Richard Feynman referred to Euler's Identity, $e^{i\pi} + 1 = 0$ as a "jewel."

I'm trying to demonstrate this jewel without recourse to a Taylor series.

Given $z = cos\theta + i sin\theta\; |\;|z| = 1$,

$$\frac{dz}{d\theta}= -sin\theta + icos\theta =i(isin\theta+cos\theta)=iz$$

Now, if I let $z=u(\theta)$, then, $$\frac{du}{d\theta}=iu(\theta)$$
Undoing my original derivative, $$\int iu(\theta) d\theta =u(\theta)+C$$ $$ \therefore z=u(\theta)=e^{i\theta}$$ which is the general case. Substituting $\pi =\theta$ for the special case, and invoking the original equation, we are left with $z=cos\pi + isin\pi =-1 = e^{i\pi}$ $$\therefore e^{i\pi}+1=0$$

When I first worked through this, the constant of integration disturbed me, like a nasty inclusion marring the jewel. But now, I think it's fair for me to excise it in the line, $\;\therefore z=u(\theta)=e^{i\theta}$

Is that correct?

  • $\begingroup$ I think its fine, but you can even do it from the differential equations (I think its a better way). And there are many other methods to do it without the Taylor series. $\endgroup$ – Kartik Sep 5 '15 at 15:58
  • 1
    $\begingroup$ It looks ok. Also, as a reference I would suggest Needham's Visual Complex Analysis. He has a great motivation for this formula in the first chapter, which is essentially the differential equation derivation if I remember correctly. $\endgroup$ – Moya Sep 5 '15 at 16:01
  • $\begingroup$ See this answer, especially the 2nd answer and the answers after it. $\endgroup$ – Kartik Sep 5 '15 at 16:01
  • $\begingroup$ It' fine if you $define$ $ e^{i x}$ as the solution to $u'=i u$ with $u(0)=1$. Or if you show that the real function $e^x$ has an analytic extension to $C$. $\endgroup$ – DanielWainfleet Sep 5 '15 at 16:37

So, just rearranging a bit the proof, the two functions: $$ f(z) = \cos(z)+i\sin(z),\qquad g(z)=e^{iz} $$ are indeed the same function since they both are solutions to the first-order ODE $$ h'(z)-i\cdot h(z) = 0 $$ with the initial condition $h(0)=1$. We are just exploiting the Cauchy-Lipschitz theorem.

By evaluating $f$ and $g$ at $z=\pi$ we get Euler's celebrated identity.

The proof is perfectly fine - but it is not really different from proving $f\equiv g$ by comparing their Taylor series, if we consider that Frobenius' power-series method is the most natural way for solving some ODEs. The same situation happens in a discrete setting, for instance. We may prove that:

$$ F_n = \sum_{0\leq k \leq \frac{n-1}{2}}\binom{n-1-k}{k} $$ holds by exploiting the binomial theorem, or by proving that both the LHS and the RHS satisfy the same recurrence relation with the same initial conditions.

  • 1
    $\begingroup$ Nice insight. Thanks. $\endgroup$ – Shailesh Sep 5 '15 at 16:21
  • 1
    $\begingroup$ This doesn't prove Euler's identity. It demonstrates the formal analogy between the exponential function and rotations (which follows from both being homomorphisms from $R$ to some group). To prove the identity one needs to justify the equation $g(z) = e^{iz}$ by showing that it follows from a uniform definition of $e^z$ that works for both real and imaginary $z$. $\endgroup$ – ASCII Advocate Oct 22 '15 at 2:22
  • 1
    $\begingroup$ In the existence and uniqueness theorem for $h'(z)=ih(z)$, the $z$ is a real number. This does not tell us anything about the existence of an extension of the solution of $E'(x) = E(x)$ from real $x$ to complex $x$, which is what would be needed in order to write equations like $g(z) = e^{iz}$ and to assert that $g'(z)=ig(z)$. There are some related remarks in my answer. $\endgroup$ – ASCII Advocate Oct 23 '15 at 0:28
  • 1
    $\begingroup$ It is not simply a (metrized) "field extension" of the C-L proof. In the iteration proof of existence and uniqueness, the complex integrals become potentially dependent on the path. If you allow arbitrary paths it spoils the Lipschitz upper bound. And if you use straight-line paths you obtain the bound, but only radial differentiability rather than a complex derivative $dy/dz$. The first two problems are solved by iterating on holomorphic functions, but one still has to show analyticity of the solution. $\endgroup$ – ASCII Advocate Oct 23 '15 at 2:32
  • 1
    $\begingroup$ I think the point is that the notation "$e^{iz}$" used in defining $g(z)$ needs to be justified (that is, related to the real exponential function) by some method. Euler essentially showed that any of the constructions he gave for $e^t$, can be applied consistently for complex $t$ (without having exactly the concept of analytic continuation, or the complex plane). Modern arguments such as convergent series or solving the complex ODE are more precise, but as in Euler's work, these constructions are deeper than the one-line computation relating Exp and Sin/Cos, and are the core of the proof. $\endgroup$ – ASCII Advocate Oct 23 '15 at 22:47

This is not a proof of Euler's discovery, but of something much weaker.

What Euler noticed is that there is a connection between the exponential function and the trigonometric functions. Your proof only shows that solutions to $F'(t)=iF(t)$ (for real $t$) are equivalent to uniform circular motion in the complex plane. This suggests some formal relation of $F(t)$ to exponentials, but it does not:

  • give a construction of $F(t)$ independent of circular motion;
  • show that $F(ix)$ exists for real $x$
  • show that $F(-ix)$, for real $x$, equals the function we know as $e^x$.

Those (or their equivalents) are the earth-shaking facts that Euler uncovered and the reason his formula is celebrated.

Starting from a more advanced formalism (such as power series) where $e^z$ is defined for both real and imaginary $z$ then an argument like this one might reproduce Euler's finding.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.