means imaginary part. Both a and b are real numbers; the imaginary term is ib.

For z = 2 + 5i, the real part is 2 and the imaginary part is 5. For z = −1 + i√3 they are −1 and √3. A fraction can be a component too: 4 − i/11 has real part 4 and imaginary part −1/11. The letter z is just a convenient name for the complete number.

A real number a can be written a + 0i. A number such as 7i can be written 0 + 7i and is called purely imaginary when its imaginary part is nonzero. Zero itself is 0 + 0i. These cases show why a complex number need not have two nonzero components.

NumberReal partImaginary partStandard form
2 + 5i252 + 5i
−1 + i√3−1√3−1 + i√3
4 − i/114−1/114 − i/11
−3i0−30 − 3i
6606 + 0i
0000 + 0i
The imaginary part of 2 + 5i is 5

The component multiplying i is the imaginary part. Writing Im(2 + 5i) = 5i confuses the imaginary part with the imaginary term. Use Im z = b when z = a + ib.

Two complex numbers are equal exactly when their real parts match and their imaginary parts match. One equality therefore supplies two real equations. This comparison is valid after both sides have been simplified into standard form and the components are known to be real.

Equality of complex numbersLaTeX
a, b, c and d are real. The symbol ⇔ says that each statement implies the other.
Example — Equating both components

Problem
If 4x + i(3x − y) = 3 − 6i, where x and y are real, find x and y.

  1. 1.Compare real parts: 4x = 3, so x = 3/4.
  2. 2.Compare imaginary parts: 3x − y = −6. Substitute x = 3/4 to get 9/4 − y = −6.
  3. 3.Add y and then add 6: y = 9/4 + 6 = 9/4 + 24/4 = 33/4.
  4. 4.Check the imaginary part: 3(3/4) − 33/4 = −24/4 = −6. Thus x = 3/4 and y = 33/4 satisfy the original equality.

Addition and subtraction keep the components separate

When adding complex numbers, combine real terms with real terms and multiples of i with multiples of i. This follows the same collection of like terms used in algebra. Subtraction means adding the negative of the entire second number, so both of its components change sign.

Addition and subtractionLaTeX

For instance, (2 + 3i) + (−6 + 5i) has real component 2 − 6 and imaginary component 3 + 5, giving −4 + 8i. Because the sums of real components remain real, the answer again has the form a + ib. This is the closure property for addition: the operation stays within the complex numbers.

Addition propertyMeaningRelationship
ClosureAdding two complex numbers gives another complex numberz₁ + z₂ is complex
CommutativityThe order of the two addends can be exchangedz₁ + z₂ = z₂ + z₁
AssociativityRegrouping three addends preserves the result(z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
Additive identityAdding zero changes neither componentz + 0 = z
Additive inverseThe negative cancels both componentsz + (−z) = 0

Here z₁, z₂ and z₃ name any complex numbers. The zero complex number is 0 + 0i. If z = a + ib, its additive inverse is −z = −a − ib. Associativity concerns where brackets are placed; commutativity concerns the order of terms. They are different properties, even though both make a long sum easier to organise.

Example — Subtraction in both orders

Problem
Find (6 + 3i) − (2 − i), and then reverse the order.

  1. 1.For the first subtraction, distribute the negative: 6 + 3i − 2 + i.
  2. 2.Collect components to obtain 4 + 4i.
  3. 3.In the reverse order, (2 − i) − (6 + 3i) = 2 − i − 6 − 3i = −4 − 4i.
  4. 4.The results are negatives of one another. Subtraction is not commutative, although addition is.
Subtract both components

In (a + ib) − (c + id), the minus sign applies to c and to id. An error often occurs when the second imaginary term is already negative: subtracting −i means adding i.

Multiplication uses distribution and i² = −1

To multiply, first distribute the factors just as for two binomials. The only new simplification is that the product of the two imaginary terms contains i². Replacing that square by −1 produces a real contribution, which must be combined with the original real product.

Expand (a + ib)(c + id) into ac + iad + ibc + i²bd. The last term becomes −bd. Collecting the remaining imaginary terms gives the product rule below. This derivation explains the minus sign in its real part.

Product of two complex numbersLaTeX
ac − bd is the real part; ad + bc is the imaginary part. All four components are real.
Example — Multiplying two numbers with both components

Problem
Express (3 + 5i)(2 + 6i) in standard form.

  1. 1.Distribute: 3×2 + 3×6i + 5i×2 + 5i×6i = 6 + 18i + 10i + 30i².
  2. 2.Use i² = −1 to obtain 6 + 28i − 30.
  3. 3.Collect the real terms: the result is −24 + 28i.
  4. 4.The same result follows from ac − bd = 6 − 30 and ad + bc = 18 + 10. The negative real part comes from the imaginary-term product.
Multiplication propertyMeaningRelationship
ClosureProducts again have two real componentsz₁z₂ is complex
CommutativityExchanging two factors preserves the productz₁z₂ = z₂z₁
AssociativityRegrouping factors preserves the product(z₁z₂)z₃ = z₁(z₂z₃)
Multiplicative identityMultiplication by 1 + 0i changes nothingz·1 = z
Multiplicative inverseEvery nonzero number has a reciprocalz·z⁻¹ = 1, for z ≠ 0
DistributivityMultiplication distributes across additionz₁(z₂ + z₃) = z₁z₂ + z₁z₃; (z₁ + z₂)z₃ = z₁z₃ + z₂z₃

The components ac − bd and ad + bc are real, explaining closure. Commutativity and associativity follow from the corresponding real-number laws when the components are expanded. Distributivity is the rule that justified our expansion in the first place. The multiplicative identity is 1, while the additive identity is 0; their names describe the operations in which they leave a number unchanged.

Reciprocals and division

A multiplicative inverse of z is a number which multiplies z to give 1. It exists whenever z is nonzero. To discover its formula, look for a factor that cancels the imaginary cross terms in a product.

For z = a + ib, multiply it by a − ib. Distribution gives a² − (ib)² = a² + b², a real number. If z ≠ 0, at least one of a and b is nonzero, so a² + b² > 0. Dividing the factor a − ib by that positive number therefore gives a reciprocal.

Multiplicative inverseLaTeX
The condition means a and b are not both zero. Either component may individually be zero.

Division by a nonzero z₂ is defined as multiplication by its reciprocal: z₁/z₂ = z₁(1/z₂). Zero has no reciprocal, because multiplying zero by any number gives zero, never one. The factor a − ib used above will later receive the name conjugate.

Example — Division by an inverse

Problem
Find (6 + 3i)/(2 − i).

  1. 1.For 2 − i, the components are a = 2 and b = −1. Its reciprocal is [2 − i(−1)]/[2² + (−1)²] = (2 + i)/5.
  2. 2.Multiply: (6 + 3i)(2 + i)/5 = [12 + 6i + 6i + 3i²]/5.
  3. 3.Replace i² by −1 to get (9 + 12i)/5 = 9/5 + (12/5)i.
  4. 4.The denominator was nonzero, so the reciprocal was legitimate. As a check, multiplying this answer by 2 − i returns 6 + 3i.
Nonzero does not require both components to be nonzero

For z = a + ib, the inverse requires a² + b² > 0. Thus 4 + 0i and 0 − i both have inverses. Only 0 + 0i is excluded. Requiring a ≠ 0 and b ≠ 0 together would wrongly exclude many valid numbers.

Powers of i form a repeating cycle

Repeated multiplication by i gives a short pattern. Rather than calculate a large power by multiplying hundreds of factors, observe the first four powers and use the return to 1. Negative powers can be handled too because i has a reciprocal.

Starting with i⁰ = 1, we obtain i¹ = i, i² = −1, i³ = i²i = −i and i⁴ = (i²)² = 1. Multiplying by another i starts the same sequence again: i⁵ = i and i⁶ = −1. A full group of four factors contributes 1, so only the remainder after dividing the exponent by four matters.

Remainder when n is divided by 40123
Value of iⁿ1i−1−i
The four-power patternLaTeX
ℤ means the integers, including negative integers and zero. These formulas cover every integer exponent.
Multiplying by i moves to the next value1i−1−i×i×i×iThe next multiplication returns −i to 1.The reverse cycle explains negative powers.
The four values of integer powers of i— Follow each multiplication by i. After four steps the starting value returns, so a whole group of four contributes 1.

For negative powers, i⁻¹ = 1/i = −i because i(−i) = −i² = 1. Then i⁻² = −1, i⁻³ = i and i⁻⁴ = 1. You can either use these reciprocals or write the negative exponent as 4k + r with remainder r in {0, 1, 2, 3}. Both methods use the same pattern.

Example — A large negative exponent

Problem
Express i⁻³⁵ in standard form.

  1. 1.Write −35 = 4(−9) + 1. The remainder is 1, even though the quotient is negative.
  2. 2.Thus i⁻³⁵ = (i⁴)⁻⁹i = 1·i = i.
  3. 3.To check by reciprocals, 35 = 4(8) + 3 gives i³⁵ = −i. Therefore i⁻³⁵ = 1/(−i) = i.
  4. 4.The standard form is 0 + 1i. Using the nonnegative remainder avoids a mistaken sign for the negative exponent.
Example — Products of imaginary factors

Problem
Express (−5i)(i/8) and (−i)(2i)(−i/8)³ in standard form.

  1. 1.For the first product, combine the real coefficients and powers: (−5/8)i² = (−5/8)(−1) = 5/8. The imaginary part is zero.
  2. 2.For the second, (−i/8)³ = (−1/8)³i³ = −i³/512.
  3. 3.The real coefficients from the full product multiply to (−1)·2·(−1/512) = 1/256. The powers of i multiply to i·i·i³ = i⁵.
  4. 4.Since i⁵ = i, the second result is i/256. This separates coefficient signs from the power cycle before combining them.

Negative square roots need careful notation

Both i and −i square to −1. More generally, multiplying either sign of √t by i gives a square root of −t when t is positive. Distinguishing the square-root symbol from all solutions of a squared equation prevents a common loss of one solution.

For t > 0, (i√t)² = i²t = −t and (−i√t)² = −t. The notation √(−t) in this chapter selects i√t. But the equation u² = −t has two roots, +i√t and −i√t. This is the same distinction between √9 = 3 and the two solutions ±3 of u² = 9.

Negative real square rootsLaTeX
The radical selects one value; ± lists the two solutions of the equation. The number t is a positive real number.
Example — A radical and a squared equation

Problem
Find √(−3), then solve u² = −3.

  1. 1.The selected radical value is √(−3) = i√3. Squaring gives i²·3 = −3.
  2. 2.The negative −i√3 also has square −3. Hence the equation has solutions u = i√3 and u = −i√3.
  3. 3.Checking both signs verifies the two roots. It would be incorrect to write √(−3) = ±i√3 as though the radical symbol itself denoted two values.

The familiar rule √a·√b = √(ab) works for nonnegative real a and b. Under the selected negative-root convention it also works when exactly one of the two real factors is negative and the other is positive. It fails when both are negative: each selected root contributes a factor i, and their product contributes i² = −1.

Example — Why a familiar square-root rule cannot be used everywhere

Problem
Compare √(−1)√(−1) with √[(−1)(−1)], and then evaluate √(−4)√(−9).

  1. 1.The first product is i·i = −1. The single radical is √1 = 1. These are different, disproving a universal square-root product rule for negative factors.
  2. 2.For the second product, convert the radicals separately: √(−4) = 2i and √(−9) = 3i.
  3. 3.Their product is (2i)(3i) = 6i² = −6. Multiplying the radicands first would wrongly give √36 = 6.
  4. 4.If either real radicand is zero, the product is zero on both sides. The failure described here concerns two strictly negative real radicands.
Convert negative radicals before multiplying

Write each negative real square root as i times a positive real square root. Then use i² = −1. Applying √a√b = √(ab) to two negative radicands loses the negative sign.

Familiar identities still work in complex algebra

Polynomial identities come from addition, multiplication and distribution. Since complex numbers obey those same laws, identities such as the square of a sum remain valid. The calculations may contain powers of i, but their algebraic structure is unchanged.

Example — Deriving the square-of-a-sum identity

Problem
Show that (z₁ + z₂)² = z₁² + 2z₁z₂ + z₂² for complex z₁ and z₂.

  1. 1.Write the square as (z₁ + z₂)(z₁ + z₂).
  2. 2.Distribute the first factor over the second to get z₁(z₁ + z₂) + z₂(z₁ + z₂).
  3. 3.Distribute again: z₁² + z₁z₂ + z₂z₁ + z₂².
  4. 4.Complex multiplication is commutative, so z₂z₁ = z₁z₂. The two middle terms combine into 2z₁z₂.
  5. 5.The result is z₁² + 2z₁z₂ + z₂². The proof depended on the algebraic laws, so it works for complex inputs as well as real ones.
Squares and difference of squaresLaTeX

For a cube, multiply the square identity by one more copy of the same binomial and collect terms. The coefficients 1, 3, 3, 1 arise from repeated distribution. Replacing z₂ by −z₂ changes the signs of the terms containing odd powers of z₂.

Cubes of a sum and differenceLaTeX
Example — A complex cube

Problem
Express (5 − 3i)³ in standard form.

  1. 1.Use the difference-cube identity with z₁ = 5 and z₂ = 3i: 125 − 3(25)(3i) + 3(5)(3i)² − (3i)³.
  2. 2.The second term is −225i. The third is 15·9i² = −135.
  3. 3.For the last term, (3i)³ = 27i³ = −27i, so subtracting it contributes +27i.
  4. 4.Collect real terms 125 − 135 = −10 and imaginary terms −225i + 27i = −198i.
  5. 5.Therefore (5 − 3i)³ = −10 − 198i. Retaining the minus sign before the cube term avoids a common sign error.
Example — A product containing a negative radical

Problem
Express (−√3 + √(−2))(2√3 − i) in standard form.

  1. 1.Replace √(−2) by i√2 before expanding. The product becomes (−√3 + i√2)(2√3 − i).
  2. 2.Distribute to obtain −6 + i√3 + 2i√6 − √2i².
  3. 3.Use i² = −1: the last term is +√2. The real part is −6 + √2.
  4. 4.The imaginary terms combine into i(√3 + 2√6) = i√3(1 + 2√2).
  5. 5.The result is (−6 + √2) + i√3(1 + 2√2). Positive real radicals can be combined here using ordinary radical rules.

Modulus and conjugate make useful relationships visible

A complex number has two real components. Its modulus combines their sizes into one nonnegative real number, while its conjugate reverses the imaginary component. These two constructions explain why the denominator in the reciprocal formula is real and positive.

Definition
Modulus

For z = a + ib, the modulus is |z| = √(a² + b²). It is a nonnegative real number and equals zero exactly when z = 0.

Definition
Conjugate

The conjugate of z = a + ib is the complex number a − ib, denoted z̄. Only the imaginary component changes sign.

Modulus and conjugateLaTeX
The vertical bars denote modulus. The bar above a number denotes conjugation. They are different operations.

For a real number a + 0i, the modulus is √(a²) = |a|, its ordinary absolute value. So the complex modulus extends the familiar notion of size on the real line. A conjugate does not generally equal the negative of the number: conjugation preserves the real part, whereas taking the negative changes both parts.

Example — Finding modulus and conjugate

Problem
Find the modulus and conjugate of 3 + i, 2 − 5i, and −5 − 3i.

  1. 1.For 3 + i, the modulus is √(3² + 1²) = √10 and the conjugate is 3 − i.
  2. 2.For 2 − 5i, the modulus is √[2² + (−5)²] = √29 and the conjugate is 2 + 5i.
  3. 3.For −5 − 3i, the modulus is √[25 + 9] = √34 and the conjugate is −5 + 3i.
  4. 4.In each case the modulus is real and nonnegative. The conjugate preserves the real part, including its sign.

Multiplying z by z̄ cancels the imaginary cross terms. The result a² + b² is the square of the modulus. This unites the two constructions and rewrites the reciprocal in compact notation.

Conjugate product and inverseLaTeX
The squared modulus is positive for nonzero z. The inverse formula therefore excludes only the zero complex number.
Example — An inverse using the conjugate

Problem
Find the multiplicative inverse of 2 − 3i.

  1. 1.The conjugate is 2 + 3i and the squared modulus is 2² + (−3)² = 13.
  2. 2.The inverse is therefore (2 + 3i)/13 = 2/13 + (3/13)i.
  3. 3.Alternatively multiply 1/(2 − 3i) by (2 + 3i)/(2 + 3i). Its denominator becomes 4 − (3i)² = 4 + 9 = 13.
  4. 4.Check: (2 − 3i)(2 + 3i)/13 = 13/13 = 1. This verifies that the number found is a multiplicative inverse.
Example — Rationalising a complex denominator

Problem
Express (5 + i√2)/(1 − i√2) in standard form.

  1. 1.The denominator is nonzero. Multiply numerator and denominator by its conjugate 1 + i√2. This multiplies the fraction by 1.
  2. 2.The denominator becomes (1 − i√2)(1 + i√2) = 1 − (i√2)² = 1 + 2 = 3.
  3. 3.Expand the numerator: 5 + 5i√2 + i√2 + 2i² = 3 + 6i√2.
  4. 4.Divide both components by 3 to get 1 + 2i√2.
  5. 5.Multiplying by the conjugate removed the imaginary component of the denominator, while keeping the quotient’s value unchanged.

Properties of modulus and conjugation

These constructions interact predictably with arithmetic. Knowing the relationships can simplify a proof before a long expansion is attempted. Each quotient relationship retains the requirement that its denominator be nonzero.

The modulus of a product equals the product of the moduli. For a quotient, it equals the quotient of the moduli. Conjugating a sum, difference or product can be done by conjugating its components first; the same is true for a defined quotient. These statements concern two different operations, so keep the bars and vertical lines distinct.

Modulus of products and quotientsLaTeX
Conjugates and arithmeticLaTeX
Conjugation also leaves modulus unchanged, and applying it twice returns z: |z̄| = |z| and the conjugate of z̄ is z.
Example — Why modulus multiplies

Problem
Derive |z₁z₂| = |z₁||z₂| from components.

  1. 1.Let z₁ = a + ib and z₂ = c + id. Their product has components ac − bd and ad + bc.
  2. 2.Its squared modulus is (ac − bd)² + (ad + bc)². Expand: a²c² − 2abcd + b²d² + a²d² + 2abcd + b²c².
  3. 3.The cross terms cancel. Factor the remaining terms as a²(c² + d²) + b²(c² + d²) = (a² + b²)(c² + d²).
  4. 4.Thus |z₁z₂|² = |z₁|²|z₂|². Taking the nonnegative square root gives |z₁z₂| = |z₁||z₂|.
  5. 5.If z₂ ≠ 0, write z₁ = (z₁/z₂)z₂. Apply the product result and divide by |z₂| > 0 to obtain the quotient-modulus property.

To see the product-conjugate property from the same components, z₁z₂ = (ac − bd) + i(ad + bc), so its conjugate changes the plus to a minus. Multiplying (a − ib)(c − id) gives precisely (ac − bd) − i(ad + bc). Addition and subtraction follow by changing the sign of the combined imaginary component. For a quotient q = z₁/z₂, conjugate qz₂ = z₁ and divide by the nonzero conjugate of z₂.

Modulus does not distribute over addition

The product and quotient rules do not imply |z₁ + z₂| = |z₁| + |z₂|. For z₁ = 1 and z₂ = −1, the left side is 0 and the sum of moduli is 2. Apply each rule only to the operation it describes.

Quiz

Quick check

For z = −4 + 7i, what is Im z?

Quick check

If 2x + i(x − y) = 6 − 4i with real x and y, what is y?

Quick check

What is (2 + 3i)(1 − i)?

Quick check

Which complex number has no multiplicative inverse?

Quick check

What is i⁻³⁹?

Quick check

What is √(−4)√(−9), using the selected radical values?

Quick check

Which pair gives the conjugate and modulus of −3 − 4i?

Quick check

What is the multiplicative inverse of −i?

Practice Problems

Practice Problems
  1. Express (5i)(−3i/5) in standard form and explain the sign of its real part.
  2. Reduce i⁹ + i¹⁹ to standard form using the four-power cycle.
  3. Express i⁻³⁹ in standard form, showing a valid remainder calculation or reciprocal method.
  4. Express 3(7 + 7i) + i(7 + 7i) in standard form.
  5. Simplify (1 − i) − (−1 + 6i), showing the distribution of the minus sign.
  6. Express (1/5 + 2i/5) − (4 + 5i/2) in standard form.
  7. Simplify [(1/3 + 7i/3) + (4 + i/3)] − (−4/3 + i).
  8. Express (1 − i)⁴ in standard form. Use an intermediate square to organise the work.
  9. Express (1/3 + 3i)³ in standard form using the cube identity.
  10. Express (−2 − i/3)³ in standard form, keeping the signs of all four cube terms visible.
  11. Find the multiplicative inverse of 4 − 3i and check its product with the original number.
  12. Find the multiplicative inverse of √5 + 3i.
  13. Find the multiplicative inverse of −i and explain why a zero real part causes no difficulty.
  14. Express [(3 + i√5)(3 − i√5)]/[(√3 + i√2) − (√3 − i√2)] in standard form, verifying that its denominator is nonzero.
  15. Solve 3x + i(2x + y) = −6 + 5i for real x and y.
  16. Evaluate √(−16)√(−25) and explain why multiplying the radicands first gives an incorrect result.
  17. Derive the square-of-a-difference identity from distribution, then apply it to (2 − 5i)².
  18. For z = −5 − 3i, compare −z, z̄ and |z| and state what each operation changes.

Key Takeaways

Key Takeaways

• The imaginary unit satisfies i² = −1; real numbers are included as a + 0i. • A complex number’s real and imaginary parts are real coefficients, and equality compares both. • Addition and multiplication obey familiar algebraic laws; every nonzero complex number has an inverse. • Integer powers of i repeat in groups of four, including negative powers. • A selected negative-real radical is i times a positive real square root, while a squared equation has two roots. • Conjugation reverses only the imaginary component; modulus is a nonnegative real size. • The relationship zz̄ = |z|² explains both the reciprocal formula and division using conjugates.

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Complex Numbers on the Argand Plane