$e^{i\theta}$ $=$ $\cos \theta + i \sin \theta$, a definition or theorem?

Question

My question is simply whether the well-known formula $e^{i \theta}$ $=$ $\cos \theta$ $+$ $i \sin \theta$ a definition or there is some proof of the result.

It seems to me that the formula is a definition (as is the case with the definition of $e$ from which the definition of $e^x$ can easily be derived). But if I try to form the definition in the same manner I have to us the definition from Complex Analysis. Is there any way to prove the result without using any ideas from Complex Analysis?

Answer 1

In Function Theory of One Complex Variable by Robert E Greene and Steven G Krantz, they define

$$e^{iy}=\cos(y)+i\sin(y)$$

And if $x$ is real

$$e^x=\sum_{j=0}^{\infty} \frac{x^j}{j!}$$

But then they prove later that their definition isn't circular while discussing zeros of holomorphic functions. The point is, this is a "formal definition" and that's fine because you won't get any contradictions and more importantly we can (if desired) prove something else that we would normally consider an axiom. In other words, it's possible to switch the role of an axiom and a corollary as long as they are a tautology in some sense and one isn't in fact so much more fundamental.

Answer 2

It's a theorem.

Assuming that $e^z$ is defined as $\sum\limits_{n=0}^{\infty}\frac{z^n}{n!}$ (remember that there are a few ways of defining $e$), we have:

$$e^{i\theta}:=1+i\theta+\frac{\theta^2i^2}{2!}+\frac{\theta^3i^3}{3!}+\frac{\theta^4i^4}{4!}+\frac{\theta^5i^5}{5!}+\cdots=$$

Simplifying this, we find that $$\boxed{e^{i\theta}\equiv1+i\theta-\frac{\theta^2}{2!}-\frac{i\theta^3}{3!}+\frac{\theta^4}{4!}+\frac{i\theta^5}{5!}-\cdots}$$

Now, $\color{blue}{\underbrace{\cos(\theta)=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+\cdots}_{\text{Taylor expansion of} \cos(\theta)}}$.

Also, $\underbrace{\sin(\theta)=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}+\cdots}_{\text{Taylor expansion of} \sin(\theta)}\iff \color{green}{i\sin(\theta)=i\theta-\frac{i\theta^3}{3!}+\frac{i\theta^5}{5!}-\frac{i\theta^7}{7!}+\cdots}$.

Now, $\color{blue}{\cos(\theta)}\color{\green}{+i\sin(\theta)}=\color{blue}{1}\color{green}{+i\theta}\color{blue}{-\frac{\theta^2}{2!}}\color{green}{-\frac{i\theta^3}{3!}}\color{blue}{+\frac{\theta^4}{4!}}\color{green}{+\frac{i\theta^5}{5!}}\color{blue}{-}\cdots=e^{i\theta}$.

Answer 3

It depends on how you define $e$, $\cos$, and $\sin$!

You can define $$ e^{i\theta} = \cos(\theta) + i\sin(\theta).$$ In that case, you need to go on and show that your other definition of the exponential for real numbers gives you an equivalent result when extended to the entire complex plane.

Alternatively, you can define $$ e^z = \sum_{n=0}^\infty \frac{z^n}{n!}.$$ In this case, you need to prove (or define!) the infinite series expansions of $\cos$ and $\sin$, show that everything always converges, then show that when restricted to a purely imaginary argument, $e^{i\theta} = \cos(\theta) + i\sin(\theta).$

Answer 4

I think if one sets up analysis for aiming at economy of definition (rather than pedagogy), it would be natural to introduce the exponential function first (through the differential equation $f'=f$, or by its series) and then define cosine and sine by $$ \cos(z)=\frac{\exp(z)+\exp(-z)}2 \qquad\text{and}\qquad \sin(z)=\frac{\exp(z)-\exp(-z)}{2\mathbf i}. $$ Now your result is a theorem, but a pretty obvious one proved by elementary algebra from the definitions.