Trigonometric form of complex numbers
A complex number has the form ${a+bi}$, where ${a}$ and ${b}$ are real numbers and ${i^2=-1}$. (They are introduced in Complex numbers in College Algebra.) This page draws complex numbers as points in a plane and describes them with a length and an angle. In that form, multiplying and dividing become simple.
The complex plane
A complex number ${a+bi}$ is made of two real numbers, ${a}$ and ${b}$. So we can plot it as the point ${(a,b)}$. The horizontal axis is called the real axis, and the vertical axis is the imaginary axis. This is the complex plane.
Modulus and argument
Draw an arrow from ${0}$ to the point ${z=a+bi}$. Its length is the modulus (or absolute value) of ${z}$:
${|z|=r=\sqrt{a^2+b^2}}$
The angle ${\theta}$ from the positive real axis to the arrow is an argument of ${z}$. It satisfies ${\tan\theta=\dfrac{b}{a}}$, with the quadrant of ${(a,b)}$ deciding which angle.
Exactly as with polar coordinates, ${a=r\cos\theta}$ and ${b=r\sin\theta}$. Putting these into ${z=a+bi}$ gives the trigonometric form (or polar form) of ${z}$:
${z=r(\cos\theta+i\sin\theta)}$
An argument is not unique: adding ${2\pi}$ gives another. We usually choose ${\theta}$ between ${0}$ and ${2\pi}$.
Example 1: Write each number in trigonometric form.
(a) ${1+i}$ (b) ${-\sqrt{3}+i}$ (c) ${-3i}$
Solution:
(a) ${r=\sqrt{1+1}=\sqrt{2}}$. The point ${(1,1)}$ is in quadrant I with ${\tan\theta=1}$, so ${\theta=\dfrac{\pi}{4}}$:
$\begin{align*}&1+i\\&=\sqrt{2}\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)\end{align*}$
(b) ${r=\sqrt{3+1}=2}$. The point ${(-\sqrt{3},1)}$ is in quadrant II with reference angle ${\dfrac{\pi}{6}}$, so ${\theta=\dfrac{5\pi}{6}}$:
$\begin{align*}&-\sqrt{3}+i\\&=2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)\end{align*}$
(c) The point ${(0,-3)}$ is on the negative imaginary axis, so ${r=3}$ and ${\theta=\dfrac{3\pi}{2}}$:
$\begin{align*}&-3i\\&=3\left(\cos\dfrac{3\pi}{2}+i\sin\dfrac{3\pi}{2}\right)\end{align*}$
Example 2: Write $4\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)$ in the form ${a+bi}$.
Solution:
$\begin{align*}&4\left(-\dfrac{1}{2}+i\cdot\dfrac{\sqrt{3}}{2}\right)\\&=-2+2\sqrt{3}\,i\end{align*}$
Multiplying and dividing
Let ${z_1=r_1(\cos\theta_1+i\sin\theta_1)}$ and ${z_2=r_2(\cos\theta_2+i\sin\theta_2)}$. Then
$\begin{align*}&z_1z_2\\&=r_1r_2\left[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\right]\end{align*}$
$\begin{align*}&\dfrac{z_1}{z_2}\\&=\dfrac{r_1}{r_2}\left[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right]\end{align*}$
In words: to multiply, multiply the moduli and add the arguments. To divide, divide the moduli and subtract the arguments. Multiplying by a complex number stretches and rotates.
Why it is true. Multiply out
$(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)$
and use ${i^2=-1}$. The real part is $\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2$, which is ${\cos(\theta_1+\theta_2)}$. The imaginary part is $\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2$, which is ${\sin(\theta_1+\theta_2)}$. These are the sum formulas from Chapter 3.
Example 3: Let
$z_1=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)$
$z_2=\dfrac{3}{2}\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)$
Find ${z_1z_2}$ and ${\dfrac{z_1}{z_2}}$.
Solution:
Product: Multiply the moduli, ${2\cdot\dfrac{3}{2}=3}$, and add the arguments, $\dfrac{\pi}{3}+\dfrac{\pi}{6}=\dfrac{\pi}{2}$:
$\begin{align*}z_1z_2&=3\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)\\&=3i\end{align*}$
Quotient: Divide the moduli, ${2\div\dfrac{3}{2}=\dfrac{4}{3}}$, and subtract the arguments, $\dfrac{\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{6}$:
$\begin{align*}\dfrac{z_1}{z_2}&=\dfrac{4}{3}\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)\\&=\dfrac{4}{3}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i\right)\\&=\dfrac{2\sqrt{3}}{3}+\dfrac{2}{3}i\end{align*}$
Example 4: Multiply ${(1+i)(-\sqrt{3}+i)}$ in two ways, and compare.
Solution:
Directly:
$\begin{align*}&-\sqrt{3}+i-\sqrt{3}\,i+i^2\\&=(-\sqrt{3}-1)+(1-\sqrt{3})i\end{align*}$
In trigonometric form (from Example 1): multiply the moduli, ${\sqrt{2}\cdot 2=2\sqrt{2}}$, and add the arguments, $\dfrac{\pi}{4}+\dfrac{5\pi}{6}=\dfrac{13\pi}{12}$:
$2\sqrt{2}\left(\cos\dfrac{13\pi}{12}+i\sin\dfrac{13\pi}{12}\right)$
With a calculator, both are about ${-2.732-0.732i}$.
For a single product, the direct way is often quicker. The trigonometric form shines for powers and roots, the topic of the next lesson.
Summary
- The complex number ${a+bi}$ is the point ${(a,b)}$ in the complex plane.
- Modulus ${r=\sqrt{a^2+b^2}}$. Argument ${\theta}$: ${\tan\theta=\dfrac{b}{a}}$, with the quadrant of ${(a,b)}$.
- Trigonometric form: ${z=r(\cos\theta+i\sin\theta)}$.
- To multiply, multiply the moduli and add the arguments. To divide, divide the moduli and subtract the arguments.