Why do we need complex numbers?
Square any real number: 3² = 9, (−3)² = 9, 0² = 0. The answer is never negative. So x² = −1, or x² + 1 = 0, has no real solution.
Mathematicians solved this by adding one new number, called iota and written i, with i² = −1. So √(−1) = i, and √(−9) = √9 · i = 3i.
A complex number is any number z = a + ib where a and b are real. We call a the real part, Re(z), and b the imaginary part, Im(z). Real numbers are complex numbers with b = 0, so nothing old is lost. See step 1 of the 3D: i is a quarter turn of 1.
Powers of i
i¹ = i, i² = −1, i³ = i²·i = −i, i⁴ = (i²)² = 1. After that the pattern repeats. To find iⁿ, divide n by 4 and use the remainder: i⁴ᵏ = 1, i⁴ᵏ⁺¹ = i, i⁴ᵏ⁺² = −1, i⁴ᵏ⁺³ = −i. Example: i¹⁰² → 102 = 4 × 25 + 2, so i¹⁰² = −1.
Careful: √a · √b = √(ab) is true only when a and b are not both negative. For example √(−4)·√(−9) = 2i·3i = 6i² = −6, not √36 = 6.
Algebra of complex numbers
- Equality: a + ib = c + id only when a = c and b = d.
- Addition: (a + ib) + (c + id) = (a + c) + i(b + d). It is closed, commutative and associative; 0 is the identity and −z is the additive inverse.
- Subtraction: (a + ib) − (c + id) = (a − c) + i(b − d).
- Multiplication: multiply like brackets and put i² = −1: (a + ib)(c + id) = (ac − bd) + i(ad + bc). It is also commutative, associative and distributive over addition; 1 is the identity.
- Multiplicative inverse: for z = a + ib ≠ 0, z⁻¹ = (a − ib)/(a² + b²).
- Division: z₁/z₂ = z₁ · z₂⁻¹. In practice, multiply top and bottom by the conjugate of the bottom.
Identities like (z₁ + z₂)² = z₁² + 2z₁z₂ + z₂² still work for complex numbers.
Conjugate and modulus
The conjugate of z = a + ib is z̄ = a − ib (change the sign of the imaginary part). The modulus is |z| = √(a² + b²). Useful facts: z · z̄ = |z|², |z₁z₂| = |z₁||z₂|, |z₁/z₂| = |z₁|/|z₂|, conj(z₁ + z₂) = z̄₁ + z̄₂, conj(z₁z₂) = z̄₁ z̄₂.
There is no ‘bigger’ or ‘smaller’ between two non-real complex numbers: 2 + i > 1 + 3i has no meaning. Only their moduli can be compared.
The Argand plane
Draw two perpendicular axes. The horizontal one is the real axis and the vertical one is the imaginary axis. The number z = a + ib is the point (a, b). This picture is called the Argand plane (or complex plane).
- Real numbers (like 4) sit on the real axis; pure imaginary numbers (like 3i) sit on the imaginary axis.
- |z| is the distance of the point from the origin O, by Pythagoras (step 5 of the 3D).
- z̄ is the mirror image of z in the real axis.
- z₁ + z₂ is the fourth corner of the parallelogram made by O, z₁ and z₂ (step 3).
- Multiplying by i turns the point 90° anticlockwise about O (step 4).
Quadratic equations with complex roots
For ax² + bx + c = 0 (a, b, c real), the roots are x = (−b ± √D)/2a with D = b² − 4ac.
- D > 0: two different real roots. D = 0: two equal real roots.
- D < 0: √D = i√(−D), so the roots are x = (−b ± i√(−D))/2a. They are complex and always a conjugate pair.
Example: x² + 2x + 5 = 0. D = 4 − 20 = −16, √D = 4i, x = (−2 ± 4i)/2 = −1 ± 2i. Try it in free play.
Board exam focus
Algebra carries about 25 marks in Class 11. From this chapter expect: powers of i (1 mark), writing a quotient in a + ib form (2–3 marks), modulus, conjugate and inverse (2 marks), and solving quadratics with D < 0 (2–3 marks). Always give the final answer in the form a + ib.
Key formulas and definitions
- i = √(−1), i² = −1, i³ = −i, i⁴ = 1
- z = a + ib; Re(z) = a, Im(z) = b
- (a + ib)(c + id) = (ac − bd) + i(ad + bc)
- z̄ = a − ib; |z| = √(a² + b²); z·z̄ = |z|²
- z⁻¹ = z̄ / |z|² = (a − ib)/(a² + b²)
- ax² + bx + c = 0, D < 0: x = (−b ± i√(−D)) / 2a
Worked examples
1. Find i⁵⁷ and i⁻³.
57 = 4 × 14 + 1, so i⁵⁷ = i¹ = i. i⁻³ = 1/i³ = 1/(−i) = −1/i. Multiply top and bottom by i: −i/i² = −i/(−1) = i. So i⁻³ = i.
2. Simplify (5 − 3i) + (−2 + 7i) and (5 − 3i) − (−2 + 7i).
Sum: real 5 + (−2) = 3, imaginary −3 + 7 = 4, so 3 + 4i. Difference: real 5 − (−2) = 7, imaginary −3 − 7 = −10, so 7 − 10i.
3. Multiply (2 + 3i)(4 − i).
2·4 + 2·(−i) + 3i·4 + 3i·(−i) = 8 − 2i + 12i − 3i². Put i² = −1: 8 + 3 + 10i = 11 + 10i.
4. Find the modulus, conjugate and multiplicative inverse of z = 3 − 4i.
|z| = √(9 + 16) = 5. z̄ = 3 + 4i. z⁻¹ = z̄/|z|² = (3 + 4i)/25 = 3/25 + (4/25)i. Check: (3 − 4i)(3 + 4i)/25 = 25/25 = 1.
5. Write (1 + 2i)/(3 − i) in the form a + ib.
Multiply top and bottom by 3 + i. Bottom: 9 + 1 = 10. Top: (1 + 2i)(3 + i) = 3 + i + 6i + 2i² = 1 + 7i. Answer: 1/10 + (7/10)i.
6. Solve 2x² − 2x + 1 = 0.
a = 2, b = −2, c = 1. D = 4 − 8 = −4 < 0, so √D = 2i. x = (2 ± 2i)/4 = 1/2 ± (1/2)i. The roots 1/2 + i/2 and 1/2 − i/2 are conjugates.
7. Find real x and y if (x + 2y) + i(2x − y) = 5 + 5i.
Equal complex numbers have equal parts: x + 2y = 5 and 2x − y = 5. From the second, y = 2x − 5. Then x + 4x − 10 = 5, so x = 3 and y = 1.
8. If (a + ib)/(c + id) = x + iy, show that (a² + b²)/(c² + d²) = x² + y².
Take the modulus of both sides: |a + ib|/|c + id| = |x + iy|. So √(a² + b²)/√(c² + d²) = √(x² + y²). Squaring both sides gives the result.
Common mistakes
- Writing i² = 1. It is i² = −1; that is the whole reason i exists.
- Using √(−4)·√(−9) = √36 = 6. When both numbers are negative, first write 2i · 3i = 6i² = −6.
- Dividing by a complex number directly. Always multiply top and bottom by the conjugate of the bottom.
- Saying 3 + 2i > 1 + i. Non-real complex numbers cannot be ordered; compare moduli instead.