Conjugate of a complex number
A mirror image across the real axis: top becomes bottom, but the distance from the axis stays the same.
The conjugate of z = a + ib is z̄ = a − ib (change the sign of the imaginary part). It is the reflection of z in the real axis.
Formula: conjugate of (a + ib) = a − ib
In the figure: z = 3 + 2i (purple) and its conjugate 3 − 2i (green) are mirror images in the real axis.
Rule 1: Find conjugates — Only the sign of the imaginary part changes.
- Easy: Conjugate of 5 + 7i is 5 − 7i.
- Medium: Conjugate of −2 − 3i is −2 + 3i.
- Exam-type: Conjugate of a real number x is x itself.
Rule 2: Properties — The conjugate respects sums, products and quotients.
- Easy: z + z̄ = 2 Re z, a real number.
- Medium: z − z̄ = 2i Im z, purely imaginary.
- Exam-type: z z̄ = a² + b².
z̄ = z exactly when z is real.
Changing the sign of the real part as well.
Conjugate: flip the Imaginary sign.
Practice questions
- 1Find the conjugate of 4 − 9i.
- 2Find the conjugate of 7i.
- 3Find z + z̄ for z = 6 + 2i.
- 4Find z z̄ for z = 3 + 4i.
- 5Find z − z̄ for z = 5 − 3i.
- 6If z = 2 + 5i, find (z̄)̄.
- 7Find the conjugate of (1 + i)².
- 8Find the conjugate of (2 + i)/(1 − i).
Show answers (try them yourself first)
- 14 + 9i
- 2−7i
- 312
- 425
- 5−6i
- 62 + 5i
- 7−2i
- 80.5 − 1.5i
Modulus of a complex number
On a map, the distance of your house from the origin (the city centre) is the length of the straight line to it.
The modulus of z = a + ib is |z| = √(a² + b²), the distance of the point (a, b) from the origin. Also |z|² = z z̄.
Formula: |z| = √(a² + b²)
In the figure: The point 3 + 4i is 5 units from the origin: √(9 + 16) = 5.
Rule 1: Find |z| — Square, add, then take the root.
- Easy: |3 + 4i| = 5.
- Medium: |5 − 12i| = 13.
- Exam-type: |1 + i| = √2 ≈ 1.4142.
Rule 2: Properties — Modulus respects products and quotients.
- Easy: |z₁z₂| = |z₁||z₂|.
- Medium: |z₁/z₂| = |z₁|/|z₂|.
- Exam-type: |(3 + 4i)(5 − 12i)| = 5 × 13 = 65.
|z| ≥ 0, and |z| = 0 only for z = 0.
Writing |a + ib| = a + b.
Pythagoras for complex numbers.
Practice questions
- 1Find |6 + 8i|.
- 2Find |−5|.
- 3Find |2 − 2i|.
- 4Find |8 + 15i|.
- 5Find |(1 + i)/(1 − i)|.
- 6If |z| = 5 and z = x + 3i with x > 0, find x.
- 7Find |1 + i|⁴.
- 8Find all real x such that |x + 2i| = 2√2.
Show answers (try them yourself first)
- 110
- 25
- 32√2 ≈ 2.828
- 417
- 51
- 6x² + 9 = 25, x = 4
- 74
- 8x² + 4 = 8, x = ±2
Argand plane
A graph paper on which the x-axis stands for real parts and the y-axis for imaginary parts.
Every complex number a + ib is shown as the point (a, b) in the Argand plane. The x-axis is the real axis and the y-axis the imaginary axis. Adding two complex numbers is the same as adding their position vectors.
In the figure: z = −2 + 3i lies in quadrant II; its conjugate −2 − 3i lies in quadrant III.
Rule 1: Plot — Real part is x, imaginary part is y.
- Easy: 3 + 4i is in quadrant I.
- Medium: −2 + 3i is in quadrant II.
- Exam-type: −1 − i is in quadrant III.
Rule 2: Special points — Real numbers sit on the x-axis, pure imaginary on the y-axis.
- Easy: 5 is on the real axis.
- Medium: −4i is on the imaginary axis.
- Exam-type: All z with |z| = 3 form a circle of radius 3.
The distance between z₁ and z₂ in the plane is |z₁ − z₂|.
Mixing up the axes.
Real → right, Imaginary → up.
Practice questions
- 1In which quadrant is 4 − 2i?
- 2Where is −6 plotted?
- 3Find the distance between 1 + i and 4 + 5i.
- 4Which points have |z| = 1?
- 5Find the point midway between 2 + 3i and 6 − i.
- 6Name the figure with vertices 0, 3, 3 + 3i and 3i.
- 7What is the area of the triangle with vertices 0, 4 and 4i?
- 8The points 0, 1 + i and 2 are joined. What is the angle at 1 + i?
Show answers (try them yourself first)
- 1Quadrant IV
- 2On the real axis, left of the origin.
- 3|3 + 4i| = 5
- 4The unit circle.
- 54 + i
- 6Square (side 3).
- 78 square units
- 890° (the triangle is right-angled and isosceles).
Polar form
To give the location of a ship you may say "5 km north-east" instead of "3 km east and 4 km north". That is polar form.
If |z| = r and the angle with the positive real axis is θ (argument), then z = r(cos θ + i sin θ), with r = √(a² + b²), cos θ = a/r, sin θ = b/r. The principal argument lies in (−π, π].
Formula: z = r(cos θ + i sin θ)
Rule 1: Find r and θ — Find r first, then the quadrant for θ.
- Easy: z = 1 + i: r = √2, θ = π/4.
- Medium: z = −1 + i: r = √2, θ = 3π/4.
- Exam-type: z = √3 − i: r = 2, θ = −π/6.
Rule 2: Back to a + ib — Evaluate cos θ and sin θ.
- Easy: 2(cos π/3 + i sin π/3) = 1 + 1.732051i.
- Medium: 4(cos π/2 + i sin π/2) = 4i.
- Exam-type: 3(cos π + i sin π) = −3.
To find θ, never use tan⁻¹(b/a) blindly; check the quadrant of (a, b).
Giving θ = tan⁻¹(b/a) for a point in quadrant II or III.
Find r by Pythagoras, θ by the quadrant.
Practice questions
- 1Write 2i in polar form.
- 2Write −3 in polar form.
- 3Find the principal argument of 1 − i.
- 4Write 1 + √3 i in polar form.
- 5Write −1 − i in polar form.
- 6Convert 6(cos π/6 + i sin π/6) to a + ib.
- 7Find |z| and arg z for z = −√3 + i.
- 8If z = r(cos θ + i sin θ), show that |z̄| = r.
Show answers (try them yourself first)
- 12(cos π/2 + i sin π/2)
- 23(cos π + i sin π)
- 3−π/4
- 42(cos π/3 + i sin π/3)
- 5√2(cos 5π/4 + i sin 5π/4), or θ = −3π/4 as principal argument
- 65.196152 + 3i (3√3 + 3i)
- 7|z| = 2, arg z = 5π/6
- 8z̄ = r(cos θ − i sin θ), whose modulus is √(r²cos²θ + r²sin²θ) = r