MathsComplex numbers › Modulus, argument and loci

Modulus, argument and loci

Length and direction take over from across and up: every complex number is a distance from the origin at an angle, multiplication becomes rotation, and equations in z draw circles and lines.

Year FMEDEXCEL 9FM0 CP1

Builds on Complex arithmetic and the Argand diagram and Radians, arcs and small angles.

IN THIS TOPIC

  • Find the modulus and argument of a complex number, handling every quadrant.
  • Convert between x + yi and modulus-argument form, and multiply and divide in it.
  • Sketch and describe the standard loci: circles, perpendicular bisectors, half-lines.

WHAT YOU PROBABLY THINK

The argument of z is tan⁻¹(y/x), whatever the quadrant z is in.

Length and angle

The modulus |z| is z's distance from the origin, √(x2 + y2), and the argument arg z is the angle from the positive real axis, measured in radians and reported in (−π, π]. Together they pin the point as surely as x and y do, and the conversion is the trigonometry of one right triangle:

z = r(cos θ + i sin θ)

with r = |z| and θ = arg z. The opening claim is the quadrant trap. tan⁻¹(y/x) only lands in the right half plane; for a number like −1 + i the calculator's −π/4 must be corrected to 3π/4 by a sketch. Always plot the point first and read which quadrant the angle belongs to.

The number 1 plus root 3 i in modulus-argument form: length 2 at an angle of pi over 3z = 1 + √3 i|z| = 2π/3
FIG. 11 + √3 i has modulus 2 and argument π/3: one right triangle converts between the two descriptions.

Multiplying in modulus-argument form

Multiplication has a clean geometric reading: moduli multiply and arguments add. Dividing divides the moduli and subtracts the arguments. A product that looks messy in x + yi form can be almost mental arithmetic in modulus-argument form.

WORKED EXAMPLE

A product done both ways

Find (1 + √3 i)(√3 + i), using modulus-argument form, and check directly.

Both factors have modulus 2; the arguments are π/3 and π/6. So the product has modulus 4 and argument π/3 + π/6 = π/2: the product is 4i.

Directly: (1 + √3 i)(√3 + i) = √3 + i + 3i + √3 i2 = (√3 − √3) + 4i = 4i, as claimed.

Multiplying by a complex number scales by its modulus and rotates by its argument; here the rotation carried the product onto the imaginary axis.

Loci: equations that draw

An equation in z picks out a set of points, a locus, and three shapes cover the syllabus. |z − a| = r says 'distance from a is r': a circle, centre a, radius r. |z − a| = |z − b| says 'equidistant from a and b': the perpendicular bisector of the segment joining them. arg(z − a) = θ says 'the direction from a is θ': a half-line from a (excluding a itself) at angle θ.

The locus |z − (2 + i)| = 2: a circle of radius 2 centred at 2 + i on the Argand diagram2 + iradius 2every z exactly 2 from 2 + i
FIG. 2The locus |z − (2 + i)| = 2: a circle of radius 2 about the point 2 + i.

YOUR TURN

Reading three loci

Describe the loci |z − 4| = 3, |z| = |z − 2i|, and arg(z − 1) = π/4.

Show the working

|z − 4| = 3: a circle, centre 4 (the point (4, 0)), radius 3.

|z| = |z − 2i|: points equidistant from 0 and 2i, the horizontal line through i, that is the line with equation y = 1 on the diagram.

arg(z − 1) = π/4: a half-line starting at 1 (excluded) heading up-right at π/4 to the real axis. Naming centre and radius, or the two fixed points, is what the marks attach to.

THE EXAM BIT

  • Report arguments in (−π, π] in radians; plot the point before trusting any inverse tan.
  • Convert to modulus-argument form before multiplying or dividing; convert back only if asked.
  • For |z − a| = r, read the centre from what is subtracted: z − (2 + i) means centre 2 + i.
  • A half-line locus excludes its endpoint; say so when describing arg(z − a) = θ.

CHECK YOURSELF

Sketch the locus |z − 3i| = 3, and state its centre and radius. Where does it meet the axes?

Show a hint

Read the centre from the subtraction, then think about how far the circle reaches.

Show the answer

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Modulus is distance from the origin; argument is the angle in (−π, π], read from a sketch.

Products multiply moduli and add arguments; |z − a| = r is a circle, centre a, radius r.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

6 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

CHECK YOUR PROGRESS

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  • Find the modulus and argument of a complex number, handling every quadrant.
  • Convert between x + yi and modulus-argument form, and multiply and divide in it.
  • Sketch and describe the standard loci: circles, perpendicular bisectors, half-lines.

Open the full revision checklist to see every objective in the course in one place.

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