1The idea in one paragraph

In 1.12 you located a complex number by walking a steps across and b steps up: z = a + bi. You can equally locate it by facing in direction θ and walking a distance r: that is its modulus–argument form, or polar form, z = r(cos θ + i sin θ), shortened to r cis θ, and written most compactly in Euler form as re^(iθ). The payoff is multiplication. In Cartesian form, multiplying means expanding brackets. In polar form, you multiply the moduli and add the arguments, and in Euler form that is just the law of indices eᵃ × eᵇ = eᵃ⁺ᵇ. So multiplying by w stretches every point by |w| and turns it by arg w. Dividing does the reverse. Adding, by contrast, is still easiest in Cartesian form.

2Modulus–argument (polar) form

Figure 1 shows z = a + bi with modulus r = |z| and argument θ = arg z. The right-angled triangle gives

z = r(cos θ + i sin θ) = r cis θ, where r = |z| > 0 and θ = arg z. The formula booklet gives this, together with the Euler form re^(iθ).

"cis" is short for cos + i sin. The IB uses it freely, and you may too.

Cartesian to polar. Find r and θ exactly as in 1.12: r by Pythagoras, θ by sketching the point, finding the acute angle and placing it in the right quadrant.

Worked example 1. Paper 1. Write z = 1 − √3 i in the form r cis θ, where −π < θ ≤ π.

Polar to Cartesian. Evaluate the cosine and the sine and multiply out.

Worked example 2. Paper 1. Write 4 cis(3π/4) in the form a + bi.

On Paper 2 the angle need not be special. 5 cis 2 = 5 cos 2 + (5 sin 2)i ≈ −2.08 + 4.55i, in radians. The GDC converts both ways directly.

Two things that are not polar form. The modulus must be positive: −2 cis(π/6) is not in polar form, even though it is a correct expression. Since −1 = cis π, it equals 2 cis(π/6 − π) = 2 cis(−5π/6). And the bracket must be exactly cos θ + i sin θ with the same angle and a plus sign: 3(cos θ − i sin θ) is not polar form, but it equals 3(cos(−θ) + i sin(−θ)) = 3 cis(−θ), because cos is even and sin is odd.

Arguments repeat every 2π. cis θ = cis(θ + 2π) = cis(θ − 2π), since turning a full circle brings you back. So 2 cis(7π/4) and 2 cis(−π/4) are the same number. When a question fixes the range, give the argument inside it.

Figure 2 gathers the four ways of writing the same number, including the Euler form of section 3.

3Euler form

Euler's formula links the exponential function to the circle:

e^(iθ) = cos θ + i sin θ, so z = r cis θ = re^(iθ).

At this level, take it as a definition of what e to an imaginary power means. It is the right definition, and AHL 5.19 gives the reason: the Maclaurin series for eˣ, with iθ put in for x, separates into the series for cos θ and i times the series for sin θ.

Figure 3 shows what it says. As θ increases, e^(iθ) moves round the circle of radius 1, because |cos θ + i sin θ| = √(cos²θ + sin²θ) = 1. At θ = π it reaches the point −1:

This is Euler's identity. It links five of the most important numbers in mathematics (0, 1, e, i and π) in one line, and the guide asks, as a TOK question, why mathematicians call it beautiful. Elegance, surprise and economy are the usual answers; whether beauty is evidence of truth is a better question.

The conjugate in Euler form. z* reflects z in the real axis, which keeps the modulus and reverses the argument: if z = re^(iθ) then z* = re^(−iθ). So 2e^(iπ/3) has conjugate 2e^(−iπ/3).

4Multiplying: moduli multiply, arguments add

Multiply two numbers in polar form and use the compound angle formulae from AHL 3.10.

With moduli in front, and in Euler form where it is simply a law of indices:

r₁ cis θ₁ × r₂ cis θ₂ = r₁r₂ cis(θ₁ + θ₂), that is, r₁e^(iθ₁) × r₂e^(iθ₂) = r₁r₂ e^(i(θ₁ + θ₂)). So |zw| = |z||w| and arg(zw) = arg z + arg w (adjusted by 2π if needed).

Worked example 3. Paper 1. Given z = 2 cis(π/6) and w = 3 cis(π/4), find zw in polar form.

That is one line, against four for the same product in Cartesian form. If the arguments add to something outside the range you are asked for, add or subtract 2π: 3 cis(3π/4) × 2 cis(π/2) = 6 cis(5π/4) = 6 cis(−3π/4).

What multiplication does geometrically. Multiplying z by w = r cis θ multiplies its length by r and adds θ to its angle. So multiplication by w is an enlargement by scale factor |w| and a rotation by arg w, both centred on the origin. Figure 4 shows it for one point, and Figure 5 for a whole triangle multiplied by 1 + i = √2 cis(π/4): it turns through π/4 and grows by √2.

Three special cases are worth knowing by heart.

• Multiplying by i = cis(π/2) rotates by π/2 anticlockwise, which is the quarter-turn you saw in 1.12.
• Multiplying by −1 = cis π is a half-turn about the origin.
• Multiplying by cis θ, which has modulus 1, is a pure rotation by θ with no stretching.

5Dividing: moduli divide, arguments subtract

Division undoes multiplication, so it divides the moduli and subtracts the arguments.

r₁ cis θ₁ ÷ r₂ cis θ₂ = (r₁/r₂) cis(θ₁ − θ₂), and in particular 1/(r cis θ) = (1/r) cis(−θ).

Geometrically, dividing by w shrinks by |w| and turns back by arg w.

Worked example 4. Paper 1. Given z = 2e^(iπ/3) and w = √2 e^(−iπ/4), find z/w in Euler form.

Watch the double negative. Subtracting −π/4 adds π/4.

6Two forms, one answer: exact trigonometric values

A product or quotient can be found in Cartesian form and in polar form. The two answers are the same number, so their real parts must match and their imaginary parts must match. That produces exact values of cos and sin at angles such as 5π/12 and 7π/12, which are not on the list of special angles. This is a favourite multi-part Paper 1 question.

Worked example 5. Paper 1. (a) Write 1 + i and 1 − √3 i in polar form. (b) Hence find (1 + i)/(1 − √3 i) in polar form. (c) Find (1 + i)/(1 − √3 i) in Cartesian form. (d) Hence find the exact value of cos(7π/12).

A sense check: 7π/12 is just past π/2, so its cosine should be small and negative, and (√2 − √6)/4 ≈ −0.259. Equating the imaginary parts gives sin(7π/12) = (√6 + √2)/4 in the same way.

7Adding in polar and Euler form

There is no neat rule for adding in polar form. Moduli do not add: |z + w| is usually less than |z| + |w|, because the arrows point in different directions. To add, convert to Cartesian form, add, and convert back if you need to. Geometrically, addition is still the parallelogram of 1.12.

One sum does have a neat answer, and it is examined: two numbers with the same modulus. Figure 6 shows 1 + e^(iθ). The two arrows 1 and e^(iθ) both have length 1, so the parallelogram they make is a rhombus, and the diagonal of a rhombus bisects the angle between its sides. The sum therefore has argument θ/2. The algebra confirms it by taking out a factor of e^(iθ/2):

So when cos(θ/2) > 0, which is true for −π < θ < π, the modulus of 1 + e^(iθ) is 2 cos(θ/2) and its argument is θ/2. Check with θ = 2π/3: the formula gives 2 cos(π/3) e^(iπ/3) = e^(iπ/3) = ½ + (√3/2)i, and directly, 1 + (−½ + (√3/2)i) = ½ + (√3/2)i. They agree.

8All the geometry on one list

This is why complex numbers are used to describe phase in electrical engineering: a voltage or current that oscillates is represented by a number re^(iθ) whose modulus is the amplitude and whose argument is the phase angle, and a component that shifts the phase multiplies it by a complex number of the right argument.