1 Series expansions for exponential and trigonometric functions

We have, so far, considered two ways of representing a complex number:

$z=a+ib\text{Cartesian form}$

or

$z=r\left(cos\theta +isin\theta \right)\text{polar form}$

In this Section we introduce a third way of denoting a complex number: the exponential form .

If $x$ is a real number then, as we shall verify in HELM booklet  16, the exponential number $\text{e}$ raised to the power $x$ can be written as a series of powers of $x$ :

${\text{e}}^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\⋯$

in which $n!=n\left(n-1\right)\left(n-2\right)\dots \left(3\right)\left(2\right)\left(1\right)$ is the factorial of the integer $n$ . Although there are an infinite number of terms on the right-hand side, in any practical calculation we could only use a finite number. For example if we choose $x=1$ (and taking only six terms) then

which is fairly close to the accurate value of $\text{e}=2.71828$ (to 5 d.p.)

We ask you to accept that ${\text{e}}^x$ , for any real value of $x$ , is the same as $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\⋯$ and that if we wish to calculate ${\text{e}}^x$ for a particular value of $x$ we will only take a finite number of terms in the series. Obviously the more terms we take in any particular calculation the more accurate will be our calculation.

As we shall also see in HELM booklet  16, similar series expansions exist for the trigonometric functions $sinx$ and $cosx$ :

$sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\⋯$

$cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\⋯$

in which $x$ is measured in radians.

The observant reader will see that these two series for $sinx$ and $cosx$ are similar to the series for ${\text{e}}^x$ . Through the use of the symbol $i$ (where $i^2=-1$ ) we will examine this close correspondence.

In the series for ${\text{e}}^x$ replace $x$ on both left-hand and right-hand sides by $i\theta$ to give:

${\text{e}}^{i\theta }=1+\left(i\theta \right)+\frac{{\left(i\theta \right)}^2}{2!}+\frac{{\left(i\theta \right)}^3}{3!}+\frac{{\left(i\theta \right)}^4}{4!}+\frac{{\left(i\theta \right)}^5}{5!}+\⋯$

Then, as usual, replace every occurrence of $i^2$ by $-1$ to give

${\text{e}}^{i\theta }=1+i\theta -\frac{{\theta }^2}{2!}-i\frac{{\theta }^3}{3!}+\frac{{\theta }^4}{4!}+i\frac{{\theta }^5}{5!}+\⋯$

which, when re-organised into real and imaginary terms gives, finally:

Example 5

Find complex number expressions, in Cartesian form, for

1. ${\text{e}}^{i\pi ∕4}$
2. ${\text{e}}^{-i}$
3. ${\text{e}}^{i\pi }$

We use Key Point 8:

Solution

1. ${\text{e}}^{i\pi ∕4}=cos\frac{\pi }{4}+isin\frac{\pi }{4}=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$
2. ${\text{e}}^{-i}=cos\left(-1\right)+isin\left(-1\right)=0.540-i\left(0.841\right)$ don’t forget: use radians
3. ${\text{e}}^{i\pi }=cos\pi +isin\pi =-1+i\left(0\right)=-1$