Polar and Modulus-Argument Form (AI HL)
Every complex number can be located by a distance from the origin (its modulus) and an angle from the positive real axis (its argument), rather than by its horizontal and vertical parts. This page covers converting between Cartesian and polar or exponential form, with worked examples and the mistakes that lose the most marks. It's part of the broader Complex Numbers topic.
13 questions on this sub-topic.
The two formulas
Covered under IB syllabus reference AHL1.13: modulus-argument (polar) and exponential (Euler) form, and conversion between Cartesian, polar and exponential forms.
Modulus, from Cartesian form
\(|z|=\sqrt{a^2+b^2}\)
Not in the formula booklet - it's just Pythagoras' theorem applied to the real and imaginary parts of \(z=a+bi\).
Polar form
\(z=r(\cos\theta+i\sin\theta)=r\,\mathrm{cis}\,\theta\)
In the formula booklet. The equivalent exponential form \(z=re^{i\theta}\) is also given, and both describe the same point using distance and angle instead of horizontal and vertical parts.
Need the full syllabus wording and formula-booklet reference table? See Complex Numbers.
Worked examples
For \(z = 4 + 3i\), find \(|z|\) and \(\arg z\) (radians, 3 significant figures).
(a)(i) Find \(|z|\).
(a)(ii) Find \(\arg z\).
Worked solution
(a)(i) \(|z| = \sqrt{4^2 + 3^2}\) M1
\(= 5.\) A1
(a)(ii) \(\arg z = \arctan(3/4) = 0.644\) rad. M1 A1
Two AC voltages are \(V_1 = 10e^{i\,0}\) and \(V_2 = 10e^{i\pi/2}\) (volts).
Find the resultant \(V_1 + V_2\) in polar form (modulus and argument).
(a)(i) State the modulus of the resultant.
(a)(ii) State the argument of the resultant.
Worked solution
(a)(i) \(V_1 = 10,\ V_2 = 10i.\) Sum \(= 10 + 10i.\) M1
Modulus \(= \sqrt{200} = 10\sqrt2\) M1
(a)(ii) \(\approx 14.1\) V; A1
argument \(= \tfrac{\pi}{4}.\) M1 A1
Describe geometrically the effect of multiplying a complex number \(z\) by \(i\), and verify for \(z=3+i.\)
Worked solution
Multiplying by \(i = \text{cis}\tfrac{\pi}{2}\). M1
This rotates \(z\) by \(90^\circ\) anticlockwise about the origin (modulus unchanged). A1
\(i(3+i) = 3i - 1 = -1 + 3i.\) M1
Indeed \((3,1) \to (-1,3)\), a \(90^\circ\) rotation. A1 AG
Common mistakes
- Adding complex numbers in polar form directly. Only multiplication, division and powers are simple in polar form - to add or subtract, convert back to Cartesian first, as in the AC voltage example above.
- Using \(\arctan(b/a)\) without checking the quadrant. A calculator's \(\tan^{-1}\) only returns angles between \(-\tfrac{\pi}{2}\) and \(\tfrac{\pi}{2}\), so for \(z\) in the second or third quadrant you must add or subtract \(\pi\) to get the correct argument.
- Giving the argument in the wrong unit. Always check whether the question wants radians or degrees, and make sure your GDC's angle mode matches before you convert or evaluate.
Ready to practise properly?
14 polar-form questions, marked instantly like the real exam.
Quick answers
How do you write a complex number in polar form?
Find the modulus \(r=\sqrt{a^2+b^2}\) and argument \(\theta=\arctan(b/a)\), adjusted for the quadrant \(a+bi\) lies in, then write \(z=r(\cos\theta+i\sin\theta)=r\,\mathrm{cis}\,\theta\), or \(z=re^{i\theta}\) in exponential form.
Why does arctan(b/a) sometimes give the wrong argument?
A calculator's inverse tangent only returns values between \(-\tfrac{\pi}{2}\) and \(\tfrac{\pi}{2}\). For a complex number in the second or third quadrant, you need to add or subtract \(\pi\) to get the correct argument.