# “philosophical” question about the transcendence of $\pi$

I don't have any knowledge on transcendence proofs. I just heard that Lindemann proved that for any $\alpha \in \mathbb R^*$ algebraic, $e^\alpha$ is transcendental. Then, since $i$ is algebraic, and so is $e^{i\pi}=-1$, $\pi$ must be transcendental, because otherwise $e^{i\pi}$ would be transcendental.

My question is the following. Of course this proof is "valid", I'm not questioning that. But, since the complex exponential function is constructed as an extension of the real exponential by means of the power series (as far as I know, although maybe it is possible to define it in another way), and although its uniqueness can be proved, does this proof of the transcendence of $\pi$ have as much force as a proof which would not use the complex exponential function ? (As, I think, is the case for the transcendence of the Liouville number and $e$).

To put it in another way : when we talk about $e^{i\pi}$ or $3e^{i\frac{{\pi}}{5}}$, is the $e$ used in the same way as in $e^5$ ; I mean, are we really talking about the number $e$ ? Or is it just a "manner of talk", something which simplifies calculations and helps us find Euler's formulas when we have forgotten them ? If the definition of $e^{i\theta}$ is $cos\space\theta+i\space\sin\space \theta$, then it is something different from "the real number e that we see in real analysis", and then how can we be sure that the property I mentioned, ie the fact that $e^\alpha$ is transcendental if $\alpha$ is algebraic, is still valid for complex values ?

I hope my question is clear enough.

• How do you define $e^{\sqrt2}$? – Mariano Suárez-Álvarez Jan 7 '14 at 21:02
• "Does this proof of the transcendence of $\pi$ have as much force as a proof which would not use the complex exponential function ?" Yes. – Andrés E. Caicedo Jan 7 '14 at 21:03
• recommend maa.org/publications/ebooks/irrational-numbers You can skip the proofs until such time as you are ready. – Will Jagy Jan 7 '14 at 21:06
• When you get a new cellphone, maybe you can bend its screen, which you couldn't with the old one, but you can still make calls. If you are using it to tall with a friend you are using the same properties your old phone had, so you can call them both cellphones. When you fold the new phone into a tiny purse and you can't do the same with the old one you make the remark in the name that the new phone is a bendable cell phone. If all you are using from $e^z$ over the complex are the properties that $e^n$ has over the naturals, for all it matters (in that context) $e^z$ is the exponential of $e$. – user119256 Jan 7 '14 at 21:13
• If you read Lindeman's proof, you will see there is nothing in it where the exponent is assumed to be real. – Gerry Myerson Jan 7 '14 at 21:39

But we're not using complex exponentiation, we're using imaginary exponentiation, which is just a convenient shorthand for circular geometry.

Circles need to come in somewhere. We could as well say this: if $\alpha\in\mathbb{R}$ is algebraic and nonzero, then $\cos\alpha$ and $\sin\alpha$ (and $e^\alpha$, for that matter) are transcendental. And that is really the intuitive geometric core of the transcendence of $\pi$: that the relationship between angles and lengths on a circle cannot be described using polynomials.

Nevertheless, the fact that $\mathbb{C}$ exists means that there will always be some algebraic connection between the exponential and the trigonometric functions. Even if we forget sometimes that circular motion is nothing more than exponential growth (where the growth coefficient is "rotation by 90 degrees"), it makes sense that one would formulate a proof over $\mathbb{C}$, rather than proving two different theorems over $\mathbb{R}$.

Can we multiply a number $\sqrt{-1}$ times with itself ? No. But since $\displaystyle e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=\exp x$ for all real exponents or arguments x, why would we want to intentionally create a self-inconsistency or self-incoherence in the function's main property or definition when it comes to complex exponents or arguments, especially since this formula actually makes sense for them as well ?

I've asked myself the same question, actually, and the idea is that, despite the obvious differences between real and imaginary numbers, all algebraics share the same fundamental properties, and the proof is not dependent upon their differences, but only on what they share in common. Hope this helps.

It is better to think of $e^x$ as the exponential function rather than "the number $e$ raised to the power of $x$", i.e. think of it as $\exp(x)$. Of course the nice thing about this function is that for rational numbers (say $\frac{m}{n}$ where m, n are positive integers) we do in fact have $$\exp\left(\frac{m}{n}\right) = e^{\frac{m}{n}} = \sqrt[n]{\underbrace{e \dotsb e}_m}$$ So when $e^x$ is extended to complex arguments using its Taylor series, think of it as a function on complex values. The number $e$ is there just as a notation.

I'm sure when Lindemann proved his theorem, he used the complex version of the exponential function. So his theorem is perfectly valid when dealing with complex numbers.

• I agree very much with you Pratyush. The notation "$e^x$ " obscures the true nature of the exponential. It is a surprising fact that there exists a real number $e$ such that the exponential can be written as $e^x$! And it is nonsensical to define the exponential as "the number $2.718...$ raised to the power $x$, where $2.718...$ is some magical number called $e$" – Bruno Joyal Jan 13 '14 at 4:06

What might help you is an explanation of why the definition of $e^x$ is equally valid for real and complex exponent. I can think of no better source than Walter Rudin's Real and Complex Analysis. If you can get hold of a copy, his Prologue is all about the exponential function, and he shows you very naturally how the Euler formula arises, for example, along with many other wonderful facts. In summary, the fact that the exponential function is an absolutely convergent power series, it applies to the whole complex plane. As Gerry Myerson points out, the Lindemann proof invokes an algebraic exponent, which is a complex number, not necessarily a real number. So go find a copy of Rudin!