PLEASE HELP ! is this proof correct ?

[abstract sail]

I don’t know what’s the problem with the proof

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[summer iron]

probably you can’t do stuff normally when you have complex

[forest sundial]

Complex logarithm is multivalued.
e^z = w
z = Ln(w) = ln(w) + 2πin, n ∈ ℤ

[icy escarp]

Nope you can’t because the complex exponential is not injective and also the complex logarithm has multiple definitions in the complex plane (multiple branches) which gives multiple values

[abstract sail]

Oh , thx

Thanks for the help

[abstract sail]

I checked and it’s correct , ln(x) for x < 0 is a complex function , and most complex functions are multivalued including ln(x) , x < 0

[forest sundial]

Well, complex logarithm is always multivalued, even for real arguments.
For example, ln(e^2) = 2, but Ln(e^2) = 2 + 2πin, n ∈ ℤ.

[abstract sail]

So , ln(x) for any value of x , is multivalued

And that Ln(x) = ln(x) + 2πin, n ∈ ℤ.

That’s what I understood

[forest sundial]

Don't confuse ln with Ln.

[abstract sail]

Yeah , I modified

[forest sundial]

ln is single-valued, Ln is multivalued.

[forest sundial]

Should be Ln(x) = ln(x) + 2πin, n ∈ ℤ.

[abstract sail]

Oh srry

[abstract sail]

Thx for the help

[barren depot]

@abstract sail has given 1 rep to @forest sundial

[forest sundial]

You're welcome!

Note that the same relationship holds for Arg(z).

Though, that's just because arg(z) = ln(z/|z|), so that's not very suprising.

[abstract sail]

Oh

Thx

Ln(x) is multivalued and ln(x) is single-valued

[forest sundial]

Yup.

[abstract sail]

And arg(z) = ln(z/absolute_value(z))

[forest sundial]

A similar thing happens for Arg and arg, Arcsin and arcsin and such.

Since they are all logarithms, technically.

[abstract sail]

Oh ok , so func(x) is a function , and Func(x) is a complex function

[forest sundial]

Well, only for some specific functions.

[abstract sail]

Yeah , but the concept is the same

ln(x) and Ln(x) ….

[forest sundial]

Ah, then yeah.

I've seen such notation used for ln, arg and inverse trigonometric functions. Well, inverse hyperbolic functions should also work the same way.

[abstract sail]

But what is the general formula for making a function complex?

Is there one ?

[forest sundial]

Sorry, don't know that, haven't studied complex analysis.

[abstract sail]

Oh , don’t worry , ngl it’s difficult

I tried to study it but it’s really hard

I’m just studying real analysis

[icy escarp]

Im not sure but I think it has something to do with series expansion, the idea is that a function that has a series expansion on an open subset of C is holomorphic on that subset ( so continuous and différentiable). So to extend the notion to the complex plane using series expansion is a sort of bridge between real and complex

Im not completely sure though

[abstract sail]

We know that Ln(x) = ln(x) + 2πin, n ∈ ℤ.

If my proof is true , it means that Ln(x) = ln(x) + 0 because i said that 2πin = 0 , so we arrive at a contradiction of Ln(x) = ln(x) , and we know it’s false , so it means that 2πin ≠ 0 , so my proof is false

Yeah , not a rigorous proof

[forest sundial]

I think I know how to correct your mistake.

[abstract sail]

How ?

[forest sundial]

Well, your mistake in understanding, rather.
Do you agree that sin(0) = sin(2π)?

[abstract sail]

Yeah , if the first one is in degrees and the second one in radians

[forest sundial]

Both in radians, as usual.

If I wanted degrees, I'd write the symbol.

So, if sin(0) = sin(2π), does that mean 0 = 2π?

[abstract sail]

No , so 0 ≠ 2*pi

[forest sundial]

Right.

That's the mistake you were making above.

e^w = e^z does not in general imply w = z.

It does imply w = z + 2πin for some integer n, though.

Same as sin(a) = sin(b) implies a = b + 2πn or a = π - b + 2πn for some integer n.

[abstract sail]

So where is the mistake in my proof ?

I used the property of natural logarithm

[forest sundial]

You assumed that 2ln(-1) is equal to ln((-1)^2), which is incorrect.

If you used Ln instead of ln, that would be correct.

[abstract sail]

Oh ok

[forest sundial]

Yes.

[abstract sail]

And Ln(x) = ln(x) + 2πin, n ∈ ℤ.

Yeah

So , if it only applies to Ln(x) , then I made a mistake

[forest sundial]

In general, yes.
If x is a positive real number, then it also applies to ln(x), of course.

[abstract sail]

Ah ok

We should replace ln(x) by Ln(x)

[forest sundial]

You just have to be a bit more careful when complex numbers are involved.

[abstract sail]

Ok thx for catching the mistake

The equality a*ln(b) = ln(b^a) applies when b > 0

[forest sundial]

Yup.

[abstract sail]

Oh thx for the help