Euler's formula proof with Calculus

I was reading this source here and it provides a proof of Euler's formula using calculus. Although I technically understand the reasoning, I can't quite wrap my head around one particular step: if $f(x)= \cos(x)+i \sin(x)$, then $f'(x)=if(x) \implies f(x)=e^{ix}$. I can kind of get this, as derivating $e^{ix}$ gives $ie^{ix}$, and I know that $e^x$ is the only non-constant function s.t. $f'(x)=f(x)$. Is there a more clear way to think about this step? Can we extend that property of $e^x$ to when it has more in the exponent? And why does it have that property anyways? Would there be a way of obtaining this step without simply knowing that $e^{ix}$ would work?

• There are a lot of questions here. At least 4. What are you specifically wanting covered here? Feb 11, 2017 at 5:46
• @BrevanEllefsen Just a general grasp of the thing, really. A better intuition for how and why this step can be and is done. Feb 11, 2017 at 5:55
• The slightly more general formula that is necessary here is $f'(x)=af(x) \leftrightarrow f(x)=Ke^{ax}$ (don't forget factor $K$) Feb 11, 2017 at 6:07
• Some important intuition in the background of this: think of differentiation as being a $\pi/2$ shift: for example, when you differentiate $\cos$: $\cos'(x)=-\sin(x)=cos(x+\pi/2)$; the same for $\sin'(x)=\cos(x)=sin(x+\pi/2).$ Feb 11, 2017 at 6:11
• @Vedvart1 I threw up some super short answers to your questions. Since your posts is so broad my answers are pretty broad, but let me know if they help at all. Feb 11, 2017 at 6:17

Can we extend that property of $e^x$ to when it has more in the exponent?

Yes we can.

And why does it have that property anyways?

Good question. This is because complex derivatives work almost exactly the same as you are used to real derivatives working. Take the formula for the complex derivative as an example $$f'(x)=\lim_{h \to 0} \frac{f(z+h)-f(z)}{h}$$ Where $h$ and $z$ are complex numbers. You can go ahead and plug $e^z$ into this function and find it is complex differentiable. The proof is pretty much exactly the same as in the real case. If you want a more geometric approach then I recommend this video.

Depends on your definition of clear. I think the proof you provided is far clearer than what I am about to post, but perhaps you are defining clarity as rigor? Here is a proof from Wikipedia that is much more formal and explanatory, but I feel it is less clear. Let me know what you want here and I can adjust.

All complex numbers can be expressed in polar coordinates. Therefore, for ''some'' $r$ and $\theta$ depending on $x$, $e^{ix} = r (\cos \theta + i \sin \theta)$ No assumptions are being made about $r$ and $\theta$; they will be determined in the course of the proof. From any of the definitions of the exponential function it can be shown that the derivative of $e^{ix}$ is $ie^{ix}$. Therefore, differentiating both sides gives $i e ^{ix} = (\cos \theta + i \sin \theta) \frac{dr}{dx} + r (-\sin \theta + i \cos \theta) \frac{d\theta}{dx}$ Substituting $r(\cos\theta + i\sin\theta)$ for $e^{ix}$ and equating real and imaginary parts in this formula gives $\frac{dr}{dx} = 0$ and $\frac{d\theta}{dx} = 1$. Thus, $r$ is a constant and $\theta$ is $x+C$ for some constant $C$. The initial values $r(0)=1$ and $\theta(0)=0$ come from $e^{0i}=1$, giving $r=1$ and $\theta = x$. This proves the formula $e^{ix} = 1(\cos x +i \sin x) = \cos x + i \sin x$

There is something important to note in this paragraph that you are neglecting; we have to solve for the constant of integration. When we solve $y'=y$ we find that $y=ke^x$ and thus we must have some initial condition to solve for $k$.

Would there be a way of obtaining this step without simply knowing that $e^{ix}$ would work?

We have to know some properties of $e^x$ to ever prove Euler's Formula. Here we just use the fact that $ke^x$ is the only non-constant solution to $y'=y$, which is a pretty natural condition if you ask me.

• This is very helpful, just what I was looking for. How exactly do we know (for the last part) that $ke^x$ is the only non constant solution to that? Continuity implying differentiability is also intuitive, but then you have the Weierstrauss function. How do we know there isn't some strange function out there that also satisfies it? Feb 11, 2017 at 6:23
• @Vedvart1 It doesn't actually matter if any other functions satisfy it; if they did, they would have to also be equal to $\cos(x) + i\sin(x)$. Knowing whether examples other than $ke^x$ is not necessary. That being said, it isn't too hard to show that $ke^x$ is the only "nice" function that satisfies the functional equation $y'=y$ (the key property being everywhere differentiability) Feb 11, 2017 at 6:29
• Brevan, what you've missed is that the logic is completely circular. How have you defined the function $e^{ix}$? We know that $f(x)=\cos(x)+i\sin(x)$ and that $f'(x)=if(x)$. The solution as given by $e^{ix}$ is just symbolic. We don't know (yet) that this is the number $e$ raised to the power $ix$. And none of its properties is yet described. And we need to do a lot of work to show that it has the same properties as the function with which we are familiar. In most introductory treatises, we define $e^{ix}$ as $\cos(x)+i\sin(x)$ and proceed from there. Feb 11, 2017 at 6:52
• @Dr.MV I'll agree that simply defining it first is a generally better way to go in introductory texts, but I prefer to follow the approach of first choosing a definition for $e$, then defining real exponentiation, then defining the real derivative and subsequently the complex derivative, and finally prove some uniqueness theorems for differential equations involving complex functions. Assuming you defined in there what a complex function is (i.e., what does $z(x,y) = f(x)+if(y)$ mean?) we can be careful to avoid circular logic... it is just a tedious process. Feb 11, 2017 at 6:57
• @Dr.MV Given the OP is just learning this stuff I think an overload of rigour would be a bad thing, perhaps even off-putting. As such, I favored an arguably circular approach in order to keep complexity down :) I could always prove my argument isn't circular, but I'll leave it to someone else to write down all the steps if he/she desires. Feb 11, 2017 at 6:58

As a complement to the explanation of @Brevan Ellefsen.

Have a look at the figure below. It explains geometrically that $e^{ix}$, imagined as a turning vector at unit speed, has a unitary "speed vector" turning at the same speed, orthogonal to it.

This illustrates the fact that differentiation is equivalent to a $\pi/2$ shift, i.e., in geometrical terms, a $\pi/2$ rotation.

If you analyse it in a separate way:

$$\begin{cases}\cos'(x)=-\sin(x)=\cos(x+\pi/2)\\ \sin'(x)=-\cos(x)=\sin(x+\pi/2)\end{cases}$$

But this boils down to say that the derivative of $e^{ix}$ is $ie^{ix}$, which is interpretated as $e^{ix}$ "turned by" $\pi/2$.

For your last question, consider a differential equation of the form $f: \mathbb{R} \to \mathbb{C}$ and $\forall x \in \mathbb{R}$ $f'(x) = kf(x)$ where $k \in \mathbb{C}$ and $k \neq 0$.

(Ah! I will still make use of the fact that $e^{kx}$ is a solution but rather, in a different manner)

Consider the following function $g : \mathbb{R} \to \mathbb{C}$, $g(x) = e^{-kx}f(x)$

Since, both exponential function and $f$ are differentiable, so $g$ is differentiable. Hence $g'(x) = -ke^{-kx}f(x)+e^{-kx}f'(x) = e^{-kx}(f'(x)-kf(x)) = 0$ $\forall x \in \mathbb{R}$. Therefore, $g$ is a constant function. Hence, $g(x) = e^{-kx}f(x) = c \implies f(x) = ce^{kx}$

Complex differentiation follows exactly the same process like with the real argument except that we now regard $i$ as an imaginary constant.

I am posting a bit different not suitable for teaching but as it appeared to me long time ago, so am ready for a bit of down-voting.

If we wish to work on basis of circular trigonometric functions,

$$f(x)= \cos(x)+i \sin(x)$$

$$f'(x)= -\sin(x)+i \cos(x) = if(x)$$

the above is rotation by $\pi/2$ in complex plane. And,

$$f''(x)=-f(x) \rightarrow f''(x)+f(x) = 0$$

The above is real rotation of period $2 \pi$ in complex plane.

But if we wish to work on basis of hyperbolic functions we can consider differentiation with respect to $i x$ instead of real $x$ and apply Chain Rule ( divide by the constant $i$).

$$f(x)= \cosh(x)+i \sinh(x)$$

$$f'(x)= \cosh(x)-i \sinh(x)$$

which is complex conjugate of $f(x).$ And

$$f''(x)= -\cosh(x)-i \sinh(x) =-f(x), \rightarrow f''(x)+f(x) = 0$$

which is still the same rotation in the complex plane, argument change notwithstanding.