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Modulus and argument
Edexcel IAL Further Mathematics · WFM01 Further Pure Mathematics 1
Plotting z = a + bi as the point (a, b) gives two new measurements: its distance from the origin (the modulus) and its angle from the positive real axis (the argument).
Modulus
The modulus of z = a + bi is |z| = sqrt(a2 + b2), the distance from O to the point (a, b). It is never negative.
|3 + 4i| = 5 and |z*| = |z| for every z. Note zz* = |z|2.
Argument
The argument arg z is the angle between the positive real axis and the line from O to z, measured anticlockwise as positive. In F1 it is given in radians.
The principal argument lies in the range -π < arg z ≤ π. Points below the real axis have a negative argument.
Use a sketch
tan-1(b/a) only gives angles between -π/2 and π/2, so it is right only in the first and fourth quadrants.
- Second quadrant (a < 0, b > 0): arg z = π - tan-1|b/a|.
- Third quadrant (a < 0, b < 0): arg z = -(π - tan-1|b/a|).
On the axes: arg(positive real) = 0, arg(negative real) = π, arg(ki) = π/2 and arg(-ki) = -π/2 for k > 0.
| Quadrant | Example | arg z |
|---|---|---|
| 1st | 1 + i | π/4 |
| 2nd | -1 + i | 3π/4 |
| 3rd | -1 - i | -3π/4 |
| 4th | 1 - i | -π/4 |
Worked example
Find the modulus and the argument of z = -2 + 2sqrt(3)i.
- |z| = sqrt(4 + 12) = 4.
- z is in the second quadrant. The angle with the negative real axis is tan-1(2sqrt(3)/2) = π/3.
- arg z = π - π/3.
Answer: |z| = 4, arg z = 2π/3
Check yourself
- Find |5 - 12i|.
Show answer
13. |5 - 12i| = sqrt(25 + 144) = sqrt(169) = 13.
- Find arg(1 + sqrt(3)i).
- π/6
- π/3
- 2π/3
- -π/3
Show answer
π/3. First quadrant: tan-1(sqrt(3)/1) = π/3.
- Find arg(-3 - 4i), in radians to 3 decimal places.
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-2.214. Third quadrant. The angle with the negative real axis is tan-1(4/3) = 0.927. So arg z = -(π - 0.927) = -2.214.
- Find arg(4 - i), in radians to 3 decimal places.
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-0.245. Fourth quadrant, so arg z = -tan-1(1/4) = -0.245.
- Find arg(-2 + i), in radians to 3 decimal places.
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2.678. Second quadrant: arg z = π - tan-1(1/2) = π - 0.464 = 2.678.
- What is arg(-5)?
- π
- 0
- -π/2
- -π
Show answer
π. -5 lies on the negative real axis. The principal argument is π (the range is -π < θ ≤ π, so -π is excluded).
- z has modulus 6 and argument π/5. Which describes z*?
- Modulus -6, argument π/5
- Modulus 6, argument π - π/5
- Modulus 6, argument -π/5
- Modulus 1/6, argument -π/5
Show answer
Modulus 6, argument -π/5. z* is the reflection of z in the real axis: the same distance from O, with the angle reflected.
Sources
- Pearson Edexcel International Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (issue 3), Unit FP1
- Pearson Edexcel IAL Further Pure Mathematics 1 scheme of work
Last updated 8 November 2026. Spotted a mistake? Tell us.