MathematicsCore20 min read

Complex Numbers in Polar Form

Modulus and argument, and why multiplying becomes rotating

This topic appears in:

1st Year Mathematics · ch.1A Level Mathematics (9709) · ch.3
01

A second address for the same point

On the Argand diagram, z = a + bi gives a point by its horizontal and vertical distances — a street address. There is a second way to reach the same point: say how far it is from the origin and in what direction. That is the polar form, and for everything except addition it is by far the more useful of the two.

The distance is the modulus r = |z| = √(a² + b²). The direction is the argument θ, the angle measured from the positive real axis, taken anticlockwise as positive.

z = r(cos θ + i sin θ)often written r cis θr = √(a² + b²)θ = tan⁻¹(b / a)a = r cos θb = r sin θthe principal argument is taken in the range −180° < θ ≤ 180°
r
the modulusnever negative
θ
the argumentdepends on the quadrant, not just on b/a

Switch to z × w. Notice the modulus of the answer is the product of the two moduli and its argument is their sum. That single observation is the whole reason polar form exists.

02

Getting the argument right

The calculator computes tan⁻¹(b/a) and always hands back an angle between −90° and 90°. That is correct only when the point is in the first or fourth quadrant. For the other two you must adjust, and the adjustment is worth a mark every time.

The safe method is to find the acute angle α = tan⁻¹|b/a| first, then place it using a quick sketch.

QuadrantSigns of a, bArgument θ
First+ , +α
Second− , +180° − α
Third− , −α − 180°
Fourth+ , −−α

Sketch before you trust the calculator

For z = −1 + i the calculator gives tan⁻¹(−1) = −45°, which points into the fourth quadrant — the wrong side of the diagram entirely. The point is plainly in the second quadrant, so the argument is 135°. A five-second sketch catches this every time; nothing else reliably does.

03

Multiplication and division become rotation and scaling

In Cartesian form, multiplying two complex numbers is an expansion with four terms and an i² to tidy up. In polar form it is two operations, and both are geometric.

To multiply, multiply the moduli and add the arguments. To divide, divide the moduli and subtract the arguments. So multiplying by a complex number stretches the plane by its modulus and rotates it by its argument — multiplying by i, which has modulus 1 and argument 90°, is exactly a quarter turn.

z₁z₂ = r₁r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]z₁/z₂ = (r₁/r₂) [cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)]moduli multiply or divide; arguments add or subtract
Worked example

Express z = −√3 + i in polar form, and hence find z².

  1. r = √(3 + 1) = 2.The modulus ignores the signs, since both parts are squared.
  2. Acute angle α = tan⁻¹(1/√3) = 30°. The point is in the second quadrant, so θ = 180° − 30° = 150°.Real part negative, imaginary part positive — second quadrant. The calculator would have said −30°.
  3. z = 2(cos 150° + i sin 150°).
  4. Squaring: modulus 2² = 4, argument 2 × 150° = 300°, so z² = 4(cos 300° + i sin 300°).Multiply the moduli, add the arguments — here both numbers are the same one.
  5. In Cartesian form: 4(0.5 − i(√3/2)) = 2 − 2√3 i.Check by squaring directly: (−√3 + i)² = 3 − 2√3 i + i² = 2 − 2√3 i ✓

z = 2 cis 150°; z² = 4 cis 300° = 2 − 2√3 i

04

De Moivre's theorem

If multiplying adds the arguments, then raising to the power n multiplies the argument by n. Written out, that is De Moivre's theorem, and it turns a problem that would need repeated expansion into a single line.

It also runs backwards. Taking an n-th root divides the argument by n — and because adding 360° to an argument gives the same number, there are n different n-th roots, spaced evenly around a circle of radius r^(1/n).

[r(cos θ + i sin θ)]ⁿ = rⁿ (cos nθ + i sin nθ)n-th roots: r^(1/n) [cos((θ + 360k)/n) + i sin((θ + 360k)/n)], k = 0, 1, …, n−1the n roots are the vertices of a regular n-gon centred on the origin

Where the trigonometric identities come from

Take (cos θ + i sin θ)² = cos 2θ + i sin 2θ and expand the left side directly: cos²θ − sin²θ + 2i sin θ cos θ. Comparing real parts gives cos 2θ = cos²θ − sin²θ and imaginary parts gives sin 2θ = 2 sin θ cos θ. Both double-angle formulae, from one line of algebra.

Before you leave this chapter

  1. r = √(a² + b²) and θ = tan⁻¹(b/a), adjusted for the quadrant. Always sketch.
  2. Principal argument lies in −180° < θ ≤ 180°.
  3. Multiplying multiplies moduli and adds arguments; dividing does the reverse.
  4. De Moivre: (r cis θ)ⁿ = rⁿ cis nθ.
  5. A complex number has exactly n distinct n-th roots, evenly spaced around a circle.
06

The square root of a complex number

Every complex number has two square roots, and there are two routes to them. The polar route is quick: halve the argument and take the square root of the modulus, then add 180° for the second root.

The algebraic route avoids trigonometry altogether and is often preferred when the answer should be exact. Set √(a + bi) = x + yi, square both sides, and compare real and imaginary parts to get two equations in x and y.

Worked example

Find the square roots of 3 + 4i.

  1. Let √(3 + 4i) = x + yi. Squaring: x² − y² + 2xyi = 3 + 4i.Remember i² = −1, which is what turns the y² term negative.
  2. Comparing parts: x² − y² = 3 and 2xy = 4, so xy = 2.Two real equations from one complex equation — that comparison is the key step.
  3. From y = 2/x: x² − 4/x² = 3, so x⁴ − 3x² − 4 = 0.Multiply through by x² to clear the fraction.
  4. Factorise as a quadratic in x²: (x² − 4)(x² + 1) = 0. Since x is real, x² = 4 and x = ±2.x² = −1 is rejected because x was defined as a real number.
  5. x = 2 → y = 1, and x = −2 → y = −1.Check: (2 + i)² = 4 + 4i − 1 = 3 + 4i ✓

±(2 + i)

The two roots are always negatives of each other

If w is one square root then −w is the other, because (−w)² = w². So finding one root gives the second for free — and any answer offering only one of them is incomplete.

07

Loci on the Argand diagram

A condition on z that is not an equation with a single answer usually describes a set of points — a curve or region in the plane. Reading these is mostly a matter of translating the notation back into the geometry it came from.

The one fact that unlocks all of them is that |z − a| is the distance between the point z and the point a. Once that is fixed in mind, each standard locus becomes an ordinary sentence about distances and angles.

ConditionIn wordsThe locus
|z − a| = rdistance from a is rcircle, centre a, radius r
|z − a| < rdistance from a is less than rthe inside of that circle
|z − a| = |z − b|equally far from a and bperpendicular bisector of ab
arg(z − a) = θthe direction from a is θa half-line from a, at angle θ
|z| = |z − 4|equally far from 0 and 4the vertical line x = 2
Worked example

Describe the locus given by |z − 3 + 2i| = 4, and find the greatest value of |z| on it.

  1. Rewrite in the standard form: |z − (3 − 2i)| = 4.The subtraction must be made explicit. The +2i in the question is −(−2i), so the centre has a negative imaginary part.
  2. So the locus is a circle with centre 3 − 2i and radius 4.Reading straight from |z − a| = r, taking care with the sign of the centre.
  3. The distance from the origin to the centre is √(3² + (−2)²) = √13 ≈ 3.606.|z| is the distance from the origin, so the origin is the point to measure from.
  4. The greatest |z| is that distance plus the radius: √13 + 4 ≈ 7.61.The furthest point of a circle from an external point lies on the line through the centre, on the far side.

A circle, centre 3 − 2i, radius 4; greatest |z| = √13 + 4 ≈ 7.61

A half-line, not a full line

The locus arg(z − a) = θ is a half-line starting at a and heading in the direction θ. Points on the opposite side have argument θ ± π, not θ, so they are excluded. The starting point a is excluded too, because arg 0 is undefined — mark it with an open circle.

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Define the modulus and the argument of a complex number.
Model answer

The modulus |z| = √(a² + b²) is the distance of the point from the origin on the Argand diagram. The argument is the angle the line from the origin to the point makes with the positive real axis, measured anticlockwise.

Examiner tip. Say "measured from the positive real axis". Without that reference direction the definition is incomplete.

SQ2[2 marks]
Find the modulus and argument of z = 1 − i.
Model answer

r = √(1 + 1) = √2. The point is in the fourth quadrant with α = tan⁻¹(1) = 45°, so θ = −45°.

Examiner tip. Positive real part, negative imaginary part means the fourth quadrant, so the argument is negative. Here the calculator happens to be right — but check the quadrant anyway.

SQ3[2 marks]
What is the geometrical effect of multiplying a complex number by i?
Model answer

i has modulus 1 and argument 90°, so multiplying leaves the modulus unchanged and adds 90° to the argument: the point is rotated a quarter turn anticlockwise about the origin.

Examiner tip. Both facts — modulus unchanged, argument plus 90° — are needed. Saying only "it rotates" scores half.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
Express z = −2 − 2i in polar form.
Full working
  1. r = √(4 + 4) = √8 = 2√2[1]
  2. Acute angle α = tan⁻¹(2/2) = 45°use the absolute values to get the acute angle[1]
  3. Both parts are negative, so the point is in the third quadrant and θ = 45° − 180° = −135°accept 225° if the range is not restricted[1]
  4. z = 2√2 (cos(−135°) + i sin(−135°))[1]

z = 2√2 cis(−135°)

Examiner tip. The principal argument is taken in −180° < θ ≤ 180°, so −135° is preferred over 225°. Both are the same direction; only one is the principal value.

N2[4 marks]
Use De Moivre's theorem to evaluate (1 + i)⁸.
Full working
  1. r = √2 and θ = 45°, so 1 + i = √2 cis 45°first quadrant, so no adjustment needed[1]
  2. De Moivre: (√2)⁸ cis(8 × 45°)[1]
  3. (√2)⁸ = 2⁴ = 16 and 8 × 45° = 360°[1]
  4. 16(cos 360° + i sin 360°) = 16a real answer, since the argument came back to a full turn[1]

16

Examiner tip. Expanding (1+i)⁸ directly would take a binomial expansion with nine terms. Converting to polar form first turns it into two short calculations.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
Consider the equation z³ = 8.
  1. Write 8 in polar form.
  2. Find all three cube roots of 8, giving your answers in polar form.
  3. Show that the three roots are equally spaced on a circle, and state its radius.
Mark scheme
  1. 8 = 8(cos 0° + i sin 0°)8 lies on the positive real axis, so its argument is 0°[1]
  2. Roots have modulus 8^(1/3) = 2[1]
  3. Arguments (0 + 360k)/3 for k = 0, 1, 2, giving 0°, 120°, 240°adding 360° each time before dividing is what produces the extra roots[1]
  4. Roots: 2 cis 0°, 2 cis 120°, 2 cis 240°the first is the obvious real root 2[1]
  5. The arguments differ by 120° = 360°/3, so the roots are equally spacedthey form an equilateral triangle[1]
  6. All three have modulus 2, so they lie on a circle of radius 2 centred at the origin[1]

(b) 2 cis 0°, 2 cis 120°, 2 cis 240° (c) equally spaced 120° apart on a circle of radius 2

Examiner tip. Every "find all n-th roots" question has the same shape: one modulus for all of them, and arguments spaced 360°/n apart. Once you have the first root the rest need no calculation.