Exponential form

Euler’s formula connects the exponential and trigonometric forms of a complex number: \(e^{i\theta}=\cos\theta+i\sin\theta\). The following series argument is an optional connection with Maclaurin series.

The exponential notation and its laws are used in AA HL. Extending the exponential function to complex inputs through an infinite series gives the additional justification below; a full complex-analysis treatment is beyond the required course.

The motivation begins from an observation of the identity

\[(\cos \theta + i \sin \theta)(\cos \phi + i \sin \phi) = \cos (\theta + \phi) + i \sin (\theta + \phi).\]

The multiplication of the complex numbers of the form \(\cos \theta + i \sin \theta\) (where \(\theta \in \mathbb{R}\)) clearly resulted in a complex number of the same form. In particular, the parameters \(\theta\) and \(\phi\) (which we know to be the arguments) got added together. This likens to a familiar situation in the following law of indices:

\[a^m \cdot a^n = a^{m+n}.\]

The addition of angles under multiplication suggests an exponential notation. Comparing the familiar Maclaurin series explains why the natural base \(e\) appears.

The Maclaurin series for \(\cos x\), \(\sin x\) and \(e^x\) have related coefficients:

\[\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots\]

and

\[\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots\]

Furthermore,

\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + \frac{x^7}{7!} + \dots\]

As an extension, define the complex exponential by its power series. It converges absolutely for every complex input, which justifies grouping its real and imaginary terms. With \(x=i\theta\), \(\theta\in\mathbb R\),

\[e^{i\theta} = \sum_{k=0}^{\infty} \frac{(i\theta)^k}{k!}.\]

We recall that

\[i^n = \begin{cases} 1 & \text{if}\hspace{0.5em} n \equiv 0 \pmod{4}; \\ i & \text{if}\hspace{0.5em} n \equiv 1 \pmod{4}; \\ -1 & \text{if}\hspace{0.5em} n \equiv 2 \pmod{4}; \\ -i & \text{if}\hspace{0.5em} n \equiv 3 \pmod{4}; \end{cases}\]

Group the indices \(k\) by their four possible remainders on division by \(4\):

\[\begin{aligned}e^{i\theta}=\sum_{r=0}^{\infty}\Bigg[&\frac{(i\theta)^{4r}}{(4r)!}+\frac{(i\theta)^{4r+1}}{(4r+1)!}\\&+\frac{(i\theta)^{4r+2}}{(4r+2)!}+\frac{(i\theta)^{4r+3}}{(4r+3)!}\Bigg].\end{aligned}\]

Note that

\[(i\theta)^{4r} = i^{4r} \theta^{4r} = (i^4)^r \theta^{4r} = \theta^{4r},\]

and

\[(i\theta)^{4r+2} = i^{4r+2} \theta^{4r+2} = (i^4)^r \cdot i^2 \theta^{4r+2} = -\theta^{4r+2},\]

which are purely real, while

\[(i\theta)^{4r+1} = i^{4r+1} \theta^{4r+1} = (i^4)^r \cdot i \theta^{4r+1} = i\theta^{4r+1},\]

and

\[(i\theta)^{4r+3} = i^{4r+3} \theta^{4r+3} = (i^4)^r \cdot i^3 \theta^{4r+3} = -i\theta^{4r+3},\]

which are purely imaginary.

So, the real part of \(e^{i\theta}\) is

\[\sum_{r=0}^{\infty} \left[ \frac{\theta^{4r}}{(4r)!} - \frac{\theta^{4r+2}}{(4r+2)!} \right],\]

of which all powers of \(\theta\) being even and consists of alternating signs, may be rewritten as

\[\sum_{n=0}^{\infty} (-1)^n \frac{\theta^{2n}}{(2n)!} = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \dots.\]

The imaginary part of \(e^{i\theta}\) is

\[\sum_{r=0}^{\infty} \left[ \frac{\theta^{4r+1}}{(4r+1)!} - \frac{\theta^{4r+3}}{(4r+3)!} \right].\]

Note that this time all the power of \(\theta\) are odd but the sign continues to alternate. So, we may rewrite the above as:

\[\sum_{m=1}^{\infty} (-1)^{m-1} \frac{\theta^{2m-1}}{(2m-1)!} = \frac{\theta}{1!} - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots.\]

These are precisely the standard Maclaurin series for cosine and sine:

\[\cos \theta = \sum_{n=0}^{\infty} (-1)^n \frac{\theta^{2n}}{(2n)!} = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \dots.\]

\[\sin \theta = \sum_{m=1}^{\infty} (-1)^{m-1} \frac{\theta^{2m-1}}{(2m-1)!} = \frac{\theta}{1!} - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots.\]

Combining the real and the imaginary parts, we obtain the famous Euler’s formula:

\[e^{i\theta} = \cos \theta + i \sin \theta, \quad \theta \in \mathbb{R}.\]

Remark. Euler’s formula explains the notation \(re^{i\theta}=r(\cos\theta+i\sin\theta)\). For AA HL calculations, use this identity together with the modulus and angle rules; the complex-series justification above is an optional extension.