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Thread: prove e^{i\theta}

  1. #1 prove e^{i\theta} 
    Forum Ph.D. Heinsbergrelatz's Avatar
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    can anyone explain and prove why

    thank you~


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  3. #2 Re: prove e^{i\theta} 
    . DrRocket's Avatar
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    Quote 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.

    What you do is start with the power series for
    and then observe that the real part is and the imaginary part is


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  4. #3  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    may i ask where the thread is?

    thx
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  5. #4  
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    Quote Originally Posted by Heinsbergrelatz
    may i ask where the thread is?

    thx
    Certainly you can ask.

    But I do not recall the title. You can do the search as easily as I can.
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  6. #5  
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    You can do the search as easily as I can
    reasonable suggestion.
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  7. #6  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    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;







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  8. #7  
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    Quote 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.
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  9. #8  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    so is my proof wrong?
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  10. #9  
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    Quote 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.
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  11. #10  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    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
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  12. #11  
    . DrRocket's Avatar
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    Quote 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).
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  13. #12  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    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
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  14. #13  
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    Quote 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 .
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  15. #14  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    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
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  16. #15  
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    Quote 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.
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  17. #16  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    just a question, but can this formula be proved by using series expansion?
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  18. #17  
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    As DrRocket mentioned several times, that's the default way of doing it.
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