Paper 3 only. Plot \(x + iy\) as the point \((x, y)\) and every complex number gets a distance from the origin and an angle. In that form, multiplying is just multiplying the lengths and adding the angles.
\(z = x + iy = r(\cos\theta + i\sin\theta) = re^{i\theta}\), where \(r = |z| = \sqrt{x^{2} + y^{2}}\) and \(\theta = \arg z\) with \(-\pi < \theta \le \pi\). Then \(|z_{1}z_{2}| = |z_{1}|\,|z_{2}|\) and \(\arg(z_{1}z_{2}) = \arg z_{1} + \arg z_{2}\); for division, divide the moduli and subtract the arguments.
Fig. 1: u = 3 + 2i, its conjugate u* (reflection in the real axis) and iu = −2 + 3i (rotation by 90° anticlockwise about O); all three are √13 from O
Section 01
The Argand diagram, modulus and argument.
On an Argand diagram the real part goes across and the imaginary part goes up. The modulus is the distance from \(O\); the argument is the angle from the positive real axis, anticlockwise positive, with \(-\pi < \arg z \le \pi\) unless a question says otherwise. ALWAYS sketch the point first to see which quadrant you are in.
Find the modulus and argument of \(-3 - 4i\), giving the argument in radians to 3 significant figures.
\(|z| = \sqrt{9 + 16} = 5\).
Third quadrant. Reference angle \(\alpha = \tan^{-1}\dfrac{4}{3} = 0.9273\).
\(\arg z = -(\pi - 0.9273) = -2.2143\).
Answer: \(|z| = 5\), \(\arg z = -2.21\) radians (\(-126.9^\circ\))
Your turn
Find the modulus and argument of \(1 - i\), and write it in exponential form.
\(|1 - i| = \sqrt{2}\); fourth quadrant with reference angle \(\tfrac{\pi}{4}\), so \(\arg = -\tfrac{\pi}{4}\). \(1 - i = \sqrt{2}\,e^{-i\pi/4}\).
Examiner trap
Your calculator's \(\tan^{-1}\left(\tfrac{y}{x}\right)\) gives \(\tan^{-1}\left(\tfrac{-4}{-3}\right) = 0.927\) for \(-3 - 4i\), which points into the FIRST quadrant. The sketch is not optional: it is how you know to adjust.
Section 02
Multiplying and dividing in polar form.
Write \(z_{1} = r_{1}e^{i\theta_{1}}\) and \(z_{2} = r_{2}e^{i\theta_{2}}\); the laws of indices do the rest. \(z_{1}z_{2} = r_{1}r_{2}e^{i(\theta_{1} + \theta_{2})}\) and \(\dfrac{z_{1}}{z_{2}} = \dfrac{r_{1}}{r_{2}}e^{i(\theta_{1} - \theta_{2})}\). If the new argument falls outside \(-\pi < \theta \le \pi\), add or subtract \(2\pi\).
Worked example Product and quotient
\(z_{1} = 6\left(\cos\dfrac{5\pi}{6} + i\sin\dfrac{5\pi}{6}\right)\) and \(z_{2} = 2\left(\cos\left(-\dfrac{\pi}{2}\right) + i\sin\left(-\dfrac{\pi}{2}\right)\right)\). Find \(z_{1}z_{2}\) and \(\dfrac{z_{1}}{z_{2}}\) in polar form, and then in the form \(x + iy\).
Polar: \(1 + \sqrt{3}\,i = 2e^{i\pi/3}\) and \(1 - i = \sqrt{2}\,e^{-i\pi/4}\), so the product is \(2\sqrt{2}\,e^{i\pi/12}\) (\(\tfrac{\pi}{3} - \tfrac{\pi}{4} = \tfrac{\pi}{12}\)).
Equate real parts: \(2\sqrt{2}\cos\dfrac{\pi}{12} = 1 + \sqrt{3}\).
\(\cos\dfrac{\pi}{12} = \dfrac{1 + \sqrt{3}}{2\sqrt{2}} = \dfrac{\sqrt{2} + \sqrt{6}}{4}\) (multiplying top and bottom by \(\sqrt{2}\)).
Moduli MULTIPLY but arguments ADD. Multiplying the arguments, or adding the moduli, are both common slips, and \(\dfrac{4\pi}{3}\) left unadjusted loses the final mark when the range \(-\pi < \theta \le \pi\) is required.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Find \(|5 - 12i|\).
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Find \(\arg(-2i)\), with \(-\pi < \arg z \le \pi\).
Quick check+10 XP
\(|z_{1}| = 3\), \(\arg z_{1} = \dfrac{\pi}{6}\), \(|z_{2}| = 2\), \(\arg z_{2} = \dfrac{\pi}{3}\). Find the modulus and argument of \(z_{1}z_{2}\).
Quick check+10 XP
On an Argand diagram, how is \(z^{*}\) related to \(z\)?
Section 03
What the operations look like.
Every operation has a picture on the Argand diagram (Fig. 1). These pictures are what make loci and many 'show that' questions quick.
Conjugate: \(z^{*}\) is the reflection of \(z\) in the real axis (same modulus, argument negated).
Add: \(z + w\) is \(z\) translated by \(w\), the fourth vertex of the parallelogram on \(O\), \(z\), \(w\). Subtract: \(z - w\) is the vector from \(w\) to \(z\), so \(|z - w|\) is the distance between the points.
Multiply by \(re^{i\theta}\): enlarge by scale factor \(r\) about \(O\) and rotate by \(\theta\) anticlockwise. In particular, multiplying by \(i\) rotates by \(90^\circ\).
Divide by \(re^{i\theta}\): enlarge by \(\tfrac{1}{r}\) and rotate by \(\theta\) clockwise.
Worked example Rotate and enlarge
The point \(P\) represents \(u = 3 + 2i\). \(P\) is rotated through \(90^\circ\) anticlockwise about \(O\) to \(Q\), and \(Q\) is then enlarged by scale factor 2 about \(O\) to \(R\). Find the numbers represented by \(Q\) and \(R\).
Rotation by \(90^\circ\) anticlockwise is multiplication by \(i\): \(iu = 3i + 2i^{2} = -2 + 3i\) (Fig. 1).
Enlargement by factor 2 about \(O\) is multiplication by 2: \(2(-2 + 3i) = -4 + 6i\).
\(|3 + 4i| = 5\) and \(|5 - 12i| = 13\), so the modulus of the product is \(5 \times 13 = 65\).
Examiner trap
Multiplying by \(-i\) (or dividing by \(i\)) rotates CLOCKWISE. Check the direction with a simple case: \(i \times 1 = i\) moves from the real axis up to the imaginary axis, which is anticlockwise.
Stage 2 · Exam
Exam-style questions
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Find \(\arg(-\sqrt{3} - i)\), with \(-\pi < \arg z \le \pi\).
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\(z = 2\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right)\) and \(w = 4\left(\cos\dfrac{3\pi}{4} + i\sin\dfrac{3\pi}{4}\right)\). Find the modulus and argument of \(zw\), with \(-\pi < \arg \le \pi\).
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Express \(4e^{-i\pi/6}\) in the form \(x + iy\).
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The point representing \(u = 3 + i\) is rotated \(90^\circ\) anticlockwise about \(O\), and then its distance from \(O\) is doubled. Which complex number represents the final point?
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Argand diagram and polar form, sorted.
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