Integral of an analytic function f over a closed curve which is homotopic to a point in the domain of f.











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The problem states that if $f$ is nonzero and analytic on a region $A$, and if $gamma$ is closed and homotopic to a point in $A$, then $displaystyleint_{gamma} frac{f'(z)}{f(z)}dz = 0$



My approach is to use the fact that since $f$ is analytic on $A$ then so too is $f'$ on $A$, and since $f$ is nonzero, $g = frac{f'(z)}{f(z)}$ is continuous and analytic on $A$ and so we can use the antiderivate theorem and deduce that the integral of this function on a closed curve is zero. Is this the right approach? Does the fact that $gamma$ is homotopic to a point in $A$ help in any way?










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  • What is "the antiderivate theorem"?
    – user 170039
    Nov 13 at 15:53















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down vote

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The problem states that if $f$ is nonzero and analytic on a region $A$, and if $gamma$ is closed and homotopic to a point in $A$, then $displaystyleint_{gamma} frac{f'(z)}{f(z)}dz = 0$



My approach is to use the fact that since $f$ is analytic on $A$ then so too is $f'$ on $A$, and since $f$ is nonzero, $g = frac{f'(z)}{f(z)}$ is continuous and analytic on $A$ and so we can use the antiderivate theorem and deduce that the integral of this function on a closed curve is zero. Is this the right approach? Does the fact that $gamma$ is homotopic to a point in $A$ help in any way?










share|cite|improve this question
























  • What is "the antiderivate theorem"?
    – user 170039
    Nov 13 at 15:53













up vote
0
down vote

favorite









up vote
0
down vote

favorite











The problem states that if $f$ is nonzero and analytic on a region $A$, and if $gamma$ is closed and homotopic to a point in $A$, then $displaystyleint_{gamma} frac{f'(z)}{f(z)}dz = 0$



My approach is to use the fact that since $f$ is analytic on $A$ then so too is $f'$ on $A$, and since $f$ is nonzero, $g = frac{f'(z)}{f(z)}$ is continuous and analytic on $A$ and so we can use the antiderivate theorem and deduce that the integral of this function on a closed curve is zero. Is this the right approach? Does the fact that $gamma$ is homotopic to a point in $A$ help in any way?










share|cite|improve this question















The problem states that if $f$ is nonzero and analytic on a region $A$, and if $gamma$ is closed and homotopic to a point in $A$, then $displaystyleint_{gamma} frac{f'(z)}{f(z)}dz = 0$



My approach is to use the fact that since $f$ is analytic on $A$ then so too is $f'$ on $A$, and since $f$ is nonzero, $g = frac{f'(z)}{f(z)}$ is continuous and analytic on $A$ and so we can use the antiderivate theorem and deduce that the integral of this function on a closed curve is zero. Is this the right approach? Does the fact that $gamma$ is homotopic to a point in $A$ help in any way?







complex-analysis






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edited Nov 13 at 15:49









user 170039

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asked Nov 13 at 6:23









deco

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185












  • What is "the antiderivate theorem"?
    – user 170039
    Nov 13 at 15:53


















  • What is "the antiderivate theorem"?
    – user 170039
    Nov 13 at 15:53
















What is "the antiderivate theorem"?
– user 170039
Nov 13 at 15:53




What is "the antiderivate theorem"?
– user 170039
Nov 13 at 15:53










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Consider the annulus with inner radius $1/2$ and outer radius $2$, $D=operatorname{Ann}(1/2, 2)$. Consider $f(z)=z^{-1}$ then clearly $f$ is analytic and nonzero on $D$, but
begin{align}
int_{|z|=1} frac{f'(z)}{f(z)} dz =& -int_{|z|=1}frac{1}{z} dz \
=& -int^{2pi}_0 e^{-itheta} ie^{itheta} dtheta = -2pi i.
end{align}



Suppose $gamma$ is homotopic to a point $p in D$, i.e.
begin{align}
H(t, z) = (1-t)gamma(z)+tp
end{align}

is continuous on $[0, 1]times D$, then we see that
begin{align}
int_gamma frac{f'(z)}{f(z)} dz = int_{H(t)} frac{f'(z)}{f(z)} dz
end{align}

for all $t in [0, 1]$. In particular, we see that
begin{align}
left|int_{H(t)} frac{f'(z)}{f(z)} dzright| leq& int_{H(t)} frac{|f'(z)|}{|f(z)|} |dz| \
leq& left(left|frac{f'(p)}{f(p)}right|+varepsilonright)operatorname{Length}left(H(t) right)\
leq& C(1-t)left(left|frac{f'(p)}{f(p)}right|+varepsilonright)
end{align}

for $t$ sufficiently close to $1$.






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    Consider the annulus with inner radius $1/2$ and outer radius $2$, $D=operatorname{Ann}(1/2, 2)$. Consider $f(z)=z^{-1}$ then clearly $f$ is analytic and nonzero on $D$, but
    begin{align}
    int_{|z|=1} frac{f'(z)}{f(z)} dz =& -int_{|z|=1}frac{1}{z} dz \
    =& -int^{2pi}_0 e^{-itheta} ie^{itheta} dtheta = -2pi i.
    end{align}



    Suppose $gamma$ is homotopic to a point $p in D$, i.e.
    begin{align}
    H(t, z) = (1-t)gamma(z)+tp
    end{align}

    is continuous on $[0, 1]times D$, then we see that
    begin{align}
    int_gamma frac{f'(z)}{f(z)} dz = int_{H(t)} frac{f'(z)}{f(z)} dz
    end{align}

    for all $t in [0, 1]$. In particular, we see that
    begin{align}
    left|int_{H(t)} frac{f'(z)}{f(z)} dzright| leq& int_{H(t)} frac{|f'(z)|}{|f(z)|} |dz| \
    leq& left(left|frac{f'(p)}{f(p)}right|+varepsilonright)operatorname{Length}left(H(t) right)\
    leq& C(1-t)left(left|frac{f'(p)}{f(p)}right|+varepsilonright)
    end{align}

    for $t$ sufficiently close to $1$.






    share|cite|improve this answer



























      up vote
      0
      down vote













      Consider the annulus with inner radius $1/2$ and outer radius $2$, $D=operatorname{Ann}(1/2, 2)$. Consider $f(z)=z^{-1}$ then clearly $f$ is analytic and nonzero on $D$, but
      begin{align}
      int_{|z|=1} frac{f'(z)}{f(z)} dz =& -int_{|z|=1}frac{1}{z} dz \
      =& -int^{2pi}_0 e^{-itheta} ie^{itheta} dtheta = -2pi i.
      end{align}



      Suppose $gamma$ is homotopic to a point $p in D$, i.e.
      begin{align}
      H(t, z) = (1-t)gamma(z)+tp
      end{align}

      is continuous on $[0, 1]times D$, then we see that
      begin{align}
      int_gamma frac{f'(z)}{f(z)} dz = int_{H(t)} frac{f'(z)}{f(z)} dz
      end{align}

      for all $t in [0, 1]$. In particular, we see that
      begin{align}
      left|int_{H(t)} frac{f'(z)}{f(z)} dzright| leq& int_{H(t)} frac{|f'(z)|}{|f(z)|} |dz| \
      leq& left(left|frac{f'(p)}{f(p)}right|+varepsilonright)operatorname{Length}left(H(t) right)\
      leq& C(1-t)left(left|frac{f'(p)}{f(p)}right|+varepsilonright)
      end{align}

      for $t$ sufficiently close to $1$.






      share|cite|improve this answer

























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        up vote
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        down vote









        Consider the annulus with inner radius $1/2$ and outer radius $2$, $D=operatorname{Ann}(1/2, 2)$. Consider $f(z)=z^{-1}$ then clearly $f$ is analytic and nonzero on $D$, but
        begin{align}
        int_{|z|=1} frac{f'(z)}{f(z)} dz =& -int_{|z|=1}frac{1}{z} dz \
        =& -int^{2pi}_0 e^{-itheta} ie^{itheta} dtheta = -2pi i.
        end{align}



        Suppose $gamma$ is homotopic to a point $p in D$, i.e.
        begin{align}
        H(t, z) = (1-t)gamma(z)+tp
        end{align}

        is continuous on $[0, 1]times D$, then we see that
        begin{align}
        int_gamma frac{f'(z)}{f(z)} dz = int_{H(t)} frac{f'(z)}{f(z)} dz
        end{align}

        for all $t in [0, 1]$. In particular, we see that
        begin{align}
        left|int_{H(t)} frac{f'(z)}{f(z)} dzright| leq& int_{H(t)} frac{|f'(z)|}{|f(z)|} |dz| \
        leq& left(left|frac{f'(p)}{f(p)}right|+varepsilonright)operatorname{Length}left(H(t) right)\
        leq& C(1-t)left(left|frac{f'(p)}{f(p)}right|+varepsilonright)
        end{align}

        for $t$ sufficiently close to $1$.






        share|cite|improve this answer














        Consider the annulus with inner radius $1/2$ and outer radius $2$, $D=operatorname{Ann}(1/2, 2)$. Consider $f(z)=z^{-1}$ then clearly $f$ is analytic and nonzero on $D$, but
        begin{align}
        int_{|z|=1} frac{f'(z)}{f(z)} dz =& -int_{|z|=1}frac{1}{z} dz \
        =& -int^{2pi}_0 e^{-itheta} ie^{itheta} dtheta = -2pi i.
        end{align}



        Suppose $gamma$ is homotopic to a point $p in D$, i.e.
        begin{align}
        H(t, z) = (1-t)gamma(z)+tp
        end{align}

        is continuous on $[0, 1]times D$, then we see that
        begin{align}
        int_gamma frac{f'(z)}{f(z)} dz = int_{H(t)} frac{f'(z)}{f(z)} dz
        end{align}

        for all $t in [0, 1]$. In particular, we see that
        begin{align}
        left|int_{H(t)} frac{f'(z)}{f(z)} dzright| leq& int_{H(t)} frac{|f'(z)|}{|f(z)|} |dz| \
        leq& left(left|frac{f'(p)}{f(p)}right|+varepsilonright)operatorname{Length}left(H(t) right)\
        leq& C(1-t)left(left|frac{f'(p)}{f(p)}right|+varepsilonright)
        end{align}

        for $t$ sufficiently close to $1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 13 at 7:21

























        answered Nov 13 at 6:47









        Jacky Chong

        17.1k21027




        17.1k21027






























             

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