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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?

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  • $\begingroup$ There's $i$. What do you expect? $\endgroup$
    – user122283
    May 31, 2014 at 17:08
  • $\begingroup$ if I'm not mistaken it comes from Taylor series expansion if the $\sin x $ is pre-multiplied by $i$. Without the complex unit $e^x$ is expanded in hyperbolic functions. $\endgroup$
    – Alex
    May 31, 2014 at 17:10
  • $\begingroup$ @Alex Do you really need to specify pre-multiplying? Multiplication is commutative in $\mathbb{C}$. $\endgroup$
    – beep-boop
    May 31, 2014 at 17:41
  • $\begingroup$ please look at these 8 videos (it is about one hour) that explains this identity: youtube.com/playlist?list=PLHZZ0otaqNsWV01h2ZssT17Tj8fbtLiuM It is worth watching! $\endgroup$
    – user777
    Aug 7, 2020 at 13:10

4 Answers 4

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

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    $\begingroup$ It can also be your definition $\endgroup$
    – Squirtle
    May 31, 2014 at 17:49
  • $\begingroup$ @Squirtle Yeah, but I prefer to start off with one definition of $e$($:=\sum\limits_{n=0}^{\infty}\frac{1}{n!}$); that way, there are no circular arguments. $\endgroup$
    – beep-boop
    May 31, 2014 at 21:37
  • $\begingroup$ There are NO circular arguments if you are careful. $\endgroup$
    – Squirtle
    May 31, 2014 at 21:44
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    $\begingroup$ @alexqwx : I changed several instances of "..." to \cdots. In standard TeX or LaTeX (as opposed to MathJax, which is used here) if you write "a+...+b", you'll see $a+\text{...}+b$ rather than either $a+\ldots+b$ or $a+\cdots+b$. That usage is considered incorrect by people who care about correct spelling and punctuation and the like. ${}\qquad{}$ $\endgroup$ May 31, 2014 at 21:52
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    $\begingroup$ I use \ldots between commas, as in $(a,\ldots,b)$ and \cdots between binary operators, as in $a+\cdots+b$ or $1\cdot2\cdot3\cdots100$. I think MathJax, and I suspect probably LaTeX, will treat \dots like \cdots if it's between plus or minus signs or the like (e.g. $a<b<\cdots<z$) and like \ldots if it's between commas. But if you write 1+2+3+\dots with nothing after that, I think it treats it as \ldots, but I use \cdots there. If you write 1+2+3+\dots=x, I think then it will treat it like \cdots. ${}\qquad{}$ $\endgroup$ May 31, 2014 at 21:57
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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, $e^{i\theta} = \cos(\theta) + i\sin(\theta).$

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  • $\begingroup$ Has 11.8k reputation.... copies answer already given. $\endgroup$
    – Squirtle
    May 31, 2014 at 17:51
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    $\begingroup$ @Squirtle I don't see how this copies any of the other answers. +1 $\endgroup$ May 31, 2014 at 18:51
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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.

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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.

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