Polar / Modulus-Argument Form (AA HL)
Once you can plot a complex number on the Argand diagram, its distance from the origin (the modulus) and its angle from the positive real axis (the argument) give an alternative way to write it - and multiplying or dividing becomes much quicker in this form than in Cartesian. This page covers converting between forms, combining complex numbers, and describing regions in the Argand plane. It's part of the broader Complex Numbers topic.
22 questions on this sub-topic.
The key form
Covered under IB syllabus reference AHL1.13: modulus-argument (polar) form and Euler form. You're expected to convert freely between Cartesian, polar and Euler forms, and to interpret sums, products and quotients geometrically.
Write in polar form
\(z=r(\cos\theta+i\sin\theta)=r\,\text{cis}\,\theta.\)
In the formula booklet. \(r=|z|\) is the modulus, \(\theta=\arg z\) is the argument, usually given in \((-\pi,\pi]\).
Modulus of \(z=a+bi\)
\(|z|=\sqrt{a^2+b^2}\)
Not in the booklet - derived directly from Pythagoras' theorem applied to the point \((a,b)\) on the Argand diagram.
Need the full syllabus wording and formula-booklet reference table? See Complex Numbers. For calculator-specific tips, see the parent topic's GDC guidance.
Worked examples
Find the argument of \(z=1+i\), giving your answer in radians.
Worked solution
\(z = 1+i\) is in the first quadrant with \(\tan\theta = \dfrac{1}{1}\) M1
\(= 1.\) A1
So \(\arg z = \dfrac{\pi}{4}.\) A1
Given \(z_1 = 2e^{i\pi/6}\) and \(z_2 = 3e^{i\pi/4}\), find \(z_1 z_2\) and \(\dfrac{z_2}{z_1}\) in the form \(re^{i\theta}.\)
Worked solution
(multiply moduli, add arguments): \(z_1 z_2 = 6e^{i(\pi/6 + \pi/4)}\) M1 \(= 6e^{i\,5\pi/12}.\) A1 A1
(divide moduli, subtract arguments): \(\dfrac{z_2}{z_1} = \tfrac32 e^{i(\pi/4 - \pi/6)}\) M1 A1 \(= \tfrac32 e^{i\,\pi/12}.\) A1
Sketch (describe) the region in the Argand plane defined by \(1\le|z-i|\le3.\)
Worked solution
\(|z - i|\) is the distance from \(z\) to the point \(i\) M1 \(= (0,1).\) A1
\(|z-i| \ge 1\) (outside the circle of radius 1) M1 and \(|z-i| \le 3\) (inside radius 3). A1
The region is an annulus (ring) centred at \((0,1)\), inner radius 1 and outer radius 3. A1 Both boundary circles are included. R1
Common mistakes
- Forgetting to check the quadrant when finding the argument. A raw \(\arctan(b/a)\) only gives the right angle directly in the first and fourth quadrants - always sketch the point first.
- Giving an argument outside the principal range. Arguments should usually be given with \(-\pi<\theta\le\pi\) - an angle like \(\tfrac{5\pi}{3}\) should be rewritten as \(-\tfrac{\pi}{3}.\)
- Treating the modulus as additive. \(|z_1+z_2|\) is not the same as \(|z_1|+|z_2|\) - modulus only behaves simply under multiplication and division, not addition.
Ready to practise properly?
26 polar-form complex-number questions, marked instantly like the real exam.
Quick answers
What is the modulus-argument (polar) form of a complex number?
\(z=r(\cos\theta+i\sin\theta)=r\,\text{cis}\,\theta\), where \(r=|z|\) is the modulus (distance from the origin) and \(\theta=\arg z\) is the argument (angle from the positive real axis).
How do you multiply complex numbers in polar form?
Multiply the moduli and add the arguments: if \(z_1=r_1\,\text{cis}\,\theta_1\) and \(z_2=r_2\,\text{cis}\,\theta_2\), then \(z_1z_2=r_1r_2\,\text{cis}(\theta_1+\theta_2)\). To divide, divide the moduli and subtract the arguments.