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Re: [math_club] re: logarithm

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  • mahbubul hasan
    Well, I don’t know mhasan sir was talking about which restrictions? But well in my letter I restricted these things: 1. Only discussion on Olympiad level
    Message 1 of 5 , Apr 1, 2005
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      Well, I don�t know mhasan sir was talking about which restrictions? But well in my letter I restricted these things:

       

      1. Only discussion on Olympiad level problems. If anyone want to talk on school college level problems then please discuss that on 22bi7 group. I said this because we have very short time in hand and in this short time we don�t want to waste our time behind those problems.

       

      2. after finishing all the previous problems we will launch to new ones. Because if problems keep storing then we will not be able to learn and then we will be under pressure!

       

      I think I have said these two facts and I am sure all the campers will agree with me (what do you say nafi and mahbub sir?)

       

      Well, for that logarithm fact:

      We will only consider natural logarithm not the common one. (I hope you know that natural logarithm is e based)

       

      ln(-a)=x. where a > 0.

       

      So -a=e^x < 0.

       

      So it is obvious that for x real e^x > 0. Contradiction!

       

      So we will consider x to be complex.

       

      First let x=i*theta.

       

      In that case: -a=cos(theta)+isin(theta).

       

      But LHS is real hence sin(theta)=0. But we know -1<=(theta)<=1. so what will happen for �a<-1? Let�s consider a factor �a�=e^y and sin(theta)=0 and cos(theta)=-1. That case theta=pi.

       

      So, -a=e^(y+i*pi)

       

      ln(-a)=y+i*pi=ln(a)+i*pi

       

      I think mahbub sir wanted to refer to this!

       

      Thanks for raising such a nice question!

       

      Anyhow, what about the campers? We are getting mails from only the outsiders? And from camp only me and na�f. Why? Where are others?

       

      Well i did not want to post this problems before the past one ends but it is related with the napolean so i am posting this:

       

      show that the difference of the areas of the two Napoleon triangles (the outer one and the inner one that can be constructed considering the equilateral triangles inside direction of ABC) is equal to the area of the original triangle ABC.

       

      Shanto

      Rajshahi new Govt Degree College

       


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    • lucid00b9
      Dear All, (1) Please write all of you. Nafi and Shanto and Suman are writing most of the messages (2) Do not get offended if someone disagrees with your
      Message 2 of 5 , Apr 1, 2005
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        Dear All,

        (1) Please write all of you. Nafi and Shanto and Suman are writing
        most of the messages

        (2) Do not get offended if someone disagrees with your solution.
        Everybody is here to learn. The smartest person is who knows
        how to benefit from others' intelligence.

        (3) I should add one thing about logarithms. There is actually an
        ambiguity about the log of a complex number. If

        z = r exp(i theta) = r [ cos(theta) + i sin(theta) ]

        then

        log z = log r + i(theta + 2 pi n)

        Here n is an integer. This is because

        exp (i theta) = exp [i (theta + 2 pi n)]

        because

        cos(theta + 2 pi n) = cos(theta)

        and

        sin(theta + 2 pi n) = sin(theta)

        Thus

        log (-1) = log 0 + i(pi + 2 pi n)
        = i(pi + 2 pi n)

        Hence log(-1) = i pi + 2 pi i n . This is very interesting because
        it says that log (-1) doesn't have a unique value. It is what is
        called a multi-valued function. We usually want log functions to
        have only a single value, so we restrict theta as

        0 =< theta =< 2 pi

        In that case log z is single-valued. This is called choosing the
        branch of a logarithm. The theory of Riemann surfaces is the study
        of complex multivalued functions.

        salam,

        mahbub majumdar
        (Bangladesh Math Olympiad Trainer)
      • mahbubul hasan
        Well it seems that the old problems are very tough for you. So it would be better to move on than halt! So mahbub sir can you give us problems?
        Message 3 of 5 , Apr 1, 2005
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          Well it seems that the old problems are very tough for you. So it would be better to move on than halt! So mahbub sir can you give us problems?

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