Proof of Euler's Formula Without Taylor Series

This is an important result in Complex Analysis. By letting z be a function that maps real numbers to complex numbers defined as z(θ) = cos(θ)+isin(θ), we can differentiate z and solve the resulting differential equation to prove Euler's Formula. This method is more rigorous than the classic Taylor Series proof as it does not involve rearranging an infinite sum.

[willassad]

First, note that this formula is a DEFINITION in complex analysis, so this argument is not a “proof”. It is, however, a very nice explanation of why this definition makes sense. Note that the step at 2:24 isn’t exactly rigorous as it requires integration over a complex domain, which may not be well-defined.

[andrewhone]

As pointed out by someone else, this is not really a proof, because you are assuming things about complex integration and complex functions that require justification. So what this really shows is the consistency of various operations that are familiar with real variables and functions. For instance, how do you define the natural logarithm ln for a complex argument z? (In fact, it is a multivalued function with infinitely many branches.) Similarly, how do you define the exponential function for a complex argument, and show that (locally, in some domain) it is the inverse of the logarithm? One way to do this is by solving the ODE dz/dt = z in the complex domain (for complex argument t). Then by integrating along a curve in the t plane you can prove that the solution is a holomorphic function of t, hence it has a Taylor series (which you can compute from the ODE); and after that you can set t = i theta and justify the rest of your argument.

[ubertziop]

The integration is not so problematic, but rather the derivative done in the second line is doubtful since it requires the definition by the Taylor series.

[ralf]

My mind has been blown. I had never seen this proof before and yet it's so elegant and simple.

[gabrielgcarvalho]

Don’t use the expression “Taylor series approximation” . The Taylor series IS the function.

[alielhajj]

The only problem here is using logarithm in the complex setting which is a bit tricky, it’s better to just use the characteristic equation of the ODE to show that this is the solution and we know that such a linear ODE has a unique solution

[marcellomarianetti]

Nice, I'm only a little bit worried of complex integrals, because they usually behave very differently from real ones. And ln(z) can exist in many domains, the usual one is C\{Re(z)<=0}, but it cannot have C as domain

[blokyk]

This is such a nice proof ! :D Thank you for explaining it so clearly and briefly :)

[echodelta]

I shared this with my high school students to lay the foundation for answering the question: "Is there a complex number z such that a nonzero real number raised to the power of z will equal zero?"

Thank you for such a wonderful argument!

[yahyabatat]

This argument reminded me of the chicken and egg dialectic.

[nexonym]

That’s such an elegant proof. Very good video!

[bscutajar]

To anyone confused about taking theta=0:

If I give you the plot of a line y=x+C and you want to find C, you can just pick any point on the line to find C. But typically you'd choose the y-intercept (x=0) since it gives you the answer directly. Picking any other point would give the same answer, since it's the same line.

[vitovittucci]

The second equation needs the first one, that you take for garanted. (actually it is an arbitrary definition). So it's a circular proof.

[chuckstarwar]

The old became new. It's not proof. It's circular around itself. I did that in my homework, and my math professor gave me 0. But it's a good way to PROVE to some special people that "Euler's identity is right".

[nickfleiwer]

I think you can’t integrate 1/z dz to ln z, because it’s not true for complex numbers. Even with the complex log it doesn’t work, I.e. at the branch cut, whereas 1/z integrated is holomorph on C without 0.

[tannhaeuserx]

The whole thing is backward or circular. The problem arises because cos and sin are ill-defined in high school and at the beginning of calculus. To properly define sin and cos, you need to define them by Taylor series or by the exponential function or by solutions to differential equations.

Here is how to do this properly and in the most elementary way. You define ln(x) = \int 1/t dt first. Then you define exp(x) as the inverse function of ln(x). At this point, you introduce the complex numbers and define cos(x) := (exp(ix) + exp(-ix))/2 and sin(x) := (exp(ix) - exp(-ix))/(2i). The Euler formula then comes out almost as the definition.

There are alternatives, but they all come down to this: These functions are all solutions to the 4-th order differential equation (d^4/dx^4) f(x) = f(x). exp(i x), cos(x) and sin(x) are all solutions to this equation. The Euler formula is just a linear dependence relation of these three solutions over the complex numbers.

[BilalAhmed-onkd]

2:28 shouldn't it be ln|z|, which results in 2 possible solutions for u, u=±e^(iθ)

[RB_Universe_TV]

how do even calcute the integral of dθ dz
You can't take integral w.r.t. z and integral w.r.t. θ on both sides.... can you?

[Stephen-cntu]

Pl ease note that in your derivation, e^C is not equal to C. This could be misleading

[mismis]

This how my friend and I managed to convince ourselves it was true when we first saw the formula. This is the first time I've ever seen someone prove it that way and it makes me happy that we were right back then.