1. can anyone explain and prove why

thank you~

2.

3. Originally Posted by Heinsbergrelatz
can anyone explain and prove why

thank you~
I think there is a thread here somewhere where I did just that.

and then observe that the real part is and the imaginary part is

thx

5. Originally Posted by Heinsbergrelatz

thx

But I do not recall the title. You can do the search as easily as I can.

6. You can do the search as easily as I can
reasonable suggestion.

7. instead going through all posts can you see if y proof makes sense, or have dobe it properly?

saying that and we can differentiate the following;

---- if we solve this differential equation, its conclusion will give the outcome of where is a constant.

so here is the working;

8. Originally Posted by Heinsbergrelatz
instead going through all posts can you see if y proof makes sense, or have dobe it properly?

saying that and we can differentiate the following;

---- if we solve this differential equation, its conclusion will give the outcome of where is a constant.

so here is the working;

You are manipulating symbols that sort of get to the correct final answer.
But there are some logical problems with the argument, that are a bit subtle if you have not done rigorous proofs routinely.

The basic problem is that z is a complex variable and therefore one cannot integrate agaiinst dz in the manner that you have done.

The exponential function is fundamental, and one really needs to define the exponential before working with ln, especially in the case of a complex variable.

While it is possible to define ln(|z|) in terms of the integral of a real variable you need ln(z), and the machinery to make sense of that requires that you first have a defnition for

While your argument doesn't quite work, it is in fact a pretty clever attempt.

9. so is my proof wrong?

10. Originally Posted by Heinsbergrelatz
so is my proof wrong?
Yes, your proof is wrong for the reasons that I gave.

But it was an interesting attempt, and understandable if you are not familiar with the calculus of one complex variable.

11. Yes, your proof is wrong for the reasons that I gave.

But it was an interesting attempt, and understandable if you are not familiar with the calculus of one complex variable.
wait i have a better approach.

let

now if we apply the working i gave above,does it make sense? as now i have given a definition in terms of how i arrived at

12. Originally Posted by Heinsbergrelatz
Yes, your proof is wrong for the reasons that I gave.

But it was an interesting attempt, and understandable if you are not familiar with the calculus of one complex variable.
wait i have a better approach.

let

now if we apply the working i gave above,does it make sense? as now i have given a definition in terms of how i arrived at
No, it doesn't work.

The problem is that you are treating as though it was an integral of a function of a real variable which evaluates to the logarithm. It is not a function of a real variable, but rather a function of a complex variable and you do not have such a clean definition of what that means.

As I said, you are manipulating symbols that somehow come up with the right answer, but the symbols don't have an unambiguous meaning here. The theory of one complex variable has some subtleties with which you are probably not familiar, and that is understandable.

One basic problem is that the exponential function is not one-to-one but is in fact periodic. That complicates the idea of an inverse (the logarithm).

13. man,.... how about you just prove it for me, despite the fact you have proved it in some other thread. i would appreciate if you did, so i can learn and move on to the next topic. thank you

14. Originally Posted by Heinsbergrelatz
man,.... how about you just prove it for me, despite the fact you have proved it in some other thread. i would appreciate if you did, so i can learn and move on to the next topic. thank you
I told you how to provie it.

Write down the power series for Then you simply have to notice that the real part is the power series for and the imaginary part is the power series for . That is all there is to it.

In fact, in a truly rigorous advanced class in complex variables that is the definition for and . It has the advantage of being completely rigorous and has no need for pictures or other fuzzy notions. The disadvantage is that you lose some of the geometric intuition. You get it back via the relationship with elementary calculus and the Taylor series for and .

15. Write down the power series for Then you simply have to notice that the real part is the power series for and the imaginary part is the power series for . That is all there is to it.
ah thank you

16. Originally Posted by Heinsbergrelatz
Yes, your proof is wrong for the reasons that I gave.

But it was an interesting attempt, and understandable if you are not familiar with the calculus of one complex variable.
wait i have a better approach.

let

now if we apply the working i gave above,does it make sense? as now i have given a definition in terms of how i arrived at
I can't say it's a correct proof. HOWEVER, if you knew a little complex analysis, I'll bet you could convert this into a correct proof.

Of course, the main problem here is that you have to define something somewhere: either e^z or log(z) must be given some sort of elementary definition. Your approach basically comes down to defining d(log z). It's an interesting and plausible way of going about things. Turning it into a complete proof though would involve a great deal more effort and subtlety than it's probably worth.

17. just a question, but can this formula be proved by using series expansion?

18. As DrRocket mentioned several times, that's the default way of doing it.

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