Polar form of a complex number
Learn the polar form of a complex number for Year 11 Specialist Mathematics in Queensland (QCAA). Polar form describes a complex number by its modulus (distance from the origin) and its argument (angle from the positive real axis), written as r cis theta.
You will learn to convert between Cartesian and polar form, quote the principal argument in its correct range, and multiply and divide complex numbers by combining moduli and arguments — the geometry that powers complex arithmetic throughout the course.
Theory
Polar form writes a complex number as \(z=r\,\operatorname{cis}\,\theta\) — its modulus \(r=|z|\) and its argument \(\theta=\arg z\) — in Year 11 Specialist Mathematics (QCAA, Queensland). This page shows how to convert between Cartesian form \(a+bi\) and polar form, keep the argument in its principal range, and multiply and divide in polar form.
Every complex number \(z=a+bi\) is a point on the Argand plane, with the real part \(a\) along the horizontal axis and the imaginary part \(b\) up the vertical axis. Instead of the coordinates \((a,b)\), we can describe that point by how far it is from the origin and which way it points.
The modulus \(|z|=r\) is the distance from the origin to \(z\), so \(r=\sqrt{a^2+b^2}\) and \(r\ge 0\). The argument \(\arg z=\theta\) is the angle the position vector makes, measured anticlockwise from the positive real axis.
Polar form puts these together as \(z=r(\cos\theta+i\sin\theta)\), abbreviated \(z=r\,\operatorname{cis}\,\theta\). The same number in Cartesian form is \(z=a+bi\); polar form and Cartesian form are two ways of naming one point.
An angle repeats every full turn, so \(\arg z\) has infinitely many values \(\theta+2k\pi\). The principal argument \(\operatorname{Arg} z\) is the one in the range \(-\pi<\theta\le\pi\), and it is the value we quote.
From Cartesian to polar, find the modulus with Pythagoras and the argument from the real and imaginary parts (choose the quadrant so \(-\pi<\theta\le\pi\)):
From polar to Cartesian, read off the real and imaginary parts:
To multiply and divide in polar form, combine the moduli and the arguments:
Converting and combining in polar form
- Plot \(z=a+bi\) on the Argand plane so you can see which quadrant it lies in.
- Cartesian to polar: find \(r=\sqrt{a^2+b^2}\), then the argument from \(\cos\theta=\dfrac{a}{r}\) and \(\sin\theta=\dfrac{b}{r}\), choosing \(\theta\) in \(-\pi<\theta\le\pi\).
- Polar to Cartesian: compute \(a=r\cos\theta\) and \(b=r\sin\theta\), keeping surds exact.
- Multiply or divide: multiply (or divide) the moduli and add (or subtract) the arguments, then adjust by \(2\pi\) into the principal range.
Find the modulus with \(r=\sqrt{a^2+b^2}\):
| \(r\) | \(=\) | \(\sqrt{3^2+3^2}\) |
| \(=\) | \(\sqrt{9+9}\) | |
| \(=\) | \(\sqrt{18}\) | |
| \(=\) | \(3\sqrt{2}\) |
Both parts are positive, so \(z\) is in the first quadrant:
| \(\tan\theta\) | \(=\) | \(\dfrac{3}{3}\) |
| \(=\) | \(1\) | |
| \(\theta\) | \(=\) | \(\dfrac{\pi}{4}\) |
Assemble the polar form:
| \(z\) | \(=\) | \(3\sqrt{2}\,\operatorname{cis}\dfrac{\pi}{4}\) |
\(z=3\sqrt{2}\,\operatorname{cis}\dfrac{\pi}{4}\).
Find the real part with \(a=r\cos\theta\):
| \(a\) | \(=\) | \(4\cos\dfrac{\pi}{6}\) |
| \(=\) | \(4\times\dfrac{\sqrt{3}}{2}\) | |
| \(=\) | \(2\sqrt{3}\) |
Find the imaginary part with \(b=r\sin\theta\):
| \(b\) | \(=\) | \(4\sin\dfrac{\pi}{6}\) |
| \(=\) | \(4\times\dfrac{1}{2}\) | |
| \(=\) | \(2\) |
Assemble the Cartesian form:
| \(z\) | \(=\) | \(2\sqrt{3}+2i\) |
\(z=2\sqrt{3}+2i\).
Find the modulus:
| \(r\) | \(=\) | \(\sqrt{(-3)^2+(\sqrt{3})^2}\) |
| \(=\) | \(\sqrt{9+3}\) | |
| \(=\) | \(\sqrt{12}\) | |
| \(=\) | \(2\sqrt{3}\) |
Here \(a<0\) and \(b>0\), so \(z\) is in the second quadrant. Place \(\theta\) with \(\cos\theta\) and \(\sin\theta\):
| \(\cos\theta\) | \(=\) | \(\dfrac{-3}{2\sqrt{3}}=-\dfrac{\sqrt{3}}{2}\) |
| \(\sin\theta\) | \(=\) | \(\dfrac{\sqrt{3}}{2\sqrt{3}}=\dfrac{1}{2}\) |
| \(\theta\) | \(=\) | \(\dfrac{5\pi}{6}\) |
Assemble the polar form:
| \(z\) | \(=\) | \(2\sqrt{3}\,\operatorname{cis}\dfrac{5\pi}{6}\) |
\(z=2\sqrt{3}\,\operatorname{cis}\dfrac{5\pi}{6}\).
Multiply the moduli:
| \(\text{modulus}\) | \(=\) | \(2\times 4\) |
| \(=\) | \(8\) |
Add the arguments:
| \(\text{argument}\) | \(=\) | \(\dfrac{3\pi}{4}+\dfrac{\pi}{2}\) |
| \(=\) | \(\dfrac{3\pi}{4}+\dfrac{2\pi}{4}\) | |
| \(=\) | \(\dfrac{5\pi}{4}\) |
\(\dfrac{5\pi}{4}>\pi\), so subtract \(2\pi\) to reach the principal value:
| \(\dfrac{5\pi}{4}-2\pi\) | \(=\) | \(-\dfrac{3\pi}{4}\) |
| \(\text{product}\) | \(=\) | \(8\,\operatorname{cis}\left(-\dfrac{3\pi}{4}\right)\) |
\(8\,\operatorname{cis}\left(-\dfrac{3\pi}{4}\right)\).
Common pitfalls
Frequently asked questions
What is the polar form of a complex number?
It writes \(z\) as \(r\,\operatorname{cis}\,\theta=r(\cos\theta+i\sin\theta)\), where \(r=|z|\) is the modulus and \(\theta=\arg z\) is the argument.
How do you find the modulus and argument of a complex number?
The modulus is \(|z|=\sqrt{a^2+b^2}\). The argument satisfies \(\cos\theta=\dfrac{a}{r}\) and \(\sin\theta=\dfrac{b}{r}\); choose \(\theta\) in the correct quadrant.
What is the principal argument?
It is the value of \(\arg z\) in the range \(-\pi<\theta\le\pi\). Because angles repeat every \(2\pi\), \(\arg z=\operatorname{Arg} z+2k\pi\); the principal value is the one we quote.
How do you multiply complex numbers in polar form?
Multiply the moduli and add the arguments: \(z_1z_2=r_1r_2\,\operatorname{cis}(\theta_1+\theta_2)\). For division, divide the moduli and subtract the arguments.
How do you convert from polar form to Cartesian form?
Use \(a=r\cos\theta\) and \(b=r\sin\theta\), then write \(z=a+bi\), keeping surds exact.
What does cis mean?
\(\operatorname{cis}\theta\) is shorthand for \(\cos\theta+i\sin\theta\), so \(r\,\operatorname{cis}\,\theta=r(\cos\theta+i\sin\theta)\).