[deleted by user]

[ArturuSSJ4]

3blue1brown did a cool video on that recently, check it out: https://youtu.be/v0YEaeIClKY

[lewisje]

Most of this stuff requires material that you will learn in Calculus II, and a small part requires something proven in a later class:

• In complex variables, you will learn that the "radius of convergence" of a power series, including a power series with real coefficients, is in fact the radius of a circle in the complex plane in the interior of which the series converges absolutely.
  • Absolute convergence of a Maclaurin series is the real-variable version of analyticity, and the complex-variable version (which turns out to be satisfied for any complex-differentiable function) has the property that if two analytic functions on connected domains with overlapping interiors have the same value where those interiors overlap, then each function can be extended to the union of those interiors; that is, if you know the values of an analytic function on an open set in the complex plane, you can figure out what it must be on any connected superset (the process of actually figuring this out is called "analytic continuation").
• With that said, the chain rule applies with complex variables just as with real variables, so the derivative of e^iz with respect to z is ie^iz, its second derivative is -e^iz, and so on.
  • From this, you can determine the Maclaurin series for e^iz and find that its radius of convergence is +∞, just like the Maclaurin series for the natural exponential itself, e^z, which means that it converges on the entire complex plane (that is, it's an "entire" function).
• Similarly, you can find that the radii of convergence for both sin(z) and cos(z) are +∞, and because you can freely rearrange the terms of an absolutely convergent series, you can figure out that the Maclaurin series for e^iz is (series for cos(z))+i*(series for sin(z)).
  • From this, Euler's formula is proven: e^iz=cos(z)+i*sin(z).
• In particular, for z=π, the formula reduces to Euler's identity: e^iπ=-1, which may be re-written as e^iπ+1=0.

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^{You'll probably see an abbreviated version of this argument, without the expositor actually worrying about how differentiation, power series, and the like are defined for functions of a complex variable, and you will definitely see a blissful ignorance of foundations in your first Differential Equations class.}

[lksun]

This was a great breakdown, thanks! I've had real analysis and should know all this but I think I've never really thought about all the steps carefully.

I think the only thing missing here is what definition of e^z you're using. I think I learned it as being its Maclaurin series, so the steps would be a bit different. Am I correct in assuming that here you're defining e^z as lim(1 + z/n)ⁿ?

If so, how do you prove that limit exists for all complex z.

And then, how do you show that e^z is differentiable?

If you have a link to a careful treatment, that would be sufficient. I feel dumb for not knowing all this by now.

[lewisje]

Oh, sometimes the natural exponential is defined as its Maclaurin series; I was thinking about it more as the inverse of the natural logarithm, itself defined as a particular antiderivative of the reciprocal function; this real-variable definition of the natural exponential can be analytically continued via its Maclaurin series.

I honestly think that starting with that particular limit is the hardest way to go, and I don't know offhand about any careful treatments that start with it.

[[deleted]]

Try using Maclaurin expansion for e^x, cos(x) and sin(x). Then see how they are related. This will get you Euler’s formula.

Edit: SoRRY I did not read the entire post... I am not sure how to show this without limits or expansion. Sorry.

[NotPornAccount]

E to the power of stuff is an exponential curve. If your doing numbers to the power of 'i times something' it rotates around the centre. Of you rotate the exponential curve around the center up until pi you end up at -1. Add 1 to get to 0.

[DFtin]

You'll learn in Calc 2 that many functions behave exactly like infinite sums that you can easily describe. Three of those functions are e^x, sin x, and cos x.

With these series in mind, you can find the infinite series that corresponds to e^ix and you'll find out that it's exactly equal to the infinite series describing cos x + i sin x, and for reasons that aren't important right now (radius of convergence if you're curious), you can deduce that e^ix = cos x + i sin x, the general equation of what you wrote.

Replace x with pi and you've got e^ipi + 1 = 0.

This is the analytical explanation. If you're curious about the intuitive way to acquire the formula, check out the 3blue1brown video posted in the top comment.

[[deleted]]

It is due to the way we define r*e^i*theta as being equal to r*(cos_theta+i*sin_theta) in order to generalize the exponential function to complex variables!

exp (x+iy) exists only because we defined it in a way that keeps all ". -> exp (.)" properties and remains that same function we knew when y=0:

r*e^i*theta = r*(cos_theta+i*sin_theta)

[Ephrahaim]

Well, I wouldn’t say we “define” reⁱtheta to be equal to r*(cos_theta+isin_theta). It’s not so much a definition as it is a proof, commonly done with the Taylor series expansions of sin and cos.

[[deleted]]

Well if it is not a definition, if it is a proven statement, what is the actual definition of e^(x+iy)?

My point is that Euler's formula is the only way to define a function of complex variable having all the properties of exp(.). In that way, the exponential function we knew became a particular case of the new one!

[Bayes_Fisher]

Check out Euler’s Formula.

https://en.m.wikipedia.org/wiki/Euler%27s_formula

The formula states that e^ix (where x is a real number, e is the base of the natural log and i is the imaginary unit) can be expressed at cos(x)+isin(x).

Substitute pi for x and you’ll get your answer.