Edit re the close vote: I guess this 1 of those questions whose on-topic-ness depends on the answer. If the answer is no, then well maybe it's off-topic. But if the answer is yes, then I believe it's very on-topic.

More specific version of: Should Euler's formula $e^{ix}=\cos x+i\sin x$ be seen as a definition rather than something to prove?

Euler's formula (...'for complex numbers'?) is simply

$$e^{iz} = \cos(z)+i\sin(z)$$

And 1 way to prove this is using complex Maclaurin series for the complex exponential, sine and cosine.

Now, there's a way to 'prove' the 'theorem' $e^{it} = \cos(t)+i\sin(t)$, which is often said to be 'Euler's formula (...for real numbers?)' without using the original formula above such as by using Maclaurin. And ok fine, I guess you can indeed get the Maclaurin series expansion of $e^z$ by defining $e^z$ simply as the unique solution of $g:\mathbb C \to \mathbb C, g'=g,g(0)=1$ without explicitly getting $e^z = e^x(\cos(y)+i\sin(y))$ then you can just let $z=it$ (i.e. let $(x,y)=(0,t)$).


There's a 'proof' that Wikipedia does, again without using the original formula, 'by differentiation' (proofwiki does this too) where we consider the function $f:\mathbb R \to \mathbb C,f(t) = \frac{\cos(t)+i\sin(t)}{e^{it}}$, where $f$ is indeed well-defined and then show $f$ is identically 1.

Question 1: So what exactly is $h:\mathbb R \to \mathbb C, h(t)=e^{it}$ here assuming $e^z$ is defined simply as the unique solution of $g:\mathbb C \to \mathbb C, g'=g,g(0)=1$?

Goal: I'm asking if this option is strictly speaking sensible. Like those probability textbooks that talk about independence but are actually strictly speaking nonsensical. I wanna make sure there's no imprecision actually. If it's actually nonsensical, then I wanna be sure to specify that the same way I teach probability without measure theory: 'It's nonsensical but just play along.' And then in this case, which parts are nonsensical, and how should they be taught? Like in probability when we teach some nonsensical parts by 'assuming the function is "well-behaved"' or something.

Question 2: Alternatively, is there some textbook that 'proves' Euler's formula for real numbers based on the unique $g$ definition?

Anyhoo, guess for Question 1: It's $h(t)=g(0,t)$. Explicitly:

I think $h(t)=g(0,t)=u(0,t)+iv(0,t)$, where $g(z)=g(x,y)=u(x,y)+iv(x,y)$ is the unique solution to $g:\mathbb C \to \mathbb C, g'=g,g(0)=1$?

And then we can indeed show that

  1. $h$ is nowhere zero? (Probably we can show $g$ is nowhere zero?)

  2. $\frac{d}{dt} h(t) = ih(t)$, i.e. we have both $\frac{d}{dt} u(0,t) = -v(0,t)$ and $\frac{d}{dt} v(0,t) = u(0,t)$? (I think this is what proofwiki assumes too in another proof.)

  3. $-\frac{d}{dt} h(-t) = -ih(-t)$ ?

  4. As for differentiability of $h$, well I guess it follows from differentiability (or at least holomorphicity) of $g$?


  1. No need for specifics. I can work it out. I just wanna know if Wikipedia is just doing nonsense or if there's something really possible behind 'proving' the 'formula' from an alternate definition of the complex exponential.

  2. Currently, my thought is that they're doing just some heuristic definition that $e^{(\text{whatever})(x)} = \text{whatever} \ e^{(\text{whatever})(x)}$. But eh maybe there's a rigorous justification that is somehow possible without getting the formula for $e^z$ explicitly.

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    $\begingroup$ As it is, your question is not related to education (so MSE could be a better place?). Are you asking because you're considering different options for introducing Euler's formula in a class? $\endgroup$ Aug 11, 2022 at 10:27
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    $\begingroup$ Once you declare that $e^z$ is going to be the analytic extension of $e^x$, it is automatically holomorphic and a lot of these issues resolve themselves automatically. e.g. $\frac{d}{dz} f(iz) = i f'(iz)$ and hence your #2 $\endgroup$
    – Adam
    Aug 11, 2022 at 12:25
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    $\begingroup$ @BCLC Please post to math.stackexchange and I will provide thorough answers to all of your confusions. $\endgroup$ Aug 11, 2022 at 20:11
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    $\begingroup$ Unfortunately it appears that the original poster will not be permitted to repost this question to math.stackexchange due to their account status; if someone else wants to move the question and possibly repost some answers over there, then someone will have to repost it. The question will stay closed here on matheducators.SE. [For the record, the question was originally closed by a vote of five users; I never unilaterally close questions, but my failed attempt to migrate the question now attributes the close vote only to me.] $\endgroup$ Aug 16, 2022 at 20:14
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    $\begingroup$ @ChrisCunningham I copied the question and answer here: math.stackexchange.com/q/4513719/34287. $\endgroup$ Aug 17, 2022 at 9:26

2 Answers 2


First develop differential calculus of curves $\mathbb{R} \to \mathbb{C}$. The cool thing is that if you define the derivative in any sensible way, everything works out nicely. For example, if you let $\gamma(t) = u(t) + iv(t)$ for real valued $u$ and $v$, and define $\gamma'(t) = u'(t) + iv'(t)$, then you can prove that all of the usual calculus rules work for derivatives including the product rule (where multiplication of complex numbers is used in the codomain), quotient rule (where division of complex numbers is used in the codomain), etc.

None of this requires you to understand the derivative of a function $\mathbb{C} \to \mathbb{C}$. None of it uses the Cauchy-Riemann equations.

If you were going to try to define $\gamma(t) = e^{it}$ as a function $\gamma: \mathbb{R} \to \mathbb{C}$ it would be reasonable to ask that $\gamma'(t) = ie^{it} = i\gamma(t)$ and that $\gamma(0) = 1$.

The "proof by differentiation" then goes through smoothly.

Let $\gamma: \mathbb{R} \to \mathbb{C}$ be a function with $\gamma'(t) = i\gamma(t)$ and $\gamma(0) = 1$.

Define $C(t) = \frac{\cos(t)+i\sin(t)}{\gamma(t)}$

Then $\begin{align*} C'(t) &= \frac{\gamma(t)(-\sin(t) + i\cos(t)) - \gamma'(t)(\cos(t)+i\sin(t))}{\gamma(t)^2}\\ &= \frac{\gamma(t)(-\sin(t) + i\cos(t)) - i\gamma(t)(\cos(t)+i\sin(t))}{\gamma(t)^2}\\ &= \frac{(-\sin(t) + i\cos(t)) - i(\cos(t)+i\sin(t))}{\gamma(t)}\\ &= 0\\ \end{align*} $

So $C$ is a constant function. Evaluating at $t=0$ shows that $C(t) = 1$, and the result follows.

Edit to reply to BCLC's comment.

The way that this calculus of complex curves interacts with the calculus of holomorphic functions is as follows:

If $\eta: \mathbb{R} \to \mathbb{C}$ is a curve, and $f: \mathbb{C} \to \mathbb{C}$ is a holomorphic function, then $f \circ \eta$ is another curve and the definitions of derivative are compatible according to the chain rule

$$ (f \circ \eta)'(t) = f'(\eta(t))\eta'(t) $$

Here $f'(\eta(t))$ is the complex derivative of $f$ at $\eta(t)$, $\eta'(t)$ is as I defined it, and the product is the product of complex numbers.

Now let $\gamma: \mathbb{R} \to \mathbb{C}$ be the function I defined above (which we proved was equal to $\cos(t) + i\sin(t)$).

Let $\exp: \mathbb{C} \to \mathbb{C}$ be defined by $\exp'(z) = \exp(z)$ and $\exp(0)=1$.

Let $\eta: \mathbb{R} \to \mathbb{C}$ be defined by $\gamma(t) = it$.

I will argue that $\exp(\eta(t)) = \gamma(t)$.

We just need to verify that $\exp(\eta(t))$ satisfies the defining conditions of $\gamma$.

$ \begin{align*} (\exp \circ \eta)'(t) &= \exp'(\eta(t))\eta'(t)\\ &=i\exp(\eta(t)) \end{align*}$

and clearly $\exp(\eta(0)) = \exp(0) = 1$.

So we must have $\exp(\eta(t)) = \gamma(t)$.

So $e^{it} = \cos(t) + i\sin(t)$.

  • $\begingroup$ Where exactly is $e^z$ here? Or do you mean when wikipedia proves it, at least in this part, it doesn't use $e^z$ at all? $\endgroup$
    – BCLC
    Aug 11, 2022 at 22:25
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    $\begingroup$ @BCLC Added content to address the relationship to $e^z$. This really is all math content though, not math ed. $\endgroup$ Aug 11, 2022 at 23:45
  • $\begingroup$ ah I guess it was a false alarm after all. Lol. thanks. $\endgroup$
    – BCLC
    Aug 12, 2022 at 3:39

Ok maybe it's a false alarm. If anyone thinks up any reason for alarm, then feel free to post your own answer.

At least for 2, this is what I got:

For $g(x,y)=u(x,y)+iv(x,y)$, Cauchy Riemann gives us $u_x=v_y,v_x=-u_y$.

From that and $$u(z)+iv(z)=g(z)=g'(z)=\frac{\partial}{\partial x} g(x,y) = u_x(z)+iv_x(z),$$ we get






Then -- This part I'm not so sure is rigorous --

$$v_y|_{y=t}(0,y)=\frac{d}{dt}v(0,t) \ ?!$$

$$-u_y|_{y=t}(0,y)=\frac{d}{dt} -u(0,t) \ ?!$$

For 3:

Similar I guess.

For 1:

Probably some contradiction with $0=g(z_0)=g'(z_0)=u_x(x_0,y_0)=u(x_0,y_0)$

For 4:



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