Laurent Series of $ln(|x|)$?












3












$begingroup$


I read in this question that no Laurent series exists for $ln(x)$, as Patch's answer shows:



"The problem with $log z$ in the complex plane is that it is a "multi-valued function", so we must specify what range of values we are considering the function to have. Because we must make this choice, the function fails to be continuous in any (punctured) disk about $z=0$, and thus is not (complex) differentiable in any neighborhood about the point."



For me, this raises the question of what if the function were differentiable on a punctured disk about $z=0,;$ that is, is $ln(|z|)$ holomorphic, and, if so, what is its Laurent series?



I believe from Owen Biesel's answer here that $ln(|z|)$ is holomorphic, but I am new to complex analysis, so I could be very wrong. If it is holomorphic, how would I go about calculating its Laurent series? I understand that it involves taking a positively oriented closed contour integral, but I am having trouble understanding how to properly take contour integrals, as I keep obtaining answers of 0. Thank you in advance!










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$endgroup$








  • 1




    $begingroup$
    Patch's answer notes that $ln$ is not holmorphic on any set containing zero. Owen Biesel notes that if $f$ is holomorphic on some domain $U$, then $ln(|f(z)|)$ is harmonic on any simply connected subdomain of $U$ (note that harmonic and holomorphic are different---though related---ideas). Your question seems to be about the holomorphicity of $ln(|z|)$---are you familiar with the Cauchy-Riemann equations? What do they tell you about whether or not $ln(|z|)$ is holomorphic?
    $endgroup$
    – Xander Henderson
    Jan 9 at 3:59










  • $begingroup$
    @Xander I was unaware of the Cauchy-Riemann equations. I see that they provide a good criteria for determining holomorphicity. I will continue my studies. Thank you!
    $endgroup$
    – Helvetican
    Jan 9 at 5:40






  • 1




    $begingroup$
    How could a real function be holomorphic ?
    $endgroup$
    – Yves Daoust
    Jan 9 at 13:22
















3












$begingroup$


I read in this question that no Laurent series exists for $ln(x)$, as Patch's answer shows:



"The problem with $log z$ in the complex plane is that it is a "multi-valued function", so we must specify what range of values we are considering the function to have. Because we must make this choice, the function fails to be continuous in any (punctured) disk about $z=0$, and thus is not (complex) differentiable in any neighborhood about the point."



For me, this raises the question of what if the function were differentiable on a punctured disk about $z=0,;$ that is, is $ln(|z|)$ holomorphic, and, if so, what is its Laurent series?



I believe from Owen Biesel's answer here that $ln(|z|)$ is holomorphic, but I am new to complex analysis, so I could be very wrong. If it is holomorphic, how would I go about calculating its Laurent series? I understand that it involves taking a positively oriented closed contour integral, but I am having trouble understanding how to properly take contour integrals, as I keep obtaining answers of 0. Thank you in advance!










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Patch's answer notes that $ln$ is not holmorphic on any set containing zero. Owen Biesel notes that if $f$ is holomorphic on some domain $U$, then $ln(|f(z)|)$ is harmonic on any simply connected subdomain of $U$ (note that harmonic and holomorphic are different---though related---ideas). Your question seems to be about the holomorphicity of $ln(|z|)$---are you familiar with the Cauchy-Riemann equations? What do they tell you about whether or not $ln(|z|)$ is holomorphic?
    $endgroup$
    – Xander Henderson
    Jan 9 at 3:59










  • $begingroup$
    @Xander I was unaware of the Cauchy-Riemann equations. I see that they provide a good criteria for determining holomorphicity. I will continue my studies. Thank you!
    $endgroup$
    – Helvetican
    Jan 9 at 5:40






  • 1




    $begingroup$
    How could a real function be holomorphic ?
    $endgroup$
    – Yves Daoust
    Jan 9 at 13:22














3












3








3





$begingroup$


I read in this question that no Laurent series exists for $ln(x)$, as Patch's answer shows:



"The problem with $log z$ in the complex plane is that it is a "multi-valued function", so we must specify what range of values we are considering the function to have. Because we must make this choice, the function fails to be continuous in any (punctured) disk about $z=0$, and thus is not (complex) differentiable in any neighborhood about the point."



For me, this raises the question of what if the function were differentiable on a punctured disk about $z=0,;$ that is, is $ln(|z|)$ holomorphic, and, if so, what is its Laurent series?



I believe from Owen Biesel's answer here that $ln(|z|)$ is holomorphic, but I am new to complex analysis, so I could be very wrong. If it is holomorphic, how would I go about calculating its Laurent series? I understand that it involves taking a positively oriented closed contour integral, but I am having trouble understanding how to properly take contour integrals, as I keep obtaining answers of 0. Thank you in advance!










share|cite|improve this question











$endgroup$




I read in this question that no Laurent series exists for $ln(x)$, as Patch's answer shows:



"The problem with $log z$ in the complex plane is that it is a "multi-valued function", so we must specify what range of values we are considering the function to have. Because we must make this choice, the function fails to be continuous in any (punctured) disk about $z=0$, and thus is not (complex) differentiable in any neighborhood about the point."



For me, this raises the question of what if the function were differentiable on a punctured disk about $z=0,;$ that is, is $ln(|z|)$ holomorphic, and, if so, what is its Laurent series?



I believe from Owen Biesel's answer here that $ln(|z|)$ is holomorphic, but I am new to complex analysis, so I could be very wrong. If it is holomorphic, how would I go about calculating its Laurent series? I understand that it involves taking a positively oriented closed contour integral, but I am having trouble understanding how to properly take contour integrals, as I keep obtaining answers of 0. Thank you in advance!







complex-analysis complex-numbers analytic-number-theory






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edited Jan 9 at 12:54









user376343

3,8483829




3,8483829










asked Jan 9 at 3:22









HelveticanHelvetican

214




214








  • 1




    $begingroup$
    Patch's answer notes that $ln$ is not holmorphic on any set containing zero. Owen Biesel notes that if $f$ is holomorphic on some domain $U$, then $ln(|f(z)|)$ is harmonic on any simply connected subdomain of $U$ (note that harmonic and holomorphic are different---though related---ideas). Your question seems to be about the holomorphicity of $ln(|z|)$---are you familiar with the Cauchy-Riemann equations? What do they tell you about whether or not $ln(|z|)$ is holomorphic?
    $endgroup$
    – Xander Henderson
    Jan 9 at 3:59










  • $begingroup$
    @Xander I was unaware of the Cauchy-Riemann equations. I see that they provide a good criteria for determining holomorphicity. I will continue my studies. Thank you!
    $endgroup$
    – Helvetican
    Jan 9 at 5:40






  • 1




    $begingroup$
    How could a real function be holomorphic ?
    $endgroup$
    – Yves Daoust
    Jan 9 at 13:22














  • 1




    $begingroup$
    Patch's answer notes that $ln$ is not holmorphic on any set containing zero. Owen Biesel notes that if $f$ is holomorphic on some domain $U$, then $ln(|f(z)|)$ is harmonic on any simply connected subdomain of $U$ (note that harmonic and holomorphic are different---though related---ideas). Your question seems to be about the holomorphicity of $ln(|z|)$---are you familiar with the Cauchy-Riemann equations? What do they tell you about whether or not $ln(|z|)$ is holomorphic?
    $endgroup$
    – Xander Henderson
    Jan 9 at 3:59










  • $begingroup$
    @Xander I was unaware of the Cauchy-Riemann equations. I see that they provide a good criteria for determining holomorphicity. I will continue my studies. Thank you!
    $endgroup$
    – Helvetican
    Jan 9 at 5:40






  • 1




    $begingroup$
    How could a real function be holomorphic ?
    $endgroup$
    – Yves Daoust
    Jan 9 at 13:22








1




1




$begingroup$
Patch's answer notes that $ln$ is not holmorphic on any set containing zero. Owen Biesel notes that if $f$ is holomorphic on some domain $U$, then $ln(|f(z)|)$ is harmonic on any simply connected subdomain of $U$ (note that harmonic and holomorphic are different---though related---ideas). Your question seems to be about the holomorphicity of $ln(|z|)$---are you familiar with the Cauchy-Riemann equations? What do they tell you about whether or not $ln(|z|)$ is holomorphic?
$endgroup$
– Xander Henderson
Jan 9 at 3:59




$begingroup$
Patch's answer notes that $ln$ is not holmorphic on any set containing zero. Owen Biesel notes that if $f$ is holomorphic on some domain $U$, then $ln(|f(z)|)$ is harmonic on any simply connected subdomain of $U$ (note that harmonic and holomorphic are different---though related---ideas). Your question seems to be about the holomorphicity of $ln(|z|)$---are you familiar with the Cauchy-Riemann equations? What do they tell you about whether or not $ln(|z|)$ is holomorphic?
$endgroup$
– Xander Henderson
Jan 9 at 3:59












$begingroup$
@Xander I was unaware of the Cauchy-Riemann equations. I see that they provide a good criteria for determining holomorphicity. I will continue my studies. Thank you!
$endgroup$
– Helvetican
Jan 9 at 5:40




$begingroup$
@Xander I was unaware of the Cauchy-Riemann equations. I see that they provide a good criteria for determining holomorphicity. I will continue my studies. Thank you!
$endgroup$
– Helvetican
Jan 9 at 5:40




1




1




$begingroup$
How could a real function be holomorphic ?
$endgroup$
– Yves Daoust
Jan 9 at 13:22




$begingroup$
How could a real function be holomorphic ?
$endgroup$
– Yves Daoust
Jan 9 at 13:22










1 Answer
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The function $log |z| = frac{1}{2} log (x^2 +y^2)$ is not holomorphic away from $0$, either by the little Picard theorem (the function $log |e^z|$ would then be real and entire) or because it fails to satisfy the Cauchy-Riemann equations. The linked question asserts that the real part of a holomorphic function is harmonic, which also follows directly from the Cauchy-Riemann equations.






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    $begingroup$

    The function $log |z| = frac{1}{2} log (x^2 +y^2)$ is not holomorphic away from $0$, either by the little Picard theorem (the function $log |e^z|$ would then be real and entire) or because it fails to satisfy the Cauchy-Riemann equations. The linked question asserts that the real part of a holomorphic function is harmonic, which also follows directly from the Cauchy-Riemann equations.






    share|cite|improve this answer









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      4












      $begingroup$

      The function $log |z| = frac{1}{2} log (x^2 +y^2)$ is not holomorphic away from $0$, either by the little Picard theorem (the function $log |e^z|$ would then be real and entire) or because it fails to satisfy the Cauchy-Riemann equations. The linked question asserts that the real part of a holomorphic function is harmonic, which also follows directly from the Cauchy-Riemann equations.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        The function $log |z| = frac{1}{2} log (x^2 +y^2)$ is not holomorphic away from $0$, either by the little Picard theorem (the function $log |e^z|$ would then be real and entire) or because it fails to satisfy the Cauchy-Riemann equations. The linked question asserts that the real part of a holomorphic function is harmonic, which also follows directly from the Cauchy-Riemann equations.






        share|cite|improve this answer









        $endgroup$



        The function $log |z| = frac{1}{2} log (x^2 +y^2)$ is not holomorphic away from $0$, either by the little Picard theorem (the function $log |e^z|$ would then be real and entire) or because it fails to satisfy the Cauchy-Riemann equations. The linked question asserts that the real part of a holomorphic function is harmonic, which also follows directly from the Cauchy-Riemann equations.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 4:02









        anomalyanomaly

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