statement interpretation in complex analysis

Discussion in 'Math' started by e-learner, Jan 29, 2016.

  1. e-learner

    Thread Starter New Member

    Apr 25, 2015
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    what is the meaning of following statement.
    f(z) = ((conjugate(z))^ 2)/z, if z ≠ 0 and f(0) = 0. check whether the function is analytic or not.
    I could not understand ,here z ≠0 is given but at the same time the value of f(z) at z=0 is given to be 0. and if the function is not defined at z=0 , then why to bother for the function to be analytic at origin?
     
  2. Papabravo

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    Feb 24, 2006
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  3. e-learner

    Thread Starter New Member

    Apr 25, 2015
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    I am new to this complex analysis. I am just making the use of continuity and differentiation definition where z→0 along different paths. And after doing that I know that this function is not analytic at z=0. But my question is
    In
    f(z) = ((conjugate(z))^ 2)/z, if z ≠ 0 and f(0) = 0

    it is given z≠0 but how come f(0) be defined?

    for example if I am given f(x)= (x^2+ 6)/(x-1) where x≠1 . Is it possible to have f(0)=5?
     
  4. Papabravo

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    Feb 24, 2006
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    To see why the method of defining a value for f(0) is allowed under these circumstances you need to go deeper. The value which is assigned to f(0) to remove the singularity is consistent with values of f(z) in the δ-neighborhood of z=0. If you take the trouble to derive the Laurent expansion and learn about residues this may all become clear. Take it from me, you can do this and it is mathematically rigorous. The value of the function at z=0 is the coefficient of the b_{1} term in the Laurent expansion. It is also the same as the residue of the function at that point.
     
    Last edited: Jan 29, 2016
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  5. WBahn

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    Mar 31, 2012
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    The function IS defined at z=0 -- they explicitly defined it to be zero when they said f(0) = 0. If z is not zero, the function is defined one way, and if z is equal to zero it is defined another. Thus the function is defined for all z.
     
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  6. WBahn

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    Mar 31, 2012
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    No, because now you have two definitions of the f(0). Since x≠1 you are saying that it is f(0) = (0² + 6)/(0-1) = -6 AND you are saying that it is f(0) = 5.

    How about if they had written their definition slightly differently and said

    f(x) is defined such that if x≠0 then f(x) = ((conjugate(z))^ 2)/z, but if x=0 then f(x) = 0.

    This is no different than saying something like

    f(x) = |x| is designed such that if x >= 0 then f(x) = x, but if x < 0 then f(x) = -x.
     
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  7. Papabravo

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    Feb 24, 2006
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    I believe you can demonstrate that the choice of f(0)=0 is consistent with the Laurent series expansion and the calculation of the residue.
     
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  8. e-learner

    Thread Starter New Member

    Apr 25, 2015
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    I am sorry but that was a typo. I wanted to write f(1)=5.
    I learnt about residue theorem and I find values at singularities, so the function does exist at singularities. Thankyou for your explanations, its more clear to me now.

    I also tried for finding residue of this function and it's coming out to be 0. BTW, I also learned finding differentiation of (conjugate of z),which was not given /asked in my book. Thanks to you. Now the only thing left is to find the Laurent expression for the same and finding this a little difficult.
     
    Last edited: Feb 3, 2016
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