AS & A Level · Practice questions · The Argand diagram, modulus-argument and exponential form

Lengths multiply, angles add.

Six original Cambridge-style questions on the Argand diagram, modulus and argument, polar and exponential forms, and the geometry of complex multiplication.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Sketch, then modulus and argument.

Sketch the point to find its quadrant before using \(\tan^{-1}\).
Modulus: \(r = \sqrt{x^{2} + y^{2}}\). Argument: adjust the reference angle for the quadrant; \(-\pi < \theta \le \pi\).
Forms: \(x + iy = r(\cos\theta + i\sin\theta) = re^{i\theta}\).
Multiply: multiply moduli, add arguments. Divide: divide moduli, subtract arguments.
Geometry: \(z^{*}\) reflects; \(+w\) translates; \(\times re^{i\theta}\) enlarges and rotates.

Recall

[4 marks]

Find the modulus and argument of (a) \(-5i\) (b) \(-2 + 2i\), giving exact answers.

(a) \(|-5i| = 5\), \(\arg(-5i) = -\tfrac{\pi}{2}\). ✓
(b) \(|-2 + 2i| = \sqrt{8} = 2\sqrt{2}\); second quadrant with reference angle \(\tfrac{\pi}{4}\), so \(\arg = \tfrac{3\pi}{4}\). ✓

Calculation

[3 marks]

Express \(z = -3 + 4i\) in the form \(re^{i\theta}\), where \(r > 0\) and \(-\pi < \theta \le \pi\), giving \(\theta\) to 3 significant figures.

\(r = \sqrt{9 + 16} = 5\). ✓
Second quadrant: reference angle \(\tan^{-1}\tfrac{4}{3} = 0.9273\), so \(\theta = \pi - 0.9273 = 2.21\). ✓
\(z = 5e^{2.21i}\). ✓

Calculation

[4 marks]

\(z_{1} = 4e^{2\pi i/3}\) and \(z_{2} = 2e^{-\pi i/6}\). Find \(z_{1}z_{2}\) and \(\dfrac{z_{1}}{z_{2}}\), giving each in the form \(x + iy\) with exact values.

\(z_{1}z_{2} = 8e^{i(2\pi/3 - \pi/6)} = 8e^{i\pi/2} = 8i\). ✓
\(\dfrac{z_{1}}{z_{2}} = 2e^{i(2\pi/3 + \pi/6)} = 2e^{5\pi i/6}\). ✓
\(= 2\left(-\tfrac{\sqrt{3}}{2} + \tfrac{1}{2}i\right) = -\sqrt{3} + i\). ✓

Analysis

[4 marks]

Let \(u = \sqrt{3} + i\). Find \(|u|\) and \(\arg u\), and hence show that \(u^{6} = -64\).

\(|u| = \sqrt{3 + 1} = 2\); first quadrant, \(\arg u = \tan^{-1}\tfrac{1}{\sqrt{3}} = \tfrac{\pi}{6}\). ✓
Multiplying six copies: \(|u^{6}| = 2^{6} = 64\) and \(\arg u^{6} = 6 \times \tfrac{\pi}{6} = \pi\). ✓
\(u^{6} = 64(\cos\pi + i\sin\pi) = -64\). ✓

Analysis

[5 marks]

On an Argand diagram, \(U\) and \(V\) represent \(u = 2 + i\) and \(v = 1 + 3i\).

(a) Show that \(v = (1 + i)u\). (b) Hence describe the transformation that maps \(U\) to \(V\), and find angle \(UOV\). (c) Find the area of triangle \(UOV\).

(a) \((1 + i)(2 + i) = 2 + i + 2i + i^{2} = 1 + 3i = v\). ✓
(b) \(1 + i = \sqrt{2}\,e^{i\pi/4}\): an enlargement with scale factor \(\sqrt{2}\) about \(O\) combined with a rotation of \(\tfrac{\pi}{4}\) anticlockwise. Angle \(UOV = \tfrac{\pi}{4}\). ✓
(c) \(OU = \sqrt{5}\), \(OV = \sqrt{2}\sqrt{5} = \sqrt{10}\); area \(= \tfrac{1}{2}\sqrt{5}\sqrt{10}\sin\tfrac{\pi}{4} = \tfrac{1}{2}\sqrt{50} \times \tfrac{\sqrt{2}}{2} = \tfrac{5}{2}\). ✓

Proof

[6 marks]

Let \(z = 1 + \sqrt{3}\,i\) and \(w = 1 - i\).

(a) Find \(zw\) in the form \(x + iy\). (b) Find the modulus and argument of \(z\) and of \(w\), and hence of \(zw\). (c) Deduce the exact value of \(\sin\dfrac{\pi}{12}\).

(a) \(zw = 1 - i + \sqrt{3}\,i + \sqrt{3} = (1 + \sqrt{3}) + (\sqrt{3} - 1)i\). ✓
(b) \(|z| = 2\), \(\arg z = \tfrac{\pi}{3}\); \(|w| = \sqrt{2}\), \(\arg w = -\tfrac{\pi}{4}\). So \(|zw| = 2\sqrt{2}\), \(\arg zw = \tfrac{\pi}{12}\). ✓
(c) Imaginary parts: \(2\sqrt{2}\sin\tfrac{\pi}{12} = \sqrt{3} - 1\). ✓
\(\sin\tfrac{\pi}{12} = \dfrac{\sqrt{3} - 1}{2\sqrt{2}} = \dfrac{\sqrt{6} - \sqrt{2}}{4}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.