How does the modulus-argument form of a complex number turn multiplication and division into operations on lengths and angles?
Express complex numbers in modulus-argument form and use it to multiply and divide, interpreting these operations geometrically as scaling and rotation
A focused answer to the HSC Maths Extension 2 dot point on modulus-argument form. Polar form, modulus and principal argument, converting between forms, and the geometry of multiplication and division as scaling and rotation, with verified worked examples.
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What this dot point is asking
NESA wants you to express a complex number by its distance from the origin and the angle it makes with the positive real axis, the modulus-argument (polar) form. You must convert between Cartesian and polar form, identify the principal argument, and use polar form to multiply and divide complex numbers, understanding these as scaling and rotation in the Argand plane.
Modulus and argument
For , the modulus is the distance from the origin,
and the argument is the angle measured anticlockwise from the positive real axis to the ray from the origin to . The principal argument is the value in the interval . The Cartesian and polar coordinates are linked by
so that
This is often abbreviated , where .
Finding the argument correctly
The relation does not by itself fix , because the calculator inverse tangent only returns angles in . You must use the signs of and to place in the correct quadrant. For example, lies in the second quadrant, so its argument is , not . Always sketch the point in the Argand plane before committing to an argument.
Multiplication: scaling and rotation
If and , then using the angle-sum identities,
So multiplying by scales the modulus by and rotates the point anticlockwise by . In particular, multiplying by is a rotation by with no change in length, and multiplying by a real is a pure scaling.
Division: subtraction of arguments
Dividing reverses the process:
The modulus of the quotient is the ratio of moduli, and the argument is the difference of arguments. These rules make repeated multiplication and powers far easier than expanding in Cartesian form, and they lead directly into de Moivre's theorem.
Exam-style practice questions
Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2024 HSC2 marksWrite the number sqrt(3) + i in modulus-argument form.Show worked answer →
Find the modulus and the principal argument, then write z = r(cos q + i sin q).
Modulus: r = |sqrt(3) + i| = sqrt((sqrt(3))^2 + 1^2) = sqrt(3 + 1) = 2.
Argument: the point (sqrt(3), 1) is in the first quadrant, so q = arctan(1/sqrt(3)) = pi/6.
Therefore sqrt(3) + i = 2(cos(pi/6) + i sin(pi/6)).
(Equivalently, in exponential form, 2 e^(i pi/6).)
Mark notes: 1 mark for the modulus 2, 1 mark for the argument pi/6 written in correct modulus-argument form.
2021 HSC2 marksThe complex numbers z = 2 e^(i pi/2) and w = 6 e^(i pi/6) are given. Find the value of zw, giving the answer in the form r e^(i q).Show worked answer →
In modulus-argument (exponential) form, multiplication multiplies the moduli and adds the arguments. This is the geometric statement that multiplying by w scales lengths by |w| and rotates by arg(w).
Moduli: 2 times 6 = 12.
Arguments: pi/2 + pi/6 = 3pi/6 + pi/6 = 4pi/6 = 2pi/3.
Therefore zw = 12 e^(i 2pi/3).
Mark notes: 1 mark for multiplying the moduli to get 12, 1 mark for adding the arguments to get 2pi/3, giving zw = 12 e^(i 2pi/3).