Complex Numbers | ISC Class 11 Maths Notes

This note covers complex numbers, equality, powers of the imaginary unit, algebraic operations, identities, conjugates, modulus, inverses, the Argand plane, arguments, polar representation, square roots of complex numbers and cube roots of unity.

What is a complex number and why is it needed?

The square of a real number is non-negative. Consequently, the equation x² + 1 = 0, where x is an unknown number, has no real solution: it would require x² = −1. Complex numbers extend the real number system so that this equation has solutions.

Definition: The imaginary unit i satisfies i² = −1. A complex number has the form z = a + ib, where z names the number and a and b are real numbers. The expressions ib and bi mean the same product.

How are its two parts identified?

The real part is a, written Re z. The imaginary part is b, written Im z. Although its name includes “imaginary”, the imaginary part is a real number. The term containing i is ib; it is not itself the imaginary part.

For z = 2 + 5i, Re z = 2 and Im z = 5. Writing an expression in standard form means collecting its real terms and its terms containing i to obtain a + ib.

A real number a can be written as a + i0. Thus real numbers are included among complex numbers. A number of the form ib with real b ≠ 0 is called purely imaginary. The symbol ≠ means “not equal to”.

What does a negative square root mean?

The notation √ denotes a square root, with √a the non-negative square root when a is non-negative. For a positive real number a, √(−a) = i√a. Both i√a and −i√a square to −a.

In particular, i and −i solve x² = −1, while the symbol √(−1) denotes i. Distinguish a specified radical from the full set of solutions of an equation. Finding both square roots requires retaining both opposite values.

Note: The rule √a × √b = √(ab) fails when both a and b are negative real numbers. For instance, √(−1) × √(−1) = i² = −1, whereas √1 = 1.

How do equality and powers of i simplify calculations?

Let z₁ = a + ib and z₂ = c + id, where the subscripts 1 and 2 distinguish two complex numbers and a, b, c and d are real. Equality requires a = c and b = d. It therefore gives two real equations.

Property: Equality of corresponding parts

To compare expressions, first collect each into standard form. Then compare real parts with real parts and imaginary parts with imaginary parts. This method applies when the coefficients being compared are real, so that each expression is genuinely in the required form.

Worked example 1. Find the real numbers x and y if 4x + i(3x − y) = 3 − 6i.

Answer: Equating real parts gives 4x = 3, hence x = 3/4. Equating imaginary parts gives 3x − y = −6. Substitution gives 9/4 − y = −6, so y = 33/4.

How does the four-power cycle work?

Starting from i² = −1 gives i³ = i²i = −i and i⁴ = (i²)² = 1. Multiplying by i again begins the same cycle. The following table summarises the powers, with k denoting any integer, including negative integers.

A negative exponent indicates a reciprocal: i⁻ⁿ = 1/iⁿ, where n is a positive integer. Since i is non-zero, these reciprocals exist. Reduce the positive power first, then simplify the reciprocal using i² = −1.

Worked example 2. Express i⁻³⁵ in standard form.

Answer: Since 35 = 4 × 8 + 3, i³⁵ = −i. Therefore i⁻³⁵ = 1/(−i). Multiplying numerator and denominator by i gives i/(−i²) = i, or 0 + i.

The power cycle and equality rule solve different tasks. Powers remove higher occurrences of i; equality then compares the real coefficients remaining. Performing these steps in that order prevents a power such as i² from being mistaken for an imaginary term.

How are complex numbers added and subtracted?

For z₁ = a + ib and z₂ = c + id, addition combines corresponding real coefficients: z₁ + z₂ = (a + c) + i(b + d). The result is again a complex number. This is the closure property of addition.

Property: Addition laws and the additive inverse

Addition is commutative, meaning z₁ + z₂ = z₂ + z₁. It is also associative, meaning (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃), where z₃ denotes a third complex number. These properties concern order and grouping respectively.

The additive identity is 0 = 0 + i0 because z + 0 = z. The additive inverse of z = a + ib is −z = −a − ib because z + (−z) = 0. Both coefficients change sign.

Worked example 3. Add 2 + 3i and −6 + 5i.

Answer: (2 + 3i) + (−6 + 5i) = (2 − 6) + i(3 + 5) = −4 + 8i. The real coefficients contribute −4 and the imaginary coefficients contribute 8.

How does subtraction use the inverse?

Subtraction means adding the additive inverse: z₁ − z₂ = z₁ + (−z₂). Hence z₁ − z₂ = (a − c) + i(b − d). When removing brackets after a minus sign, change every sign inside those brackets.

Worked example 4. Subtract 2 − i from 6 + 3i.

Answer: (6 + 3i) − (2 − i) = (6 + 3i) + (−2 + i) = 4 + 4i. The negative of −i is +i, so the imaginary coefficient is 3 + 1.

Reversing this subtraction gives (2 − i) − (6 + 3i) = −4 − 4i. Thus the commutative law for addition does not transfer to subtraction. Keep the order specified in the question throughout the calculation.

How are products and algebraic identities used?

Complex multiplication follows ordinary expansion together with i² = −1. For z₁ = a + ib and z₂ = c + id, expansion gives ac + iad + ibc + i²bd. Replacing the last term by −bd yields z₁z₂ = (ac − bd) + i(ad + bc).

Property: Multiplication and distribution

Multiplication is closed, commutative and associative. Its multiplicative identity is 1 = 1 + i0, since z × 1 = z. The distributive law is z₁(z₂ + z₃) = z₁z₂ + z₁z₃; multiplication distributes over addition.

There is also right distribution: (z₁ + z₂)z₃ = z₁z₃ + z₂z₃. These laws allow brackets to be expanded and like terms collected. The essential additional step, compared with real algebra, is replacing powers of i by their cycle values.

Worked example 5. Multiply 3 + 5i by 2 + 6i.

Answer: (3 + 5i)(2 + 6i) = 6 + 18i + 10i + 30i². Using i² = −1 gives 6 − 30 + 28i = −24 + 28i.

Identity: Square and cube expansions

An identity is an equality valid for every allowed value of its variables. The square identity (z₁ + z₂)² = z₁² + 2z₁z₂ + z₂² follows by writing the square as a product and applying distribution twice.

Similarly, (z₁ − z₂)² = z₁² − 2z₁z₂ + z₂², and (z₁ − z₂)(z₁ + z₂) = z₁² − z₂². These identities remain valid for complex numbers. Many other identities valid for real numbers can also be proved for complex numbers.

Worked example 6. Express (5 − 3i)³ in standard form.

Answer: Using the cube expansion gives 5³ − 3 × 5² × 3i + 3 × 5 × (3i)² − (3i)³. This becomes 125 − 225i − 135 + 27i = −10 − 198i.

In the final example, (3i)² = −9 and (3i)³ = −27i. The minus sign before the cubed term changes −27i to +27i. Keeping that outer sign visible is as important as reducing the power correctly.

What do the conjugate and modulus tell us?

Definition: For z = a + ib, the conjugate is z̄ = a − ib, where the bar denotes conjugation. The modulus is |z| = √(a² + b²), where the vertical bars denote a non-negative real magnitude.

Conjugation preserves the real part and reverses the imaginary part. For example, the conjugate of 3 + i is 3 − i, while the conjugate of 2 − 5i is 2 + 5i. The conjugate of −5 − 3i is −5 + 3i.

Result: Multiplication by the conjugate

The product z z̄ = (a + ib)(a − ib) = a² − (ib)² = a² + b² = |z|². Thus multiplying a number by its conjugate produces a non-negative real number. This product is central to division and inverses.

Conjugating twice returns the original number. Also, |z̄| = |z| because squaring −b gives the same result as squaring b. These observations distinguish changing a point's imaginary coordinate from changing its distance from the origin.

Worked example 7. Find the modulus and conjugate of 2 − 5i.

Answer: The real part is 2 and the imaginary part is −5. Hence |2 − 5i| = √(2² + (−5)²) = √29. Changing the sign of the imaginary part gives the conjugate 2 + 5i.

How do these operations behave with products?

For complex z₁ and z₂, |z₁z₂| = |z₁||z₂|. If z₂ ≠ 0, then |z₁/z₂| = |z₁|/|z₂|. The condition on the denominator is necessary because division by zero is undefined.

The conjugate of a sum, difference or product is the corresponding sum, difference or product of the conjugates. For z₂ ≠ 0, the conjugate of z₁/z₂ is z̄₁/z̄₂. Conjugation can therefore be performed before or after these operations.

The modulus is a magnitude, so it is not obtained by simply discarding i. Include both real coefficients in the sum of squares. For instance, |3 + i| = √(3² + 1²) = √10, with an imaginary coefficient of 1.

How do we find inverses and divide complex numbers?

The multiplicative inverse of a non-zero complex number z is the number 1/z, also written z⁻¹, whose product with z is 1. Unlike an additive inverse, it is defined by multiplication rather than addition.

Result: The reciprocal formula

Since z z̄ = |z|², the reciprocal is 1/z = z̄/|z|². For z = a + ib, this becomes (a − ib)/(a² + b²). The requirement z ≠ 0 means that a and b are not both zero.

One coefficient may be zero without preventing an inverse. What matters is that a² + b² is non-zero. The number zero has no multiplicative inverse because multiplying zero by any complex number still gives zero, rather than 1.

Worked example 8. Find the multiplicative inverse of 2 − 3i.

Answer: Its conjugate is 2 + 3i and its squared modulus is 2² + (−3)² = 13. Therefore its inverse is (2 + 3i)/13 = 2/13 + (3/13)i. Multiplying by 2 − 3i gives 13/13 = 1.

How is a quotient put into standard form?

The quotient z₁/z₂, for z₂ ≠ 0, means z₁ multiplied by the inverse of z₂. Multiply numerator and denominator by the denominator's conjugate. This changes the denominator into a real number without changing the value of the fraction.

1. Identify the real and imaginary parts of the denominator.
2. Form its conjugate by reversing the imaginary part's sign.
3. Multiply both numerator and denominator by this conjugate.
4. Expand, replace i² by −1, and separate the real and imaginary terms.

Worked example 9. Express (6 + 3i)/(2 − i) in standard form.

Answer: Multiply by (2 + i)/(2 + i). The numerator becomes (6 + 3i)(2 + i) = 12 + 12i + 3i² = 9 + 12i. The denominator is (2 − i)(2 + i) = 5. The quotient is 9/5 + (12/5)i.

The real denominator divides both parts of the numerator. Retain brackets until the last step so that the division applies to the entire expression. A check is to multiply the quotient by the original denominator and recover the original numerator.

How does the Argand plane represent complex numbers?

The Argand plane, also called the complex plane, represents z = x + iy by the point P(x, y), where x and y are real coordinates and P names the point. The horizontal coordinate is the real part; the vertical coordinate is the imaginary part.

How are the axes interpreted?

The perpendicular coordinate axes meet at the origin, O(0, 0). The horizontal x-axis is the real axis and the vertical y-axis is the imaginary axis. A point on the real axis has y = 0; a point on the imaginary axis has x = 0.

The horizontal and vertical axes meet at O. The plotted points are A(2, 4), B(−2, 3), C(0, 1), D(2, 0), E(−5, −2) and F(1, −2), representing their corresponding complex numbers.

See Fig. 4.1 in your NCERT textbook

An ordered pair records the horizontal coordinate first and the vertical coordinate second. Thus 2 + 4i corresponds to (2, 4), while −5 − 2i corresponds to (−5, −2). The order matters because the two coordinates perform different roles.

What do modulus and conjugation look like?

A line segment joins O(0, 0) to P(x, y). Its length is labelled r = √(x² + y²), where r denotes the distance from the origin and therefore the modulus.

See Fig. 4.2 in your NCERT textbook

The modulus is the length OP, the line segment from O to P. This geometric meaning explains why |z| is non-negative. The zero complex number is represented at O itself and has modulus zero.

P(x, y) lies above the real axis and Q(x, −y) below it. Segments join both points to O. Q represents the conjugate and is the mirror image of P in the real axis.

See Fig. 4.3 in your NCERT textbook

Reflection preserves the horizontal coordinate and reverses the vertical coordinate. It also preserves distance from the origin, matching |z̄| = |z|. The algebraic rules for conjugation therefore have a direct interpretation in the geometry of the complex plane.

How are arguments and polar form determined?

For a non-zero complex number z = x + iy, let r = |z|. An argument θ, read “theta”, is an angle from the positive real axis to the ray from the origin through z. Anticlockwise angles are positive and clockwise angles are negative.

What is the principal argument?

Angles here are measured in radians, with one complete turn equal to 2π radians; π is the circle constant, the ratio of circumference to diameter. Arguments differing by a whole number of turns describe the same ray.

If θ is one argument, all arguments are θ + 2kπ for integer k. The principal argument, written Arg z, is the unique argument in the interval −π < Arg z ≤ π. The symbols < and ≤ mean “less than” and “less than or equal to”.

The zero complex number has no defined argument because it supplies no ray direction from the origin. A quadrant is one of the four regions between the coordinate axes. Identify the quadrant from the signs of x and y before selecting an angle.

How does polar form use trigonometry?

The trigonometric functions cos θ and sin θ, called cosine and sine, give x/r and y/r respectively. Thus x = r cos θ and y = r sin θ. Substituting these relations gives the polar form z = r(cos θ + i sin θ).

For x ≠ 0, tan θ = y/x, where tangent, tan θ, means sin θ/cos θ. A tangent value alone does not determine the correct direction because opposite rays have the same ratio. Check the signs of both sine and cosine.

To convert back to standard form, evaluate r cos θ and r sin θ separately. They are the real and imaginary parts. On an axis, determine the angle directly from the direction of the ray instead of dividing by a zero coordinate.

Property: Arguments of products and quotients

Let z₁ and z₂ be non-zero, with moduli r₁ and r₂ and arguments θ₁ and θ₂. Their product has modulus r₁r₂ and argument θ₁ + θ₂. Their quotient has modulus r₁/r₂ and argument θ₁ − θ₂.

These angle rules follow by expansion and the sine and cosine addition formulae. If a principal argument is required, adjust the resulting angle by whole turns into the chosen interval. Conjugation reverses an argument's sign, again allowing adjustment by whole turns.

How are square roots of a complex number found?

A square root of z is a complex number w such that w² = z. Here w names an unknown root. If w is a root, −w is also a root because (−w)² = w². A non-zero complex number has two distinct square roots.

How does comparing parts find the roots?

Write z = a + ib and w = u + iv, where a, b, u and v are real. Squaring gives w² = (u² − v²) + 2uvi. Comparing parts gives u² − v² = a and 2uv = b.

Let r = √(a² + b²), the modulus of z. Squaring and adding the two equations gives (u² + v²)² = a² + b². Since u² + v² is non-negative, u² + v² = r.

Adding and subtracting u² + v² = r and u² − v² = a gives u² = (r + a)/2 and v² = (r − a)/2. These determine the magnitudes of the two coefficients; 2uv = b determines their sign pairing.

1. Calculate r = √(a² + b²) from the given number.
2. Find the non-negative quantities (r + a)/2 and (r − a)/2.
3. Take their real square roots to obtain the magnitudes of u and v.
4. Choose signs satisfying 2uv = b, then include the opposite root and verify by squaring.

Why must the signs be linked?

If b > 0, u and v have the same sign. If b < 0, they have opposite signs. The symbol > means “greater than”. Choosing signs independently can give a square with the wrong imaginary part.

When b ≠ 0, one convenient choice is u = √((r + a)/2) and v = b/(2u). The second root is −u − iv. Here u is non-zero, so dividing by 2u is valid.

If b = 0 and a > 0, the roots are √a and −√a. If b = 0 and a < 0, they are i√(−a) and −i√(−a). If a = b = 0, the root is zero.

The sign check completes the method. Recovering u² and v² alone is insufficient: substituting into 2uv = b ensures that the proposed roots square to the given number rather than to its conjugate.

What are the cube roots of unity and their properties?

Unity means the number 1. A cube root of unity is a complex number z satisfying z³ = 1. Factorisation gives z³ − 1 = (z − 1)(z² + z + 1), so one root is 1 and the others solve z² + z + 1 = 0.

How are the three roots obtained?

Completing the square in the remaining equation gives (2z + 1)² = −3. Consequently 2z + 1 = ±i√3, where ± means taking both the plus and minus alternatives. The two remaining roots are (−1 + i√3)/2 and (−1 − i√3)/2.

Define ω, read “omega”, as (−1 + i√3)/2. Squaring this value gives ω² = (−1 − i√3)/2. The three cube roots of unity are therefore 1, ω and ω². The superscript in ω² denotes the square of ω.

Result: The fundamental cube-root relations

Since ω solves z² + z + 1 = 0, 1 + ω + ω² = 0. Also, ω³ = 1, with ω ≠ 1. These statements must be used together: ω³ = 1 alone does not distinguish ω from the real root.

The non-real roots are conjugates. Their sum is ω + ω² = −1 and their product is ωω² = ω³ = 1. Therefore each is the multiplicative inverse of the other: 1/ω = ω² and 1/ω² = ω.

All three roots have modulus 1. For either non-real root, the squared modulus is (−1/2)² + (√3/2)² = 1. Their points therefore lie on the unit circle, the circle centred at the origin with radius 1.

How are higher powers simplified?

For any integer k, ω³ᵏ = 1, ω³ᵏ⁺¹ = ω and ω³ᵏ⁺² = ω². Reduce a power using this three-term cycle, then use ω + ω² = −1 where appropriate. This cycle differs from the four-term cycle for i.

In polar form the three points have arguments 0, 2π/3 and −2π/3. The two non-real roots are reflections in the real axis. Their equal moduli, conjugate relationship and reciprocal relationship express consistent algebraic and geometric properties of the same numbers.

Glossary

• Complex number — A number expressible as a + ib, where a and b are real and i² = −1.
• Imaginary unit — The number denoted by i, defined by the relation i² = −1.
• Real part — The real coefficient a in the standard form z = a + ib.
• Imaginary part — The real coefficient b multiplying i in the standard form z = a + ib.
• Conjugate — The number obtained by preserving the real part and reversing the imaginary part's sign.
• Modulus — The non-negative magnitude √(a² + b²), equal to the point's distance from the origin.
• Additive inverse — The number which, when added to the original complex number, gives zero.
• Multiplicative inverse — The reciprocal of a non-zero number, whose product with that number is one.
• Argand plane — A coordinate plane representing complex numbers by real and imaginary coordinates.
• Argument — An angle from the positive real axis to the ray representing a non-zero complex number.
• Principal argument — The unique argument chosen in the interval greater than −π and at most π.
• Polar form — The representation r(cos θ + i sin θ), using modulus r and an argument θ.
• Square root — A complex number whose square equals the given complex number.
• Cube root of unity — Any complex number whose cube equals one, namely 1, ω or ω².

Common errors and misconceptions

• Misconception: The imaginary part of a + ib is ib. Correct: It is the real coefficient b. Separate the coefficient from the term containing the imaginary unit.
• Misconception: The additive inverse and conjugate are the same. Correct: The inverse −a − ib reverses both coefficients; the conjugate a − ib reverses the imaginary coefficient.
• Misconception: i² can be treated as +1 during expansion. Correct: Replace i² by −1, including when it appears inside a squared denominator or a product.
• Misconception: √(−1) × √(−1) equals √1. Correct: The left side equals i² = −1; the radical multiplication rule fails when both real radicands are negative.
• Misconception: A complex reciprocal requires both parts to be non-zero. Correct: The parts must not both be zero. The condition is z ≠ 0, equivalently a² + b² ≠ 0.
• Misconception: The tangent ratio uniquely specifies an argument. Correct: Opposite directions share that ratio. Check both coordinate signs and select the required principal argument.
• Misconception: The signs of u and v in a square root can be selected independently. Correct: Their product must satisfy 2uv = b, preserving the given imaginary part.
• Misconception: Every solution of z³ = 1 satisfies 1 + z + z² = 0. Correct: That relation holds for the two non-real roots, ω and ω², and fails for z = 1.

Exam-style questions with model answers

Q1. For z = 2 + 5i, where i² = −1, state its real and imaginary parts and its conjugate. [2 marks]

1. The real part is Re z = 2 and the imaginary part is Im z = 5.
2. The conjugate is z̄ = 2 − 5i, obtained by reversing the sign of the imaginary part.

Q2. Given 4x + i(3x − y) = 3 − 6i, where x and y are real and i² = −1, find x and y. [3 marks]

1. Equality of the real parts gives 4x = 3, so dividing by 4 gives x = 3/4.
2. Equality of the imaginary parts gives 3x − y = −6. Substitute the value of x to obtain 9/4 − y = −6.
3. Rearranging yields y = 6 + 9/4 = 33/4. Thus the required real numbers are x = 3/4 and y = 33/4.

Q3. Find the multiplicative inverse of 2 − 3i in standard form, where i² = −1. [3 marks]

1. The conjugate of the given non-zero number 2 − 3i is 2 + 3i, obtained by reversing the imaginary coefficient's sign.
2. Its squared modulus is 2² + (−3)² = 13. Equivalently, multiplying the given number by its conjugate gives 13.
3. The inverse is the conjugate divided by the squared modulus: (2 + 3i)/13 = 2/13 + (3/13)i.

Q4. Express (6 + 3i)/(2 − i) in the form a + ib, where a and b are real and i² = −1. [4 marks]

1. The denominator's conjugate is 2 + i. Multiply the numerator and denominator by this same non-zero number to preserve the quotient.
2. The denominator becomes (2 − i)(2 + i) = 4 − i² = 5, a real number.
3. The numerator becomes (6 + 3i)(2 + i) = 12 + 12i + 3i² = 9 + 12i.
4. Divide both terms by 5 to obtain 9/5 + (12/5)i. Therefore a = 9/5 and b = 12/5.

Q5. Let z = a + ib with real a and b, b ≠ 0 and i² = −1. Derive a method for finding both square roots of z, including the condition on their signs. [5 marks]

1. Write a square root as w = u + iv, where u and v are real. Expansion gives w² = (u² − v²) + 2uvi.
2. Compare this with a + ib to obtain u² − v² = a and 2uv = b. These equations must both hold.
3. Squaring and adding gives (u² + v²)² = a² + b². Define r = √(a² + b²); then u² + v² = r because the sum is non-negative.
4. Adding and subtracting the equations for u² + v² and u² − v² gives u² = (r + a)/2 and v² = (r − a)/2.
5. Choose u = √((r + a)/2) and v = b/(2u). Since b ≠ 0, u ≠ 0. The two roots are u + iv and −u − iv; their linked signs ensure 2uv = b.

Q6. Find all complex solutions of z³ = 1, where i² = −1. Define ω as the root with positive imaginary part and show that 1 + ω + ω² = 0 and ω³ = 1. [5 marks]

1. Rearrange to z³ − 1 = 0 and factorise as (z − 1)(z² + z + 1) = 0. This gives the root z = 1.
2. The remaining roots satisfy z² + z + 1 = 0. Completing the square gives (2z + 1)² = −3, hence 2z + 1 = ±i√3.
3. Therefore the other roots are (−1 + i√3)/2 and (−1 − i√3)/2. Define ω = (−1 + i√3)/2, the root with positive imaginary part.
4. Squaring ω gives ω² = (−1 − i√3)/2. Adding these expressions gives ω + ω² = −1, so 1 + ω + ω² = 0.
5. Multiplying the expressions for ω and ω² gives (1 + 3)/4 = 1. Thus ω³ = 1, and the three solutions are 1, ω and ω².

Q7. Let x, y, a and b be real, with a and b not both zero, and suppose x + iy = (a + ib)/(a − ib), where i² = −1. Prove that x² + y² = 1. [5 marks]

1. The denominator a − ib is non-zero because a and b are not both zero. We can therefore multiply the numerator and denominator by its conjugate, a + ib.
2. This gives x + iy = (a + ib)²/(a² + b²). Expanding the numerator gives a² − b² + 2abi, with a² + b² positive.
3. Comparing real and imaginary parts yields x = (a² − b²)/(a² + b²) and y = 2ab/(a² + b²).
4. Squaring and adding these real expressions gives x² + y² = ((a² − b²)² + 4a²b²)/(a² + b²)².
5. The numerator expands to a⁴ + 2a²b² + b⁴ = (a² + b²)². Cancelling the equal non-zero numerator and denominator gives x² + y² = 1.

Key takeaways

• A complex number has real coefficients a and b in a + ib; its imaginary part is b, and i² = −1.
• Equal complex numbers have equal real parts and equal imaginary parts, providing two real equations when their coefficients are real.
• Add corresponding parts, expand products carefully, and reduce powers of i using the cycle 1, i, −1, −i.
• The conjugate reverses the imaginary part; multiplying a number by its conjugate gives the square of its modulus.
• A non-zero number's reciprocal is its conjugate divided by its squared modulus; division uses this reciprocal of the denominator.
• The Argand point has coordinates equal to the real and imaginary parts, and its distance from the origin equals the modulus.
• Polar form uses modulus and argument; choose the correct quadrant and adjust angles into the principal interval when required.
• Square-root coefficients must satisfy both equations from comparison of parts, including the sign condition on their product.
• The cube roots of unity are 1, ω and ω², with ω³ = 1 and 1 + ω + ω² = 0.

Test yourself

What are the real and imaginary parts of z = 2 − 5i?

The real part is 2 and the imaginary part is −5. The imaginary part is the coefficient of i, not the whole term −5i.

Why does subtracting 2 − i require adding −2 + i?

Subtraction means addition of the additive inverse. Reversing both coefficients of 2 − i gives −2 + i, whose sum with the original number is zero.

What are i³ and i⁴, given i² = −1?

Multiplying i² by i gives i³ = −i. Squaring i² gives i⁴ = (−1)² = 1, completing the repeating cycle.

For z = a + ib, with real a and b, why is z z̄ real?

The imaginary cross terms cancel: (a + ib)(a − ib) = a² + b². This is the real, non-negative quantity |z|².

Where is the conjugate of the Argand point P(x, y), with real coordinates x and y?

It is at (x, −y), the reflection of P in the real axis. The horizontal coordinate stays fixed and the vertical coordinate changes sign.

Why does the zero complex number have no argument?

Zero is located at the origin, so it does not determine a ray from the origin. There is consequently no defined direction angle.

If (u + iv)² = a + ib for real u, v, a and b, what links the signs of u and v?

Comparing imaginary parts gives 2uv = b. A positive b requires matching signs, while a negative b requires opposite signs.

For ω = (−1 + i√3)/2, why is its reciprocal ω²?

The product ω × ω² equals ω³ = 1. By the definition of multiplicative inverse, ω² is therefore the reciprocal of ω.