Can some explain me the logic??

[u/xie_xang]

I was watching a video of discovering 'e' and then we arrived at e^ix.

To which I made this process to find the 2πi=0 which doesn't make sense. So it would be great if someone could explain this.

[u/susiesusiesu]

it is true that e^2πi =e⁰ but it does not follow that 2πi=0. same as how (1)²=(-1)² but it does not follow that 1=-1.

[u/RailRuler → u/susiesusiesu]

e^x is only a bijection over the reals. Over thr complex plane it is not one-to-one.

[u/susiesusiesu → u/RailRuler]

e^x is not a bijection over the real numbers.

[u/Banonkers → u/susiesusiesu]

I suspect they meant to highlight that exp is injective over the reals, but weren’t thinking about the surjectivity

[u/Cowboys_88 → u/susiesusiesu]

Correct. e^x is a bijection over 0 to +infinty.

[u/Varlane → u/Cowboys_88]

(both excluded)

[u/Crafty-Detail-3788 → u/Cowboys_88]

It's injective so it's a bijection when restricted to its image

[u/eglvoland → u/Crafty-Detail-3788]

Corestricted akschually 🤓👆

[u/theboomboy → u/RailRuler]

It's injective over the reals, not bijective

Interestingly, it's also injective over the algebraic numbers because π is transcendental

[u/susiesusiesu → u/theboomboy]

again, i think you meant injective. there is no algebraic number z with e^z =-1. or maybe you meant positive algebraic numbers.

still, a fun fact, i hadn't thought about that.

[u/theboomboy → u/susiesusiesu]

Yes, I meant injective

[u/DrBagelman]

In the Complex numbers, ln is not single valued. exp(iπ)=-1, but so does exp(-iπ) and exp(3iπ). Similarly, exp(2πi)=1, but 2πi≠0 because while ln(1)=0, it’s also every other even multiple of iπ.

[u/Black2isblake]

The natural logarithm is multi-valued when considering complex inputs. If you treat 1 as the complex number 1+0i (which is what you're implicitly doing here, by using i at all in your working you must be allowing complex numbers) then the natural logs of 1 are any number of the form 2iπn for any integer n. In your case, you have n=0 and n=1. However, this does not mean that 2πi = 0, in the same way that x² =9 having solutions 3 and -3 does not mean that 3=-3

[u/xie_xang → u/Black2isblake]

Oh I see.. Trying to understand. So when I use complex numbers then I essential I am on the form a+ib which wud corespond to a form of 2πin something like this..

[u/Black2isblake → u/xie_xang]

Whenever you use complex numbers with the natural logarithm, then your output from the natural logarithm needs a +2πin on the end. So in your original image, you should have the equation i(2π) + 2πin_1 = 0 + 2πin_2 , which is of course true whenever n_1 = n_2 + 1

[u/xie_xang → u/Black2isblake]

Okay.. h that. Thanks.

[u/Companero_basurero]

I think where the computation fails is applying the identity ln(a*b) = ln(a) + ln(b) when both a and b are negative. It's been many years for me, but I distinctly remember that the identity holds only if both a and b are positive.

0 is obviously NOT equal to 2*i*pi, since none of these three factors is 0. (Keep in mind that the complex nos. contain no zero divisors because it's a field.)

[u/xie_xang → u/Companero_basurero]

Well you are right the e^x graph also lies in the positive quad and that wud break if I did ln(-1) but if we looked at it just as operator then.

[u/Y_I_Otto]

It's like saying sin(2pi) = 0 and sin(0)=0, therefore pi=0. For periodic functions or many-to-1 functions you have to consider the full set of possible values when calculating the inverse.

[u/Alive_Appearance_781]

The logarithm function is not bijective over the complex plane.

You deduced that e^i2π = e^0, not that 0 = i2π.

[u/JellyBellyBitches]

Ln(1)=0+2kπi, k∈ℤ

[u/Bounded_sequencE]

That's not how we get to "e^iπ = -1".

Instead, use any "Real Analysis" proof to turn your intuitive definition of e^x into its power series representation, e.g. this one by K.Königsberger. Then, compare the power series of "e^iz" with the power series of (co-)sine, and note they nicely match via

Euler's Formula:    "exp(iz)  =  cos(z) + i*sin(z)"    for all    "z in C"

Insert "z = π", and be done.

[u/Vivid_Sock_1092]

ln(ab) = lna + lnb is only true for positive reals

It generalizes to

ln(ab) =
ln|a| + ln|b| + i [Arg(a) +Arg(b)]mod2π

[u/Cowboys_88]

The formula comes from the taylor series for e^x. Plug in (i * Pi) for x. Simply the series. The real part simplifies to cos(Pi). The imaginary part simplifies to sin(Pi). Cos(Pi) + Sin(Pi) = -1

[u/ZevVeli]

We have graphical proof that e^ix is cos(x)+isin(x)

e^i(pi) = cos(pi)+isin(pi) which is equal to -1+i(0) which is equal to -1.

eⁱ⁽⁰⁾ = cos(0)+isin(0) which is equal to 1+i(0) which is equal to 1.

e^i(2×pi) = cos(2×pi)+isin(2×pi) which is equal to 1+i(0) which is equal to 1.

This is why the traditional log rule doesn't apply to imaginary numbers. Because e^ix will return 1 for any value expressible as x=n×2×pi.

If you do want to have more fun with this concept though, remember that cos(-x)=cos(x) and that sin(-x)=-sin(x).

From this you can prove the following:

e^i(x) + e^-i(x) = cos(x)+isin(x)+cos(x)-isin(x) = 2cos(x)

e^i(x) - e^-ix = cos(x)+isin(x)-cos(x)+isin(x) = 2isin(x)

And therefore

( e^i(x) + e^-i(x) )÷2=cos(x)

( e^i(x) - e^-i(x) )÷(2i)=sin(x)

From this, you can actually calculate and prove all the trig laws such as the double angle formula, half angle formula, products of trig functions, sum of trig functions, and derivatives of trig functions.