From Euler we've learned that $z=re^{i\theta}$.

And it's easy to see that $|z|^2=r^2$, since $re^{i\theta}\times re^{-i\theta}=r^2$.

Why must we use e to represent these numbers correctly? It seems that I could arbitrarily choose a different exponent $z=r\pi^{i\theta}$ and get the same size for $z$ as I did before: $|z|^2=r\pi^{i\theta}\times r\pi^{-i\theta}=r^2$

What did I miss?

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    $\begingroup$ To start: What's your definition of $\pi^{i\theta}$? $\endgroup$
    – GFauxPas
    Commented Apr 13, 2015 at 18:58
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    $\begingroup$ I think it is a mistake to think that Euler just happened to choose $e$ to represent complex numbers. Rather, he found that $e$ was the number that would allow him to represent complex numbers in polar coordinates. $e$ has a number of important properties that allow it to represent complex numbers. For one, that $re^{ia\theta} = r\cos(a \theta )+ir\sin(a \theta )$. It is not true in general that $r\pi^{ia \theta} = r\cos(a \theta )+ir\sin(a \theta )$ $\endgroup$
    – graydad
    Commented Apr 13, 2015 at 19:03
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    $\begingroup$ @GFauxPas What do you mean? take $\pi$ to the power of $i\theta$... It's the same definition as it was when we used e as the base... What I'm missing is why do we have to use e as the base to get the correct representation... $\endgroup$ Commented Apr 13, 2015 at 19:04
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    $\begingroup$ $e^{i \theta} = \displaystyle \lim_{n \to \infty} \left({1 + \frac {i \theta}{n}}\right)^n = \sum_{n = 0}^{\infty}\frac{(i\theta)^n}{n!}$. What about for $\pi^{i \theta}$? $\endgroup$
    – GFauxPas
    Commented Apr 13, 2015 at 19:05
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    $\begingroup$ @graydad: Euler representing complex numbers using polar coordinates is a gross anachronism. The geometric representation of complex numbers was only discovered in 1799, more than a decade after Euler's death. $\endgroup$ Commented Apr 14, 2015 at 9:48

9 Answers 9


If we wish to express $\pi^{i\theta}$ as a series then we have:

$$\pi^{i\theta} = e^{i\ln(\pi)\theta} = \sum_{n=0}^\infty i^n \frac{(\ln(\pi)\theta)^n}{n!} = \cos(\ln(\pi)\theta)+i\sin(\ln(\pi)\theta).$$

Calculating precisely $\ln(N)$ for $N \in \mathbb{N}$ can be difficult, not to mention $\ln(\pi)$. This would add more complications than it would be worth. Moreover, $\pi^{i\theta}$ has period $2\pi/\ln(\pi)$, which is not compatible with polar coordinates.

On the other hand, since we can write $$e^{i\theta} = \cos(\theta) + i \sin(\theta),$$ we can express $e^{i\theta}$ by calculating the already well known trigonometric functions.

I would like to add that the use of $e^{i\theta}$ is because of the nice representation found by Euler. If you were to approach the polar representation for the first time, you would approach it more like this:

Let $z=x+iy$ be a complex number, which we can visualize as a vector in $\mathbb{R}^2$, $z=(x,y)$. The magnitude of $z$ is $\|z\|= \sqrt{x^2+y^2}$. We can write the real part as $x=\|z\| \cos(\theta)$ where $\theta$ is the angle formed between the real axis and the vector at the origin. Similarly $y=\|z\| \sin(\theta)$. Thus $$z= \|z\|\cos(\theta)+i \|z\|\sin(\theta) = \|z\|(\cos(\theta)+i\sin(\theta)).$$

Until now, our reasoning was completely geometric. Independently we can work out the expression, due to Euler, $e^{i\theta} = \cos(\theta)+i\sin(\theta)$. This now naturally leads to $$z=\|z\|e^{i\theta}.$$ If it turned out that $\pi^{i\theta} = \cos(\theta)+i\sin(\theta)$ then we would use that instead. However, we know that this is not the case.

I would also like to point out that there is an intuitive reason to think that $e^{i\theta}$ should be of the form $\cos(\theta)+i\sin(\theta)$.

Notice that if we write $f(\theta) = e^{i\theta} = u(\theta)+iv(\theta)$, then $$f''(\theta) = i^2 f(\theta) = - f(\theta).$$

Hence $$u''(\theta) = -u(\theta) \text{ and } v''(\theta) = - v(\theta).$$

Thus from differential equations, we can express $u$ and $v$ as a linear combination of $\sin(\theta)$ and $\cos(\theta)$.

This motivates the investigation into the series of the exponential function. From this perspective, it is not surprising to discover $\cos(\theta)$ and $\sin(\theta)$ inside the series for $e^{i\theta}$.

One final edit: If we let $A$ and $B$ be complex numbers, then my previous statement can be expressed as: $$e^{i\theta} = A\cos(\theta)+B\sin(\theta)$$

Setting $\theta=0$ we see that $e^{0}=1=A\cdot 1 = A$. And $\theta = \pi/2$ yields $e^{i\pi/2} = B$.

Therefore, $$e^{i\theta} = \cos(\theta) + e^{i\pi/2} \sin(\theta).$$ What is left is to determine $e^{i\pi/2}$. Since $e^{i\theta}$ is $2\pi$ periodic, $e^{0}=e^{i2\pi}$. Thus we can see that $(e^{i\pi/2})^4 -1 = 0$, which means $e^{i\pi/2}$ satisfies the polynomial $x^4-1=0$. Thus $e^{i\pi/2} = \pm 1 \text{ or } \pm i$.

Taking the derivative of both sides of $e^{i\theta} = \cos(\theta) + e^{i\pi/2} \sin(\theta)$ we find: $$ie^{i\theta} = -\sin(\theta) + e^{i\pi/2} \cos(\theta)$$ and therefore by setting $\theta = 0$ we have: $$i = e^{i\pi/2}.$$ Thus we conclude $$e^{i\theta} = \cos(\theta)+i\sin(\theta).$$ All without Taylor series.

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    $\begingroup$ I find this to be circular reasoning because it relies on the presupposition that $e^x$'s derivatives are "nicer" than $n^x$ where $n$ is some arbitrary positive real number. You (basically) state as an assumption that $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ and then (essentially) assert that this is mathematically more pleasing than $\pi^{i\theta}$ using the definition that $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. That is you start with the assumption that $e^{i\theta}$ is "best" and then show why if you don't use what's best, it's more complicated to write out the result. $\endgroup$
    – Jared
    Commented Apr 14, 2015 at 5:18
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    $\begingroup$ @Jared: The reason it looks like this is what is happening is because all of the various ways of thinking about $e$ are essentially the same, you can start at one definition and get to all of the others easily, and there is no established convention about which one is right. I think the most straightforward theory to Joel's answer would use the definition that $e$ is $f(1)$ for the [unique] $f$ which solves the differential equation $f'(x)=f(x)$ subject to $f(0)=1$. From this we can use Taylor, analytic continuation [which admittedly is just formalized aesthetics, yes], and recover Euler. $\endgroup$ Commented Apr 14, 2015 at 6:41
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    $\begingroup$ @Jared That "$e^x$'s derivatives are nicer than $n^x$ for $n$ an arbitrary positive real number" is not a 'presupposition'. It is a simple fact which is clear to anyone who knows how to differentiate functions like this. This makes it obvious that $e$ is 'special', without appeal to complex numbers, and was known well before Euler. $\endgroup$
    – jwg
    Commented Apr 14, 2015 at 8:00
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    $\begingroup$ @Jared It might not be obvious, but it is not an assumption or a presupposition. It is a mathematical fact. $\endgroup$
    – jwg
    Commented Apr 14, 2015 at 8:41
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    $\begingroup$ I did not derive the fact that $e^{i\theta} = \cos(\theta)+i\sin(\theta)$ because the OP is asking a question in complex analysis and from the context of his post, the OP likely knows this already. This is an exercise in many differential equations classes. It is simple to derive this from the fact that $d/dx f(x) = f(x)$ (with $f(0)=1$) has $e^x$ as a unique solution. Every answer on this site does not have to be a build up from first principles. $\endgroup$
    – Joel
    Commented Apr 14, 2015 at 15:29

There is a nice formula for $e^x$, and only $e^x$:
If you calculate $e^{0.1},e^{0.01}$, you can see the first couple of terms are correct.
So $$e^{ix}=1+ix+\frac{(ix)^2}2+\frac{(ix)^3}6+...\\=(1-\frac{x^2}2+\frac{x^2}{24}+....)+i(x-\frac{x^3}6+\frac{x^5}{120}-...)$$
Now, in radians, $$\cos x = 1-\frac{x^2}2+\frac{x^4}{24}...\\ \sin x = x-\frac{x^3}6+\frac{x^5}{120}...$$ You can check those for small values of $x$ as well. So the series for $e^{ix}$ and the series for $\sin$ and $\cos$ match (at least for small $x$).
If you do the calculus, you can find they match all the way down.


I think the answer is a matter of aesthetics. First, you have to make some assumptions, the most important of which is the following:

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

You can "justify" this through a Maclaurin series expansion of $e^x$, $\sin(x)$ , and $\cos(x)$. But that really begs the question--you really need to prove that $\frac{d}{dx}e^x = e^x$ for the specific base of $e$ and $e$ alone or, conversely, that $\frac{d}{dx}\log_e(x) = \frac{1}{x}$ (the derivative of $\ln(x)$) for the specific logarithm with base $e$--you need to prove one on its own because using one to prove the other is circular! In my opinion, this proof--that $\frac{d}{dx}e^x = e^x$ or $\frac{d}{dx}\ln(x) = \frac{1}{x}$--is the only real answer to the question because that proof and that proof alone actually explains what is so special about the value of $e$.

I'm going to assume that we now agree that $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. A computer doesn't care if you prefer the base $\pi$ (i.e. $\pi^{i\theta} = \cos(\ln(\pi)\theta) + i\sin(\ln(\pi)\theta)$. However if we choose such a base for our complex numbers, then we no longer get the nice trigonometric properties.

Specifically, let's say that we have $z = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$. If we use the original formulation of $z = e^{i\theta}$ then we immediately recognize (if we are good at trigonometry) that $\theta = 45^\circ = \frac{\pi}{4}$. On the other hand, what is $\theta$ if we prefer $\pi^{i\theta}$? It's not clear what the value is--in fact we basically have to re-engineer the value--because we know the following:

\begin{align} z = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} =&\ \cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) \\ =&\ \cos\left(\ln(\pi)x\right) + i\sin\left(\ln(\pi)x\right) \end{align}

Leading to:

$$ x = \frac{\pi}{4\ln(\pi)} \approx 0.68609911657 $$

This value has no geometric meaning--but it is no less correct mathematically. The approximate value of $0.68609911657$ rad is $\approx 0.21\pi$. Now, if I know that this is a bad base, then I can take that value and multiply by $\ln(\pi)$ to get: $0.68609911657 * \ln(\pi) \approx 0.78539816339$ which is $\approx 0.24999999999 \pi$ (which is clearly $\frac{\pi}{4}$).

With the initial representation I can quickly find where to place my coordinate on a complex plane where $x$ is the real values and $y$ are the imaginary values using polar coordinates. I can "quickly" do the same using $\pi^{i\theta}$ but it requires some intermediate steps to convert to the "nice" representation (which is $e^{i\theta}$). If, to use polar coordinates, I need to "convert" $\pi^{i\theta}$ then it makes more sense to choose the "natural" coordinates which are achieved using rather $e^{i\theta}$--where no conversion is necessary.

And the reason I say it's about aesthetics is because it's not like $e^{i\theta}$ makes all calculations easier--it doesn't. Try to find $r$ and $\theta$ from $z = 1 + 2i$:

\begin{align} 1 + 2i = r \cos(\ln(\pi)\theta) + ri\sin(\ln(\pi)\theta) \\ 1 + 2^2 = r^2 \rightarrow r = \sqrt{5} \\ \tan(\ln(\pi)\theta) = 2 \rightarrow \ln(\pi)\theta \approx 1.10714872 + 2\pi n\\ \theta \approx 0.96717027631 + \frac{2\pi n}{\ln(\pi)} \\ 1 + 2i \approx \sqrt{5}\pi^{\left(0.96717027631 + \frac{2\pi n}{\ln(\pi)}\right)i} \end{align}


\begin{align} 1 + 2i = r \cos(\theta) + ri\sin(\theta) \\ 1 + 2^2 = r^2 \rightarrow r = \sqrt{5} \\ \tan(\theta) = 2 \rightarrow \theta \approx 1.10714872 + 2\pi n\\ 1 + 2i \approx \sqrt{5}e^{\left(1.10714872 + 2\pi n\right)i} \end{align}

The only difference in the above calculations is the divide by $\ln(\pi)$--and that is why we prefer $e^{i\theta}$ over any other base--because all of the others require this "unnecessary" step--it's more aesthetically pleasing but no more mathematically correct.

And before you say, well at least using $e^{i\theta}$ you immediately know that the trigonometric angle is approximately $1.10714872$ radians, I would argue that I am actually not familiar with what that angle represents--in fact I need to convert to degrees to show that $1.10714872 \approx 63.4349489493^\circ$ before I really "know" where that value lies. And I can get that value just the same using the $\pi^{i\theta}$ it just requires an additional multiply by $\ln(\pi)$: $\theta \approx 0.96717027631 * \ln(\pi) * \frac{180^\circ}{\pi} \approx 63.4349489492^\circ$.

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    $\begingroup$ Just proving that $d/dx (e^x) = e^x$ isn't enough, we should also prove that there can be no other solution to $ dy/dx = y ; y(0) = 1$, or just mention that it can be proved. $\endgroup$
    – GPerez
    Commented Apr 14, 2015 at 10:22
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    $\begingroup$ @GPerez Well, there are a lot of things that need to be proved that I don't mention (like the derivatives of sine and cosine)--I just think the root (and hardest one) is the derivative of $e^x$. You don't necessarily need to prove that $e^x$ is the unique solution to your posed diff. eq.--proving the derivative is enough to produce a Taylor Series which can be used to show Euler's formula. So that alone, shows that $e$ is a "good" choice--although I agree, you need to show that it's the only base that works to fully appreciate how "special" $e$ is. $\endgroup$
    – Jared
    Commented Apr 14, 2015 at 16:37
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    $\begingroup$ I only remarked on it because you said and e alone, but didn't mention that this is a fundamental result. No worries though, I just thought somebody should say something about it, so I did. $\endgroup$
    – GPerez
    Commented Apr 14, 2015 at 17:36

Once we know $e^{i\theta} = \cos\theta + i \sin\theta$, the expansion of a complex number into $re^{i\theta}$ then becomes natural. What is the idea behind the polar form? It is to express a complex $z$ in terms of a magnitude and direction. What do we mean by direction? Well, in general a "direction" is just a unit vector. We use this term because multiplying by "just a direction" shouldn't change any magnitudes. This tells us that unit vector is a good definition of "direction". How can we write unit vectors then? We just said how! The unit circle is parametrized by $\cos\theta + i \sin\theta$. We now have no choice. We must pick $e$ and not $\pi$ as you question. The math has decided for us that $\cos\theta + i\sin\theta = e^{i\theta}$, not $\pi^{i\theta}$. Once we have settled on how we are going to express directions, now what does our polar form dictate the corresponding magnitude should be? The polar magnitude of $z$ has to be $|z|$, how convenient!


As you can read on Wikipedia, Euler's formula was found by comparing the series expansions of the exponential function $$ \exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!} $$ with those of the trigonometric functions $$ \cos(x)=\sum_{n=0}^\infty [n\textrm{ even}](-1)^{n/2}\,\frac{x^n}{n!} \quad\text{and}\quad \sin(x)=\sum_{n=0}^\infty [n\textrm{ odd}](-1)^{(n-1)/2}\,\frac{x^n}{n!} $$ The exponential function given by the above series, which can be deduced from the condition that the function is its own derivative and has constant term$~1$, can be (and unfortunately usually is, because it is somewhat more compact) written as $x\mapsto\mathrm e^x$ where $\mathrm e\stackrel{\textrm{def}}=\exp(1)\approx2.718281828$. But it is the exponential function, not this constant, that is of interest. The reason that this "base of the exponential function" must be used is similar to the reason that for the trigonometric functions angles must be measured in radians; if one does not do that, the series get weird constants in their coefficients.

The graphic representation of the complex numbers, and therefore the realisation that Euler's formula can be interpreted as describing complex numbers in polar coordinates, is of more recent date, and was unknown to Euler.


The choice of $e$ comes from Euler's proof that $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. This was not an arbitrary choice. The formula comes out of a mathematical technique called analytical continuation. It would not have held true for any arbitrary base raised to an imaginary power. His efforts basically showed that not only did $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ make some sense, but it was actually the only valid answer which maintained some key properties.

Euler's formula turned a mathematical quirk (the idea of the square root of -1 having a value) until something that is meaningful (a model of rotation in two dimensions). The fact that 'e' happens to be the correct exponent to tie things together has been considered a marvel for some time.

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    $\begingroup$ Analytic continuation in Euler's days, really? One didn't even realise complex numbers corresponded to points in a (real) plane then. You should not attribute to complex function theory what can easily be explained by naive manipulation of formal power series. The same goes for Newton's binomial series by the way. $\endgroup$ Commented Apr 15, 2015 at 7:48
  • $\begingroup$ Also I cannot see what is so "marvellous" about '$\mathrm e$'. I think people were aware fairly early that the ("exponential") power series $\exp(x)=\sum_n\frac{x^n}{n!}$ gives exponential growth (even though they were really more focussed on logarithms initially); if one posits that $\exp(x)=b^x$ for some base$~b$, then it does not require much genius to deduce that $b=\exp(1)$. There is no more mystery about $\exp(1)$ being exactly the right base for the exponential function than that there is mystery about $\pi/2$ being exactly the first zero if $\cos$; it is basically their definition. $\endgroup$ Commented Apr 15, 2015 at 7:58
  • $\begingroup$ @MarcvanLeeuwen Hmm, you're right. I had to go re-sort my history in my head. I do remember analytic continuation being used to show that there was only one possible valid complex natural-logarithm, but apparently that was done much later than I remembered. As for $e$, I think the part that is so fascinating is how persistently it re-appears in formula after formula. I attribute much of that to it being one of a very small number of situations where $\frac{df}{dt} = f$. I think the only other equation that has that attribute is f(x) = 0. $\endgroup$
    – Cort Ammon
    Commented Apr 15, 2015 at 15:44

For the same reason that we use radians. This is a natural base, such that $(e^z)'=e^z$.

  • $\begingroup$ Why these unexplained downvotes ? $\endgroup$
    – user65203
    Commented Aug 5, 2015 at 8:14


The choice of $e$ as the base normalizes $\theta$. In other words, $\theta$ is an angle expressed in radians.

We could choose a number I will call $p$ where:


Then $z=rp^{i\theta}$ normalizes $\theta$ in degrees, rather than radians.


Remember how in high school you had to take the cube root or fourth of some number and only getting one answer when in taking the square root you always had at least two roots? If you use Euler's formula, there's an ingenious to come up with all n distinct roots of an nth root.

take a number, real or complex, i.e., $ a+ib $ set that equal to $re^{i\theta}$ where $ r$ represents the distance from the origin to where $ a+ib $ would be on the complex plane, i.e. $r = \sqrt{a^2 +b^2}$ and $\theta $ is the angle that lies between that you have to rotate clockwise from the x axis until you hit the point $a+ib$ on the complex plane. One could look at this like the point $(a,b)$ on a regular $xy$ plane.

Now if you want to take the nth root of $re^{i\theta}$ that would be the same as $(re^{i\theta})^{1/n}$ which equals $r^{1/n}e^{i\theta/n}$. However, that's not the only answer that will work. You can also have $r^{1/n}e^{i\theta/n}e^{[2\pi i/n]}$ , i.e. , $r^{1/n}e^{i(\theta + 2\pi) /n}$ because if you take that to the nth power, you get $$ (r^{1/n}e^{[i\theta +2\pi i]/n})^n = re^{i(\theta + 2\pi)} $$ where, since $2\pi$ on the unit circle would just give us a full revolution on the unit circle, giving us $2\pi + \theta = \theta$, so $ re^{i[\theta + 2\pi]} = re^{i\theta} $ and you can take any even number, $2k$, replace it with 2 in the above equation, i.e. $$ (r^{1\over n}e^{{i\theta +2k\pi i}\over n})^n = re^{i(\theta + 2\pi k)} $$

and that will give you n distinct solutions for k = 0 to n-1 because if you say substituted values greater than n-1 then at n you'd have $ r^{1/n}e^{[i\theta +2n\pi i]/n} $ which is just $ r^{1/n}e^{i\theta /n} $. Every value of k beyond that would give you a repeat. The formula that we're looking at, i.e. before we took the power to get our initial value, is then

$$ r^{1/n}e^{[i\theta +2k\pi i]/n} $$

where k is any integer from zero to n-1

Now if we take the 4th root of 81, we know that 3 is an answer right off the bat because 3*3*3*3 = 9*9 = 81, but what about the rest of the roots? if we convert it to the complex polar form, I forgot to mention this but we must also know that $ 0 \lt \theta \le 2\pi$ so $ 81 = 81e^{2\pi i}$. so we get $[81]^{1/4} = [81e^{2\pi i}]^{1/4},[81e^{[2\pi+2\pi] i }]^{1/4},[81e^{[2\pi+4\pi] i }]^{1/4},[81e^{[2\pi+6\pi] i }]^{1/4} $ corresponding to $k = 0,1,2,3. $ this then breaks down to $ 3e^{\pi/2 i}, 3e^{\pi i}, 3e^{3 \pi/2 i}, 3e^{2\pi i} $ which when you look at the complex plane these are the points where a circle of radius 3 hits the real and imaginary axes of the complex plane, namely 3i, -3, -3i and 3.


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