My question is too simple. We know all that if we define the exponential function on $\mathbb{C}$ then we define the real part and imaginary part of $\exp{it}$ as $\cos{t}$ and $\sin{t}$.

So if we assume that we know the definition of the trigonometric functions and the exponential function not as a series but for exponential as solution of differential equation $y'-y=0$ then my question is

How can I explain this relation (intrinsically)in a simple way to a high-school student, they is any physical interpretation of this formula ?

or we can give another equivalent question (in my sens)

How can one prove the Euler relation? and What was the original proof of Euler himself?

  • 3
    $\begingroup$ This Post may help you a little. :) $\endgroup$ Jun 25, 2016 at 16:52
  • 2
    $\begingroup$ Related, but with less focus: math.stackexchange.com/questions/3510/… $\endgroup$
    – Mark S.
    Jun 25, 2016 at 21:44
  • $\begingroup$ I'm curious-what motivates this question?Are you teaching honors calculus to strong high school students or honors physics and the formula came up? $\endgroup$ Jul 4, 2016 at 4:04
  • $\begingroup$ both of them, first because i want to prove the formula for a high school student simply (without using series), but for the physical meaning, after reading the second proof of @MarkS. i want to know if we can deduce it from a physical phenomena $\endgroup$
    – Hamza
    Jul 4, 2016 at 4:14
  • 1
    $\begingroup$ I wrote the approach in this answer for a friend's high school class back in the late 70s. It uses trigonometry and limits, but no calculus. Calculus was not being taught in high school then. We had to drive to the local junior college to take calculus. $\endgroup$
    – robjohn
    Jul 8, 2022 at 21:00

7 Answers 7


I have two favorite arguments that we should have $\exp (i\theta)=\cos \theta +i\sin \theta$ for real $\theta$. The first is closely related to Mathologer's video e to the pi i for dummies, and the second is discussed in slightly more detail in II.2 “Moving Particle Argument” in Visual Complex Analysis. Finally, I conclude with a summary of how Euler did it, from How Euler Did It by Ed Sandifer for MAA Online.

1. Limit Argument

Many high school students are aware that $e=\displaystyle{\lim_{n\to\infty}}\left(1+\dfrac{1}{n}\right)^n$. For real $r$, some may be acquainted with the fact that $e^r=\displaystyle{\lim_{n\to\infty}}\left(1+\dfrac{r}{n}\right)^n$. We can declare by fiat that this will serve as a definition for all complex $r$. Then we have $e^{i\theta}=\displaystyle{\lim_{n\to\infty}}\left(1+\dfrac{i\theta}{n}\right)^n$. Now we just need to use geometric properties of complex multiplication to argue that $e^{i\theta}$ has magnitude $1$ and argument/angle $\theta$.

Magnitude $1$

$\left|e^{i\theta}\right|=\displaystyle{\lim_{n\to\infty}}\left|1+\dfrac{i\theta}{n}\right|^n=\sqrt{\displaystyle{\lim_{n\to\infty}}\left(1+\dfrac{\theta^2}{n^2}\right)^{n}}$. This is certainly no less than $1$. However, since for $n>\dfrac{\theta^2}{r}$ we have $\dfrac{r}{n}>\dfrac{\theta^2}{n^2}$, it can be no more than $\sqrt{e^r}$ for any positive $r$, so the limit is $1$ (at least if it exists). Therefore, $e^{i\theta}$ lies on the unit circle.

Angle $\theta$

When calculating $\left(1+\dfrac{i\theta}{n}\right)^n$ for fixed $n$ geometrically, we can draw a right triangle with vertices at $0$, $1$, and $1+\dfrac{i\theta}{n}$. Then a triangle on the hypotenuse, with a new vertex at $(1+\dfrac{i\theta}{n})^2$. Then a new triangle... This yields something that looks like n=5 theta=2.7

You can manipulate this spiral at a little geogebra applet I made.

Assuming $\theta>0$, the $k^{\text{th}}$ triangle has a leg away from the origin of length $\dfrac{\theta}{n}\left|1+\dfrac{i\theta}{n}\right|^{k-1}$. For $n$ large, the outer perimeter of this spiral is intuitively close to $\theta$ since $\left|1+\dfrac{i\theta}{n}\right|$ is very close to $1$, so each factor like $\left|1+\dfrac{i\theta}{n}\right|^{k-1}$ is fairly close to $1$, so that it is as if we are adding $n$ terms of $\dfrac{\theta}{n}$. Arguing that more formally may require calculus or clever algebra after the sum of the finite geometric series is calculated to be $\dfrac{\dfrac{\theta}{n}\left(\left|1+\dfrac{i\theta}{n}\right|^{n}-1\right)}{\left|1+\dfrac{i\theta}{n}\right|-1}$.

If $0\le\theta<2\pi$, then the perimeter approaching $\theta$ gives you an arc length, and hence an angle, of $\theta$ around the unit circle, as desired. If $\theta>2\pi$, then if we hope or check that $e^{i\theta}e^{i\rho}=e^{i(\theta+\rho)}$ with our definition, then you can get the desired angle by adding/subtracting the relevant multiple of $2\pi$, we get the desired result.

2. Moving Particle Argument

A student who's had most of a first year of Calculus may be able to appreciate this argument based on a vector function and derivatives. (For this moving particle argument, I'm largely just quoting another MSE answer of mine.)

Consider a particle moving counterclockwise around the unit circle in the complex plane (starting at $1+0i$) at unit speed. By the definition of radians and sine and cosine, its position in the complex plane at time $t$ is given by $\mathbf{s}\left(t\right)=\cos\theta+i\sin\theta$. Since a tangent to a circle forms a right angle and multiplication by $i$ rotates things counterclockwise by a right angle ($x+iy$ is sent to $-y+ix$) we have $\mathbf{s}'\left(t\right)=ki\mathbf{s}\left(t\right)$ for some positive real number $k$. Since it's going at unit speed, we have $\left|\mathbf{s}'\left(t\right)\right|=1$ so that $k=1$ as $\left|\mathbf{s}\left(t\right)\right|=\sqrt{\cos^{2}t+\sin^{2}t}=1$.

Now we just need to find a complex function where $\mathbf{s}\left(0\right)=1$ and $\mathbf{s}'\left(t\right)=i\mathbf{s}\left(t\right)$. The exponential function is its own derivative for real inputs, and we can declare by fiat that it should still work for complex inputs. Then the chain rule for differentiation tells us we can use $\mathbf{s}\left(t\right)=e^{it}$.

3. How Euler Did It

This is just a paraphrasing of some of How Euler Did It by Ed Sandifer - in particular, the parts where he paraphrases from Euler's Introductio. Note that Euler's work was in Latin, used different variables, and did not have modern concepts of infinity.

I'll use $\mathrm{cis}\theta$ to denote $\cos\theta+i\sin\theta$. Euler derived (possibly based on DeMoivre's work) that $(\mathrm{cis}z)^n=\mathrm{cis}(nz)$ for (positive?) integer $n$. Then he derives $$\cos(nz)=\dfrac{(\mathrm{cis}(z))^n+(\mathrm{cis}(-z))^n}{2}\text{ and }\sin(nz)=\dfrac{(\mathrm{cis}(z))^n-(\mathrm{cis}(-z))^n}{2i}\text{.}$$

Then he uses the Maclaurin series expansions for $\sin$ and $\cos$ to turn these into something along the lines of $$\cos(\theta)=\displaystyle{\lim_{n\to\infty}}\dfrac{\left(1+\dfrac{i\theta}{n}\right)^n+\left(1-\dfrac{i\theta}{n}\right)^n}{2}\text{ and }\sin(\theta)=\displaystyle{\lim_{n\to\infty}}\dfrac{\left(1+\dfrac{i\theta}{n}\right)^n-\left(1-\dfrac{i\theta}{n}\right)^n}{2i}\text{.}$$

Then he used the "$e^r=\displaystyle{\lim_{n\to\infty}}\left(1+\dfrac{r}{n}\right)^n$" idea to turn these into $$\cos(\theta)=\dfrac{e^{i\theta}+e^{-i\theta}}{2}\text{ and }\sin(\theta)=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}\text{.}$$ Then it's just a bit of algebra to get to $e^{i\theta}=\mathrm{cis}\theta$.

  • $\begingroup$ I like this heuristic idea , but for the first proof why $|e^{i\theta}|\leq \sqrt{e^r}$ for all $r>0$ ? $\endgroup$
    – Hamza
    Jun 25, 2016 at 18:13
  • $\begingroup$ can you please @Mark S give a picture to the geometric construction in the argument proof. $\endgroup$
    – Hamza
    Jun 25, 2016 at 18:18
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    $\begingroup$ @Hamza I have added the things you requested, plus an interactive applet to allow you to reproduce the sorts of pictures in the Mathologer video that I tried to describe. $\endgroup$
    – Mark S.
    Jun 25, 2016 at 20:29
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    $\begingroup$ @Hamza I just added a summary, but a good little piece about the history of Euler and others discovering this is given in "How Euler Did it" by Ed Sandifer $\endgroup$
    – Mark S.
    Jun 25, 2016 at 22:07
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    $\begingroup$ I just found this question. The Limit Argument is similar to the answer I wrote for this question. $\endgroup$
    – robjohn
    Dec 7, 2019 at 23:37

One way is to define the function $$f(x)=e^{-ix}\left(\cos x+i\sin x\right)$$ Differentiating by $x$ yields $$f'(x)=-ie^{-ix}\left(\cos x+i\sin x\right)+e^{-ix}\left(-\sin x+i\cos x\right)=0$$ (where we assume that $i$ acts like a real scalar in differentiation). That means that $f$ is constant, and of course $f(0)=1$. So $f(x)=1$ for all $x$, implying $e^{ix}=\cos x+i\sin x$.

  • $\begingroup$ there is a small problem of differentiation of a complex function, and they properties, but the proof is good. $\endgroup$
    – Hamza
    Jun 25, 2016 at 16:59
  • $\begingroup$ I like this approach. Not sure if Taylor series are covered in high school in the US, but this is still neater even if it's less natural in development. $\endgroup$
    – Deepak
    Jun 25, 2016 at 17:01
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    $\begingroup$ @zhw. Well, proving a function is constant is rather easy -- you derive it, and the function is constant iff its derivative vanishes everywhere. So, in my opinion, this is a nice way to take an identity and prove it. $\endgroup$
    – Guy
    Jul 1, 2016 at 5:48
  • 2
    $\begingroup$ It seems very natural to me, when trying to prove two functions equal, to look either at their difference or their quotient, and see whether the derivative of this is zero. $\endgroup$
    – Lubin
    Jul 5, 2016 at 4:37
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    $\begingroup$ This technique assumes that there is some definition of $e^{z}$ for complex $z$ such that if $k$ is a complex number and $x$ is a real variable then $g(x) = e^{kx}$ is differentiable and $g'(x) = ke^{kx}$. The assumption is not trivial to prove. $\endgroup$
    – Paramanand Singh
    Jul 16, 2016 at 9:03

Let $n$ be the number of elementary particles in the universe. Then $$ \begin{aligned} e^{i\theta}& = \left(1+\frac{i\theta}{n}\right)^n \\&= 1+ n \frac{i\theta}{n}+ {n \choose 2} \left(\frac{i\theta}{n}\right)^2 + {n \choose 3} \left(\frac{i\theta}{n}\right)^3 +{n \choose 4} \left(\frac{i\theta}{n}\right)^4 +\ldots \\&= \left(1 - {n \choose 2} \left(\frac{\theta}{n}\right)^2 +{n \choose 4} \left(\frac{\theta}{n}\right)^4 +\ldots\right) + i\left(n \frac{\theta}{n}- {n \choose 3} \left(\frac{\theta}{n}\right)^3+\ldots \right) \\&= 1-\frac{\theta^2}{2}+\frac{\theta^4}{4!}-\ldots \quad + i\left(\theta - \frac{\theta^3}{3!}+\ldots \right) \end{aligned} $$ because $n$ is HUGE. Therefore $e^{\theta}=\cos \theta+i\sin \theta$.

This can be easily explained to a highschool student provided you teach him binomial coefficients and series first.

To explain the relation intrinsically as you put it, note that the ordinary exponential function is a homomorphism from $\mathbb{R}$ with its additive structure to $\mathbb{R}^+$ with its multiplicative structure. Now notice that the unit complex numbers form an abelian group that is very similar to the two groups above locally. Therefore it is natural to conjecture that there should be a homomorphism from the reals with their additive structure to the unit complex numbers with their multiplicative structure (namley, you just add their Arguments!). This is precisely Euler's identity.

  • $\begingroup$ ...and also provided (at least) he doesn't ask (1) why can we apply binomial expansion and (2) why can we rearrange terms without altering the value of the series? $\endgroup$
    – user170039
    Jul 1, 2016 at 3:36
  • $\begingroup$ what that mean the fact that you take elementary particles in your calculation ? it exist some physical interpretation of Euler Formula ??? $\endgroup$
    – Hamza
    Jul 1, 2016 at 17:36
  • 1
    $\begingroup$ @Hamza it's a poetic way of saying the following is approximately correct for very large n" $\endgroup$
    – Mark S.
    Jul 2, 2016 at 16:11
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    $\begingroup$ @Hamza, to formalize these ideas it is convenient to use a hyperreal extension of the reals. In this extension one has infinite natural numbers, or hypernaturals for short. We can let $n$ be an infinite hypernatural and then the calculation I gave is exact provided equality is replaced by the more general (and still precise) relation of being infinitely close (i.e., the difference of the two sides is an infinitesimal). See Elementary Calculus for a gentle introduction. $\endgroup$ Jul 3, 2016 at 7:02
  • $\begingroup$ Thank You @MikhailKatz because I have never hear about hyperreal number and it's more clear now $\endgroup$
    – Hamza
    Jul 4, 2016 at 3:04

My favorite method by far for proving $e^{ix}=\cos x+i\sin x$, is to use Taylor- MacLaurin series for the natural exponential, sine and cosine, expand both sides and prove equality.All these series are known to converge for all real $x$ in basic calculus, which I hope wouldn't be too difficult for your students to understand. Judging from some of the other answers here,it shouldn't. The one tricky part that involves some slight of hand is we have to assume that the MacLaurin series for $e^z$ converges for pure-imaginary $z$. It does,but to prove it requires basic complex analysis, which I'm pretty sure is beyond the reach of your audience.

The MacLaurin series: \begin{align} \sin x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \\\\ \cos x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots \\\\ e^z&=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots \end{align}

Substitute $z=ix$ in the last series: \begin{align} e^{ix}&=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots \\\\ &=1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\cdots \\\\ &=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots +i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right) \\\\ &=\cos x+i\sin x \end{align}

I love this way of proving the result because it's so straightforward and clean--just a calculation. But as I said, it involves the assumption regarding the convergence of the $e^ {z}$ series.So technically, this isn't a completely rigorous proof. You'll have to be the judge of whether or not that's a problem or with your particular audience, it can be taken on faith.

  • $\begingroup$ Why are you copying my answer? $\endgroup$ Jul 4, 2016 at 7:09
  • $\begingroup$ @MikhailKatz I disagree respectfully-my answer is NOT the same as yours,although they both involve the MacLaurin series of the exponential, sine and cosine functions. Your answer involves a binomial coefficient expansion to achieve the same result. In fact,I think my direct calculus answer complements yours nicely. $\endgroup$ Jul 4, 2016 at 16:26

The best way to prove Euler's relation $$\exp(i\theta) = \cos \theta + i\sin \theta\tag{1}$$ is to use the following definition of $\exp(z)$: $$\exp(z) = \lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n}\tag{2}$$

We will use the following simple lemma:

Lemma: If $a_{n}$ is a sequence of real or complex terms such that $n(a_{n} - 1) \to 0$ as $n \to \infty$ then $a_{n}^{n} \to 1$ as $n \to \infty$.

Proof: Let $a_{n} = 1 + b_{n}$ so that $nb_{n} \to 0$ as $n \to \infty$. We now have \begin{align} |a_{n}^{n} - 1| &= |(1 + b_{n})^{n} - 1|\notag\\ &= \left|nb_{n} + \frac{n(n - 1)}{2!}b_{n}^{2} + \cdots\right|\notag\\ &\leq |nb_{n}| + \dfrac{1 - \dfrac{1}{n}}{2!}|nb_{n}|^{2} + \cdots\notag\\ &\leq |nb_{n}| + |nb_{n}|^{2} + \cdots\notag\\ &= \frac{|nb_{n}|}{1 - |nb_{n}|}\notag \end{align} It thus follows that $a_{n}^{n} \to 1$ as $n \to \infty$.

Let $z = i\theta$ where $\theta$ is real. Consider the following sequence \begin{align} a_{n} &= \dfrac{1 + \dfrac{i\theta}{n}}{\cos\dfrac{\theta}{n} + i\sin\dfrac{\theta}{n}}\notag\\ &= \left(1 + \frac{i\theta}{n}\right)\left(\cos\frac{\theta}{n} - i\sin\frac{\theta}{n}\right)\notag\\ &= \cos\frac{\theta}{n} + \frac{\theta}{n}\sin\frac{\theta}{n} + i\left(\frac{\theta}{n}\cos\frac{\theta}{n} - \sin\frac{\theta}{n}\right)\notag\\ \end{align} We have $$ n(a_{n} - 1) = n\left(\cos\frac{\theta}{n} - 1\right) + \theta\sin\frac{\theta}{n} + i\left(\theta\cos\frac{\theta}{n} - n\sin\frac{\theta}{n}\right)$$ and it is clear that $n(a_{n} - 1) \to 0$ as $n \to \infty$. Hence $a_{n}^{n} \to 1$ as $n \to \infty$ and therefore $$\lim_{n \to \infty}\dfrac{\left(1 + \dfrac{i\theta}{n}\right)^{n}}{\left(\cos\dfrac{\theta}{n} + i\sin\dfrac{\theta}{n}\right)^{n}} = 1$$ or $$\lim_{n \to \infty}\left(1 + \frac{i\theta}{n}\right)^{n} = \cos \theta + i\sin \theta$$ as was desired.


I started from the question: "Can I in a few steps get from powers-of-$i$ to Euler's formula?"
The goal is not to create an elegant or beautiful reasoning, but to get there in steps everyone can follow.

The multiplication of a complex number with $i$ corresponds to rotation over $\frac{\pi}{2}$ on the complex plane.
Raising a complex number to natural power corresponds to linear growth of the angle.

$$i=0+1i=\ 1\left(\cos{\left(\frac{\pi}{2}\right)}+i\ \sin{\left(\frac{\pi}{2}\right)}\right)=1\angle\frac{\pi}{2}\ \ (1)$$

De Moivre states:

Applying De Moivre to (1) provides:
$$i^n=1\left(\cos{\left(n\frac{\pi}{2}\right)}+i\ \sin{\left(n\frac{\pi}{2}\right)}\right)=1\angle\left(n\frac{\pi}{2}\right)\ ,\ n\in\mathbb{N} \ (2)$$

What happens if we raise $i$ to a real $(\in\mathbb{R})$ power instead of a natural $\left(\in\mathbb{N}\right)$ power?

Extending (2) to $\mathbb{R}$ results in:

$$i^x=1\left(\cos{\left(x\frac{\pi}{2}\right)}+i\ \sin{\left(x\frac{\pi}{2}\right)}\right)=1\angle\left(x\frac{\pi}{2}\right)\ ,\ x\in\mathbb{R}$$ $$i^{\theta\frac{2}{\pi}}=1\left(\cos{\theta}+i\ \sin{\theta}\right)=1\angle\theta\ \ x,\ \theta\in\mathbb{R},\ \theta=x\ \frac{\pi}{2}$$

We no longer rotate in discrete steps $n\in\mathbb{N}$, but over ‘any angle’ $\theta$.

Raising $i$ to a real power $\theta\frac{2}{\pi}$, corresponds to a rotation over an angle $\theta$ from $i^0=1+0i$ or $1\angle0$.
Now we can rotate over any angle $\theta$ on the complex plane.
Now let us look at a property below: $$a^y=\ \left(e^{\ln{a}}\right)^y=\ \ e^{y\ln{a}}$$
A power of any number, including $i$, can be rewritten as a power of $e$.

$$i^{\theta\frac{2}{\pi}}=\ e^{\ln{\left(i\right)\theta\frac{2}{\pi}}}$$ $$i^{\theta\frac{2}{\pi}}=\ e^{\ln{\left(i\right)\theta\frac{2}{\pi}}}=\ 1\left(\cos{\theta}+i\ \sin{\theta}\right) \ (3)$$

Now let us see whether we can use the properties below $$\frac{d}{dx}\left(\cos{x}+i\ \sin{x}\right)=i\ \left(\cos{x}+i\sin{x}\right)$$


If we take the derivative to $\theta$ from either side of (3)

$$\left(\ln{\left(i\right)\frac{2}{\pi}}\right)\ e^{\ln{\left(i\right)\theta\frac{2}{\pi}}}=i\left(\cos{\theta}+i\ \sin{\theta}\right)$$ $$\left(\ln{\left(i\right)\frac{2}{\pi}}\right)=i$$ $$\ln{\left(i\right)}=i\frac{\pi}{2}$$ Bringing everything together results in: $$i^{\theta\frac{2}{\pi}}=e^{\ln{\left(i\right)\theta\frac{2}{\pi}}}=e^{i\frac{\pi}{2}{\theta\frac{2}{\pi}}}=\ e^{i\theta}=\ 1\left(\cos{\theta}+i\ \sin{\theta}\right)$$

Raising $i$ to a power $\theta\frac{2}{\pi}$, corresponds to a rotation over an angle $\theta$ on the complex plane.
A power of any number, including $i$, can be rewritten as a power of $e$.

$$ri^{\theta\frac{2}{\pi}}=\ r\ e^{\ln{\left(i\right)\theta\frac{2}{\pi}}}=re^{i\frac{\pi}{2}{\theta\frac{2}{\pi}}}=\ re^{i\theta}=\ r\left(\cos{\theta}+i\ \sin{\theta}\right)=a+bi$$

You can find this reasoning and more on www.heavisidesdinner.com.
If the reasoning above is not really clear, just add a comment and I will gladly elaborate it more.


There is also another way to show that the $f(x)=\cos(x)+i\sin(x)$ is an exponential function using trigonometric identities but I cannot proceed further.

Start with the expression $$\cos(x+y) + i\sin(x+y)$$ $$= \cos x\cdot\cos y-\sin x\cdot\sin y + i(\sin x\cdot \cos y+\cos x\cdot\sin y)$$ $$= i^2\sin x\cdot \sin y + i(\sin x\cdot\cos y+\cos x\cdot\sin y) + \cos x\cdot\cos y$$ Factoring out and you will get $$=(\cos x+i\sin x)(\cos y+i\sin y)$$ Now if we define $f(x)=\cos x+i\sin x,$ $$f(x+y)=f(x)f(y)$$ holds for all x and y. This functional equation holds for $f(x)=a^x$ for some constant $a$. Thus we can write

$$a^x=\cos x+i\sin x.$$


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