2.2 The polar form of complex numbers and Euler’s formula
It turns out that to discuss the geometric meaning of multiplication it is useful to switch the viewpoint to the polar coordinates representation of complex numbers.
Take a complex number \(z=x+\mathrm{i}y \neq 0\), which has associated point \(v(z)=(x,y)\) in \(\mathbb{R}^2\). By Theorem 1.18 we can write \[v(z) = \lambda u(\theta) = \lambda (\cos \theta, \sin \theta)\] where \(\lambda = \sqrt{x^2+y^2}=|z|\) is the distance from \((x,y)\) to the origin, and \(\theta\) is the angle that the line connecting the origin with \((x,y)\) makes with the \(x\)-axis. As in Theorem 1.18, to determine \(\theta\) we solve \(\cos\theta = \frac{x}{\sqrt{x^2+y^2}}\) and \(\sin\theta = \frac{y}{\sqrt{x^2+y^2}}\). We call \(\theta\) the argument (or phase) of \(z\). Writing this in terms of complex numbers, \[z=x+\mathrm{i} y = \sqrt{x^2+y^2} \left( \frac{x}{\sqrt{x^2+y^2}} + \mathrm{i}\frac{y}{\sqrt{x^2+y^2}}\right) = |z|(\cos \theta + \mathrm{i}\sin \theta).\] Note that \(z=0\) is special: its argument is not defined.
There is a different way to view this through an important connection with the exponential function. The exponential function \(\mathrm{e}^z\) for a complex number \(z\) is defined by the series \[\begin{equation} \mathrm{e}^z:=1+z+\frac{1}{2} z^2+\frac{1}{3!} z^3+\frac{1}{4!}z^4+\cdots =\sum_{n=0}^{\infty} \frac{1}{n!}z^n\tag{2.1}\end{equation}\] This definition covers the case \(z\in \mathbb{C}\), since we can compute powers \(z^n\) of \(z\) and we can add complex numbers.2
We will use that for arbitrary complex \(z_1,z_2\) the exponential function satisfies \[\begin{equation} \mathrm{e}^{z_1}\mathrm{e}^{z_2}=\mathrm{e}^{z_1+z_2}.\tag{2.2}\end{equation}\] (The proof of this fact for real numbers also works for complex numbers.)
This leads to the following important relationship between different forms of complex numbers.
For any \(\theta \in \mathbb{R}\) we have \[\label{eq:cossinexp} \mathrm{e}^{\mathrm{i}\theta}=\cos\theta +\mathrm{i}\sin\theta.\]
This is basically a calculus result, we will sketch the proof, but you need more calculus to fully justify it. Similarly to (2.1), the sine function and the cosine function can be defined by the following power series \[\begin{aligned} \sin(z)&=z-\frac{1}{3!} z^3+\frac{1}{5!} z^5 -\cdots =\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!} z^{2k+1} \quad \text{and}\\ \cos(z)&=1-\frac{1}{2}z^2+\frac{1}{4!}z^4-\cdots =\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k)!}z^{2k} . \end{aligned}\] Now we use (2.1) with \(z=\mathrm{i}\theta\), and since \((\mathrm{i}\theta)^2=-\theta^2\), \((\mathrm{i}\theta)^3=-\mathrm{i}\theta^3\), \((\mathrm{i}\theta)^4=\theta^4\), \((\mathrm{i}\theta)^5=\mathrm{i}\theta^5\), and so on, we find by comparing the power series \[\begin{aligned} \mathrm{e}^{\mathrm{i}\theta}&=1+\mathrm{i}\theta -\frac{1}{2}\theta^2-\mathrm{i}\frac{1}{3!}\theta^3+\frac{1}{4!}\theta^4+\mathrm{i}\frac{1}{5!} \theta^5 +\cdots\\ &=\bigg[1-\frac{1}{2}\theta^2+\frac{1}{4!}\theta^4+\cdots\bigg]+\mathrm{i}\bigg[\theta -\frac{1}{3!}\theta^3+\frac{1}{5!} \theta^5+\cdots\bigg] =\cos\theta+\mathrm{i}\sin\theta . \end{aligned}\]
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Using Euler’s formula, in the notation of Theorem 1.18, we see that \[v(\mathrm{e}^{\mathrm{i}\theta})=u(\theta) .\] In particular we can write any complex number \(z\), \(z\neq 0\), in polar coordinate form \[z=\lambda \mathrm{e}^{\mathrm{i}\theta} ,\] where \(\lambda=\lvert z\rvert\) is the modulus (or amplitude) and \(\theta\) is the argument (or phase).
For the multiplication of nonzero complex numbers we find then that if \(z_1=\lambda_1\mathrm{e}^{\mathrm{i}\theta_1}\) and \(z_2=\lambda_2\mathrm{e}^{\mathrm{i}\theta_2}\) then \[z_1z_2=\lambda_1\lambda_2\mathrm{e}^{\mathrm{i}(\theta_1+\theta_2)} \quad \text{and} \quad \frac{z_1}{z_2}= \frac{\lambda_1}{\lambda_2}\mathrm{e}^{\mathrm{i}(\theta_1-\theta_2)} ,\] so multiplication corresponds to adding the arguments and multiplying the modulus. In particular if \(\lambda=1\), then multiplying by \(\mathrm{e}^{\mathrm{i}\theta}\) corresponds to rotation by \(\theta\) in the complex plane.
Using the polar form of complex numbers also makes taking large powers much easier. By Equation (2.2) we have for \(n \in \mathbb{N}\) that \((\mathrm{e}^{\mathrm{i}\theta})^n = \mathrm{e}^{\mathrm{i}n \theta}\), and since \(\mathrm{e}^{\mathrm{i}\theta} = \cos \theta + \mathrm{i}\sin \theta\) and \(\mathrm{e}^{\mathrm{i}n \theta} = \cos (n\theta) + \mathrm{i}\sin (n\theta)\) this gives us the following identity which is known as de Moivre’s Theorem .
Let \(n \in \mathbb{N}\) and \(\theta \in \mathbb{R}\). Then we have \[(\cos\theta+\mathrm{i}\sin\theta)^n=\cos(n\theta)+\mathrm{i}\sin(n \theta).\]
These results have some nice applications to trigonometric functions.
We can use de Moivre’s Theorem to derive trigonometric identities. For instance, if we let \(n=2\), and multiply out the left hand side, we obtain \(\cos^2\theta+2\mathrm{i}\sin\theta \cos \theta-\sin^2\theta=\cos(2\theta)+\mathrm{i}\sin(2 \theta)\) and separating real and imaginary part leads to the two angle doubling identities \[\cos(2\theta)=\cos^2\theta-\sin^2\theta,\quad\sin(2\theta)=2\sin\theta\cos\theta.\] Similar identities can be derived for larger \(n\).
If we use \(\mathrm{e}^{\mathrm{i}\theta}\mathrm{e}^{-\mathrm{i}\varphi}=\mathrm{e}^{\mathrm{i}(\theta-\varphi)}\) and apply Euler’s formula to both sides we obtain \((\cos \theta+\mathrm{i}\sin \theta)(\cos \varphi-\mathrm{i}\sin\varphi)=\cos(\theta-\varphi)+\mathrm{i}\sin(\theta-\varphi)\) and multiplying out the left hand side gives the two relations \[\label{eq:ang_diff} \cos(\theta-\varphi)=\cos \theta\cos \varphi+\sin\theta\sin\varphi ,\quad \sin(\theta-\varphi)=\sin \theta\cos \varphi-\cos \theta \sin \varphi .\]
Euler’s formula can be used to obtain the following standard representations for the sine and cosine functions: \[\sin\theta=\frac{\mathrm{e}^{\mathrm{i}\theta}-\mathrm{e}^{-\mathrm{i}\theta}}{2\mathrm{i}} ,\quad \cos\theta=\frac{\mathrm{e}^{\mathrm{i}\theta}+\mathrm{e}^{-\mathrm{i}\theta}}{2} .\]
Polar form also lets us deduce some results about complex numbers that may feel surprising, such as below.
We ignore the issue of convergence here, but this sum is convergent for all \(z \in \mathbb{C}\).↩︎