Exponential form

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Section 7.4 Exponential form

De Moivre's Theorem allows us to calculate powers \(z^n\) for \(n\in\mathbb{N}\text{,}\) but what about more a general exponential such as \(e^z\) when \(z\in\CC\text{.}\) Euler's formula provides the answer.

Theorem 7.4.1. Euler's formula.

\begin{equation*} \exp(i \theta) = e^{i\theta} = \cos(\theta)+i\sin(\theta)\text{.} \end{equation*}

Euler's formula is one of the most remarkable and important formulae in mathematics. E.g. the special case \(\exp(i\pi)=-1\) links together \(i\) (algebra), \(\pi\) (geometry) and \(e\) (calculus) in a single equation. Euler's formula justifies the definition of exponential form of a complex number \(z\) as.

\begin{equation} z= r\exp(i\theta) = r(\cos \theta + i \sin \theta)\text{.}\label{e_expform}\tag{7.4.1} \end{equation}

Exponential notation is a very convenient form for writing complex numbers.

There are many approaches to proving Euler's formula. The easiest is to use the power series representation for \(\exp(z)=\sum_{n=0}^{\infty}z^n/n!\text{,}\) and observing that

\begin{equation*} \exp(i\theta)=\sum_{n=0}^{\infty}\frac{1}{n!}(i\theta)^n=\sum_{j=0}^{\infty}\frac{(-1)^{j}}{(2j)!}\theta^{2j}+i\sum_{k=0}^{\infty}\frac{(-1)^{2k+1}}{(2k+1)!}\theta^{2k+1}\text{,} \end{equation*}

and recognizing the power series for \(\cos\) and \(\sin\text{.}\) In order to justify this manipulation, you'll some ideas from analysis; so a complete proof will only come later.

Exponential notation makes taking a power of a complex number much easier:

Example 7.4.2.

What is \((1+i)^{6}\text{?}\)

We could multiply this out by using the Binomial Theorem, but we'll use polar coordinates instead: To write \(1+i\) in polar form \(re^{i\theta}\text{,}\) we first find \(r\text{:}\)

\begin{equation*} r=|1+i|=\sqrt{1^2+1^2}=\sqrt{2}\text{.} \end{equation*}

Then

\begin{equation*} \exp(i\theta) = \frac{1+i}{r} =\frac{1+i}{\sqrt{2}} = \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4} + i\sin\frac{\pi}{4}=\exp(i\frac{\pi}{4})\text{.} \end{equation*}

Finally,

\begin{equation*} (1+i)^{6}= (\sqrt{2})^{6} \exp(i\frac{6\pi}{4}) = 8\left(\cos \frac{6\pi}{4}+i\sin \frac{6\pi}{4}\right) =8\left(\cos \frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right)= -8i\text{.} \end{equation*}