$\pi$ is a real number $\mathbb R$ and can be calculated using an infinite product.

As far as I know, $\mathrm{e}$ is a real number $\mathbb R$, too.

There is an exponential function which is $\mathrm{exp}(x) = e^x$ , and can be defined by an infinite sum, too.

But, I do not understand why this equivalence is correct:

$\mathrm{e}^{\mathrm{i}\,\pi} = \mathrm{exp}(\mathrm{i}\pi) = \cos\left(\pi \right) + \mathrm{i}\,\sin\left( \pi\right) = -1$

I know that there is a definition of $\mathrm{exp(x)}$ where $\mathrm{exp(x) = e^x} \ \forall \ x \in \mathbb R$ , and which has additionally the characteristic of being $\mathrm{exp(\mathrm{i}\pi)} = -1$ by allowing $x \in \mathbb C$ .

However, how can we know that $e^x = \mathrm{exp(x)}$ for $x \in \mathbb C$ ?

My opinion is that $e$ is a number which is defined as

$e = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n$ , $e \in \mathbb R$

and therefore is $\mathrm{e}^{\mathrm{i}\,\pi} = 2.71828182845904...^{\mathrm{i}\,\pi}$ , but $\mathrm{exp(\mathrm{i}\pi)} = -1$ . Therefore $\mathrm{exp(x)} \ne e^x$ for $x \in \mathbb C$ . I think $e$ is a number, so $e^x$ is simply a power equivalation, while $\mathrm{exp(x)}$ is a function.

Can you help me clarify this, so I can understand?


2 Answers 2


Well, I suppose you could leave $e^z$ undefined for complex $z$. The fact that $\exp(z)$ agrees with $e^x$ on the real line, and is the only so-called "analytic" function which agrees with $e^x$ on the real line, is the reason we define $e^z=\exp(z)$. For example, $$\exp(z)=\lim_{n\to\infty}\left(1+\frac{z}{n}\right)^n$$

But it really is worth thinking of $\exp(z)$ as the definition of $e^z$ for complex $z$. It has the nice properties that we'd want: $\exp(w+z)=\exp(w)\exp(z),$ specifically, and the complex derivative of $\exp(z)$ is $\exp(z)$.

But if you object to a really useful definition, you are stuck writing the above formula as:



$$\lim_{n\to\infty} \left(1+\frac{i\pi}{n}\right)^n = -1$$

or whatever other format you want. The above formula though is capturing what is deeply surprising about the formula - that there is some relation between the exponential function $e^{x}$ and the circle.

  • $\begingroup$ Does this definition mean, that the power-operation gets re-defined like this? $b^z = pow(b, z) := exp(z) \ \ \mathrm{if} \ \ b=e \wedge z \in \mathbb C$? I am stunned that an operator can get re-defined, because the result of the re-defined operation (-1) would differ from the result of the original operator's result (an unsolveable $e^i\pi$) . So, is is completely wrong to interprete $e \in \mathbb R$, so that $e^{i\pi}$ is a term that cannot be further solved? $\endgroup$ Commented Dec 1, 2015 at 7:23

First, "$\exp(x)$" is notation. It is defined to be identical to $\mathrm{e}^x$, whatever that is.

Second, there is nothing challenging about using complex exponents: $a^{x+ \mathrm{i} y} = a^x a^{\mathrm{i}y}$. However, to simplify further, we need to know what to do with those imaginary exponents.

When you study calculus of a real variable, you eventually construct the power series representation of $\mathrm{e}^x$, namely, $$\begin{align*} \mathrm{e}^x &= 1 + x + \frac{1}{2!} x^2 + \frac{1}{3!} x^3 + \cdots \\ &= \sum_{j=0}^\infty \frac{x^j}{j!}. \end{align*}$$

It is a happy conincidence (not really) that the same syntactic recipe yields a well behaved function if we pretend that $x$ is a complex number rather than a real number. The tradition is to indicate this change in point of view by changing the variable to $z \in \Bbb{C}$ instead of $x \in \Bbb{R}$. To wit, $$\begin{align*} \mathrm{e}^z &= 1 + z + \frac{1}{2!} z^2 + \frac{1}{3!} z^3 + \cdots \\ &= \sum_{j=0}^\infty \frac{z^j}{j!}. \end{align*}$$

Now we can compute $\mathrm{e}^{\mathrm{i} x}$ (or at least try). $$\begin{align*} \mathrm{e}^{\mathrm{i}{x}} &= 1 + \mathrm{i}x + \frac{1}{2!} \mathrm{i}^2 x^2 + \frac{1}{3!} \mathrm{i}^3 x^3 + \cdots \\ &= \sum_{j=0}^\infty \frac{\mathrm{i}^j x^j}{j!}. \end{align*}$$ If you've followed the path described above, you will have also computed power series expansions of $\sin(x)$ and $\cos(x)$. If we manipulate our series, these will pop out... $$\begin{align*} \mathrm{e}^{\mathrm{i}{x}} &= 1 + \mathrm{i}x + \frac{1}{2!} \mathrm{i}^2 x^2 + \frac{1}{3!} \mathrm{i}^3 x^3 + \cdots \\ &= \sum_{j=0}^\infty \frac{\mathrm{i}^{2j} x^{2j}}{{2j}!} + \sum_{j=0}^\infty \frac{\mathrm{i}^{2j+1} x^{2j+1}}{{2j+1}!} \\ &= \sum_{j=0}^\infty \frac{(-1)^j x^{2j}}{{2j}!} + \mathrm{i} \sum_{j=0}^\infty \frac{(-1)^j x^{2j+1}}{{2j+1}!} \\ &= \cos x + \mathrm{i} \sin x. \end{align*}$$ (There are still technical details to work out in the above, like ensuring that the rearrangement of the series is valid (follows from absolute convergence), but these are not the important milestones to recognizing the result you are asking about.)


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