- The modulus argument form writes a complex number as \(z = r(\cos\theta + i\sin\theta)\), where \(r = |z|\) is its distance from the origin and \(\theta = \arg z\) is its angle on an Argand diagram.
- To multiply, multiply the moduli and add the arguments. To divide, divide the moduli and subtract the arguments.
- De Moivre’s theorem: \(\left[r(\cos\theta + i\sin\theta)\right]^n = r^n(\cos n\theta + i\sin n\theta)\), so powers become easy.
- In our case, a 230 V supply across an impedance \(Z = 30 + 40i\) ohms gives a current of 4.6 A that lags the voltage by 53.1°, and the circuit uses 634.8 W.
- Exam point: the principal argument lies in \(-\pi < \theta \le \pi\). Always sketch the point first so you choose the right quadrant.
The case: current and phase in an alternating-current circuit
A small workshop motor is connected to a 230 V alternating-current supply. Electrically, the motor behaves like a resistor of 30 ohms in series with a coil whose reactance is 40 ohms. Engineers combine these into a single complex number called the impedance:
The real part is the resistance, which turns energy into useful work and heat. The imaginary part is the reactance of the coil, which stores energy and hands it back each cycle. (Electrical engineers write \(j\) instead of \(i\), because \(i\) is already used for current. The mathematics is the same.) (The figures are illustrative.)
The electrician needs three numbers. How large is the current? By how much does it lag behind the voltage? And how much power does the motor actually use? The rule is Ohm’s law, \(I = \frac{V}{Z}\), but now \(Z\) is complex. Dividing \(230\) by \(30 + 40i\) in Cartesian form is possible, but the size and the angle of the answer are what matter, and those come straight out of the modulus argument form.
The concept: a length and an angle
Any complex number \(z = x + iy\) can be plotted as the point \((x, y)\) on an Argand diagram, with the real part across and the imaginary part up. Instead of giving the point by its two coordinates, you can give it by its distance from the origin and the angle it makes with the positive real axis. That is the modulus argument form, also called polar form.[3]
The two forms are linked by right-angled triangle trigonometry. The horizontal side is \(x = r\cos\theta\), the vertical side is \(y = r\sin\theta\), and the hypotenuse is \(r\).
Here \(x\) and \(y\) are the real and imaginary parts, \(r\) is the modulus (always positive or zero) and \(\theta\) is the argument in radians. The equation \(\tan\theta = \frac{y}{x}\) has two solutions in \((-\pi, \pi]\), so you must use the quadrant of the point to pick the correct one. Many textbooks shorten \(\cos\theta + i\sin\theta\) to \(\operatorname{cis}\theta\), and at A Level and university you will also see it written as \(e^{i\theta}\).
Mathematics Made Simple explains complex numbers in modulus argument form with worked examples and Argand diagrams.Get the textbook · $12
The real power of this form shows up when you multiply. Using the compound-angle formulas,
So multiplication scales lengths and adds angles. Division does the reverse: divide the moduli and subtract the arguments. Applying the multiplication rule \(n\) times gives De Moivre’s theorem, \(\left[r(\cos\theta + i\sin\theta)\right]^n = r^n(\cos n\theta + i\sin n\theta)\), for any positive integer \(n\).[4]
Why this method fits the case
In an AC circuit, the voltage and the current are waves with the same frequency. What differs is their size and their timing, the phase. A complex number in modulus argument form carries exactly these two pieces of information: the modulus gives the size, and the argument gives the phase. Engineers call this a phasor.
Ohm’s law then becomes a division of complex numbers. In modulus argument form, that division is just one division and one subtraction: the size of the current is \(\frac{|V|}{|Z|}\), and its phase is \(\arg V – \arg Z\). The answer arrives in the form the electrician needs, without any messy algebra.
The modulus argument form is not the best tool for adding or subtracting complex numbers. To add two impedances in series, add the real parts and the imaginary parts separately, in Cartesian form. A good rule: add and subtract in Cartesian form; multiply, divide and take powers or roots in modulus argument form. Switch between them as the calculation demands.
Setting it up
Here are the quantities for the motor.
| Quantity | Symbol | Value | Meaning |
|---|---|---|---|
| Supply voltage | \(V\) | 230 V, argument 0 | Reference phasor on the real axis |
| Resistance | \(R\) | 30 Ω | Real part of \(Z\) |
| Reactance of the coil | \(X\) | 40 Ω | Imaginary part of \(Z\) |
| Impedance | \(Z\) | \(30 + 40i\) Ω | Complex “resistance” |
| Current | \(I\) | \(\frac{V}{Z}\) | To be found: size and phase |
| Power used | \(P\) | \(|V||I|\cos\phi\) | \(\phi\) is the phase difference |
The point \(Z = 30 + 40i\) lies in the first quadrant, so its argument is between 0 and \(\frac{\pi}{2}\). Taking the voltage as the reference, with argument 0, keeps the arithmetic simple.
Solving it
Write \(Z = 30 + 40i\) in modulus argument form. Hence find the current \(I = \frac{230}{Z}\) and the power used.
The current is 4.6 A, lagging the voltage by 53.1°, and the motor uses 634.8 W.
Given \(z_1 = 6\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)\) and \(z_2 = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\), find \(z_1z_2\) and \(\frac{z_1}{z_2}\) in modulus argument form, using principal arguments.
\(z_1z_2 = 12\left(\cos\left(-\frac{5\pi}{6}\right) + i\sin\left(-\frac{5\pi}{6}\right)\right)\) and \(\frac{z_1}{z_2} = 3i\).
Use De Moivre’s theorem to evaluate \(\left(1 + \sqrt{3}\,i\right)^6\).
\(\left(1 + \sqrt{3}\,i\right)^6 = 64\), a real number, found without expanding six brackets.
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What the answer tells us
The motor draws 4.6 A, and the current reaches its peak a little later in each cycle than the voltage, by 53.1° of the cycle. Only part of the current does useful work: the share is \(\cos 53.13° = 0.6\), which engineers call the power factor. So the supply delivers \(230 \times 4.6 = 1058\) volt-amperes, but the motor uses only 634.8 W of it.
Figure 1 shows the impedance and the current on an Argand diagram. The current points the same angle below the real axis as the impedance points above it, and its length, 4.6, is the supply voltage divided by the modulus of the impedance.
Reasonableness check. Multiply back: \(I \times Z = (2.76 – 3.68i)(30 + 40i) = 82.8 + 110.4i – 110.4i + 147.2 = 230\). The product is exactly the supply voltage, with no imaginary part, as it should be. Also, a pure 30 Ω resistor would draw \(\frac{230}{30} \approx 7.7\) A; the coil adds opposition, so 4.6 A is sensibly smaller.
Sensitivity. Electricians often add a capacitor to cancel part of the coil’s reactance. Suppose it reduces the reactance from 40 Ω to 10 Ω, so \(Z = 30 + 10i\). Then \(|Z| = \sqrt{1000} \approx 31.6\) Ω and \(\arg Z \approx 18.4°\). The current rises to about 7.27 A, the power factor rises from 0.6 to about 0.95, and the motor can now use \(7.27^2 \times 30 \approx 1587\) W from the same supply. Small changes to the argument have a large effect on how efficiently the supply is used, which is why phase matters so much in power engineering.
For a full treatment of Argand diagrams, modulus argument form and De Moivre’s theorem, with every diagram drawn, see the Mathematics Made Simple textbook. To practise converting between forms, choosing principal arguments and using De Moivre’s theorem quickly, work through the Mathematics Made Simple Practice Questions book.
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Practice questions
Sketch each point on an Argand diagram before you start; it prevents quadrant errors.
EasyQ1. Find the modulus and principal argument of \(z = -3 + 3i\). Show worked answer
Second quadrant: \(\arg z = \pi – \tan^{-1}\frac{3}{3} = \pi – \frac{\pi}{4} = \frac{3\pi}{4}\).
So \(z = 3\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)\).
EasyQ2. Write \(z = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)\) in the form \(x + iy\). Show worked answer
\(z = 2\left(\frac{1}{2} – \frac{\sqrt{3}}{2}i\right) = 1 – \sqrt{3}\,i\).
MediumQ3. Given \(z_1 = 4\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\) and \(z_2 = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\), find \(z_1z_2\) and \(\frac{z_1}{z_2}\) in the form \(x + iy\). Show worked answer
\(\frac{z_1}{z_2} = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = \sqrt{3} + i\).
MediumQ4. Use De Moivre’s theorem to find \((1 – i)^{10}\). Show worked answer
\((1 – i)^{10} = (\sqrt{2})^{10}\left(\cos\left(-\frac{10\pi}{4}\right) + i\sin\left(-\frac{10\pi}{4}\right)\right) = 32\left(\cos\left(-\frac{5\pi}{2}\right) + i\sin\left(-\frac{5\pi}{2}\right)\right)\).
\(-\frac{5\pi}{2}\) is the same angle as \(-\frac{\pi}{2}\): \(\cos = 0\), \(\sin = -1\).
Answer: \(-32i\).
ExamQ5. Let \(z = 1 + i\) and \(w = 1 – \sqrt{3}\,i\). (a) Find the modulus and argument of \(z\) and of \(w\). (b) Hence find \(\left|\frac{z}{w}\right|\) and \(\arg\frac{z}{w}\). (c) By writing \(\frac{z}{w}\) in the form \(x + iy\), show that \(\cos\frac{7\pi}{12} = \frac{\sqrt{2} – \sqrt{6}}{4}\). [8 marks] Show worked answer
(b) \(\left|\frac{z}{w}\right| = \frac{\sqrt{2}}{2}\). [B1] \(\arg\frac{z}{w} = \frac{\pi}{4} – \left(-\frac{\pi}{3}\right) = \frac{7\pi}{12}\). [M1 A1]
(c) \(\frac{z}{w} = \frac{(1 + i)(1 + \sqrt{3}\,i)}{(1 – \sqrt{3}\,i)(1 + \sqrt{3}\,i)} = \frac{(1 – \sqrt{3}) + (1 + \sqrt{3})i}{4}\). [M1]
Real parts: \(\frac{\sqrt{2}}{2}\cos\frac{7\pi}{12} = \frac{1 – \sqrt{3}}{4}\). [M1]
So \(\cos\frac{7\pi}{12} = \frac{1 – \sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{2} – \sqrt{6}}{4}\). [A1]
Frequently asked questions
Should the argument be in degrees or radians?
Use radians in pure mathematics exams unless the question says otherwise. Degrees are common in engineering, as in the circuit example. Whichever you use, be consistent, and make sure your calculator is in the matching mode.
What is the difference between the argument and the principal argument?
A complex number has infinitely many arguments, because adding any multiple of \(2\pi\) gives the same direction. The principal argument is the one value in the range \(-\pi < \theta \le \pi\). Exam questions usually want the principal argument.
How does \(e^{i\theta}\) relate to modulus argument form?
Euler’s formula states that \(e^{i\theta} = \cos\theta + i\sin\theta\). So \(z = r(\cos\theta + i\sin\theta)\) can be written \(z = re^{i\theta}\). The multiplication rule then follows from the index law \(e^{i\theta_1}e^{i\theta_2} = e^{i(\theta_1 + \theta_2)}\), and De Moivre’s theorem is simply \(\left(e^{i\theta}\right)^n = e^{in\theta}\).
Where else is the modulus argument form used?
It is used in signal processing, where sound and radio waves are written as phasors, in control engineering, in quantum physics, and in finding the \(n\)th roots of a complex number. The roots of \(z^n = 1\), for example, sit at equal angles of \(\frac{2\pi}{n}\) around the unit circle.
References
- Cambridge Assessment International Education. Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. cambridgeinternational.org
- International Baccalaureate Organization (2019). Mathematics: analysis and approaches guide. Cardiff: IBO.
- Abramson, J. et al. (2021). Algebra and Trigonometry, 2nd edn. Houston, TX: OpenStax. openstax.org
- Abramson, J. et al. (2022). Precalculus, 2nd edn. Houston, TX: OpenStax. openstax.org
