Proving theorem connecting the inverse of a holomorphic function to a contour integral of the function.












3












$begingroup$


I am asked to prove this theorem:



If $f:U rightarrow C$ is holomorphic in $U$ and invertible, $Pin U$ and if $D(P,r)$ is a sufficently small disc about P, then



$$f^{-1}(w) = frac{1}{2pi i} oint_{partial D(P,r)}{frac{sf'(s)}{f(s)-w}}ds$$



The book says to "imitate the proof of the argument principle" but I am not seeing the connection.






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should assume that $|f^{-1}(w)-P|<r.$
    $endgroup$
    – P..
    Nov 12 '12 at 7:06
















3












$begingroup$


I am asked to prove this theorem:



If $f:U rightarrow C$ is holomorphic in $U$ and invertible, $Pin U$ and if $D(P,r)$ is a sufficently small disc about P, then



$$f^{-1}(w) = frac{1}{2pi i} oint_{partial D(P,r)}{frac{sf'(s)}{f(s)-w}}ds$$



The book says to "imitate the proof of the argument principle" but I am not seeing the connection.






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should assume that $|f^{-1}(w)-P|<r.$
    $endgroup$
    – P..
    Nov 12 '12 at 7:06














3












3








3


4



$begingroup$


I am asked to prove this theorem:



If $f:U rightarrow C$ is holomorphic in $U$ and invertible, $Pin U$ and if $D(P,r)$ is a sufficently small disc about P, then



$$f^{-1}(w) = frac{1}{2pi i} oint_{partial D(P,r)}{frac{sf'(s)}{f(s)-w}}ds$$



The book says to "imitate the proof of the argument principle" but I am not seeing the connection.






complex-analysis







share|cite|improve this question











$endgroup$




I am asked to prove this theorem:



If $f:U rightarrow C$ is holomorphic in $U$ and invertible, $Pin U$ and if $D(P,r)$ is a sufficently small disc about P, then



$$f^{-1}(w) = frac{1}{2pi i} oint_{partial D(P,r)}{frac{sf'(s)}{f(s)-w}}ds$$



The book says to "imitate the proof of the argument principle" but I am not seeing the connection.






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 12 '12 at 5:56









Gerry Myerson

146k8147299




146k8147299










asked Nov 12 '12 at 5:33









MikeMike

166239




166239












  • $begingroup$
    You should assume that $|f^{-1}(w)-P|<r.$
    $endgroup$
    – P..
    Nov 12 '12 at 7:06


















  • $begingroup$
    You should assume that $|f^{-1}(w)-P|<r.$
    $endgroup$
    – P..
    Nov 12 '12 at 7:06
















$begingroup$
You should assume that $|f^{-1}(w)-P|<r.$
$endgroup$
– P..
Nov 12 '12 at 7:06




$begingroup$
You should assume that $|f^{-1}(w)-P|<r.$
$endgroup$
– P..
Nov 12 '12 at 7:06










3 Answers
3






active

oldest

votes


















5












$begingroup$

Hint: Since $f$ is holomorphic and invertible, for each $win f(D(P,r))$, $f(z)-w$ has a unique zero $f^{-1}(w):=z_0$ in $U$; moreover, $z_0in D(P,r)$. Therefore, $f(z)=w+(z-z_0)h(z)$ on $U$, where $h$ is holomorphic and has no zero on $U$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    I had a similar problem:




    1. $f(z)$ has local inverse $f^{-1}(w) = z$. Write down an integral formula that gives $f^{-1}(w)$ in terms of $f(z)$.


    My solution:




    1. Apply Cauchy Integral formula: Since $f^{-1}(w)$ is analytic by the inverse function theorem, we can say:


    $$f^{-1}(w) = frac{1}{2pi i }oint_{f(partial D(P,r))}frac{f^{-1}(u)}{u-w}du$$
    2. Using Substitution: $u=f(z)$ for $z$ on $partial D(P,r) implies du = f'(z)dz$



    3.Rewrite: $$f^{-1}(w) = frac{1}{2pi i }oint_{partial D(P,r)}frac{zf'(z)}{f(z)-w}dz$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      After some thought, it makes sense after applying the Cauchy Integral Formula to the inverse function and then making the substitution $ f^{-1}(s) = t $ where t is taken along the path given by $f^{-1}(s)$ where s is along $partial D(P,r)$.



      The problem is being able to deform that curve into a circle which goes around $f^{-1}(w)$. This can be done if we can show the path only goes around $f^{-1}(w)$ once.



      I still have no idea how the argument principle is involved






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        It's just something very similar to the argument principle (in particular the proof), eg, see the answer. Or just look at the special case $w = 0$, and compare the statements and the proofs of this and argument principle.
        $endgroup$
        – user27126
        Nov 15 '12 at 17:59










      • $begingroup$
        I would but I still don't have a proof of this, can you direct me to one?
        $endgroup$
        – Mathstudent
        Nov 15 '12 at 18:15










      • $begingroup$
        What's wrong with the other answer?
        $endgroup$
        – user27126
        Nov 15 '12 at 18:57










      • $begingroup$
        The other answer is a hint, not a proof of the the above, which I am still confused on.
        $endgroup$
        – Mathstudent
        Nov 19 '12 at 2:17










      • $begingroup$
        OK. Use Cauchy's integral formula. There is only one pole to be picked up on the right hand side, which is at $z_0 = f^{-1}(w)$. So we only need to compute the residue. Near $z_0$, write $f(z)$ as $w + (z-z_0)^k h(z)$, where $h(z_0) neq 0$. Then $f'(z) = k(z-z_0)^{k-1} (h(z) + (z-z_0)h'(z)/k)$, where the thing inside the bracket does not vanish at $z_0$. Can you see what the residue at $z_0$ is now?
        $endgroup$
        – user27126
        Nov 19 '12 at 6:53













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Hint: Since $f$ is holomorphic and invertible, for each $win f(D(P,r))$, $f(z)-w$ has a unique zero $f^{-1}(w):=z_0$ in $U$; moreover, $z_0in D(P,r)$. Therefore, $f(z)=w+(z-z_0)h(z)$ on $U$, where $h$ is holomorphic and has no zero on $U$.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        Hint: Since $f$ is holomorphic and invertible, for each $win f(D(P,r))$, $f(z)-w$ has a unique zero $f^{-1}(w):=z_0$ in $U$; moreover, $z_0in D(P,r)$. Therefore, $f(z)=w+(z-z_0)h(z)$ on $U$, where $h$ is holomorphic and has no zero on $U$.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          Hint: Since $f$ is holomorphic and invertible, for each $win f(D(P,r))$, $f(z)-w$ has a unique zero $f^{-1}(w):=z_0$ in $U$; moreover, $z_0in D(P,r)$. Therefore, $f(z)=w+(z-z_0)h(z)$ on $U$, where $h$ is holomorphic and has no zero on $U$.






          share|cite|improve this answer









          $endgroup$



          Hint: Since $f$ is holomorphic and invertible, for each $win f(D(P,r))$, $f(z)-w$ has a unique zero $f^{-1}(w):=z_0$ in $U$; moreover, $z_0in D(P,r)$. Therefore, $f(z)=w+(z-z_0)h(z)$ on $U$, where $h$ is holomorphic and has no zero on $U$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 12 '12 at 6:42









          23rd23rd

          13.9k2957




          13.9k2957























              2












              $begingroup$

              I had a similar problem:




              1. $f(z)$ has local inverse $f^{-1}(w) = z$. Write down an integral formula that gives $f^{-1}(w)$ in terms of $f(z)$.


              My solution:




              1. Apply Cauchy Integral formula: Since $f^{-1}(w)$ is analytic by the inverse function theorem, we can say:


              $$f^{-1}(w) = frac{1}{2pi i }oint_{f(partial D(P,r))}frac{f^{-1}(u)}{u-w}du$$
              2. Using Substitution: $u=f(z)$ for $z$ on $partial D(P,r) implies du = f'(z)dz$



              3.Rewrite: $$f^{-1}(w) = frac{1}{2pi i }oint_{partial D(P,r)}frac{zf'(z)}{f(z)-w}dz$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                I had a similar problem:




                1. $f(z)$ has local inverse $f^{-1}(w) = z$. Write down an integral formula that gives $f^{-1}(w)$ in terms of $f(z)$.


                My solution:




                1. Apply Cauchy Integral formula: Since $f^{-1}(w)$ is analytic by the inverse function theorem, we can say:


                $$f^{-1}(w) = frac{1}{2pi i }oint_{f(partial D(P,r))}frac{f^{-1}(u)}{u-w}du$$
                2. Using Substitution: $u=f(z)$ for $z$ on $partial D(P,r) implies du = f'(z)dz$



                3.Rewrite: $$f^{-1}(w) = frac{1}{2pi i }oint_{partial D(P,r)}frac{zf'(z)}{f(z)-w}dz$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  I had a similar problem:




                  1. $f(z)$ has local inverse $f^{-1}(w) = z$. Write down an integral formula that gives $f^{-1}(w)$ in terms of $f(z)$.


                  My solution:




                  1. Apply Cauchy Integral formula: Since $f^{-1}(w)$ is analytic by the inverse function theorem, we can say:


                  $$f^{-1}(w) = frac{1}{2pi i }oint_{f(partial D(P,r))}frac{f^{-1}(u)}{u-w}du$$
                  2. Using Substitution: $u=f(z)$ for $z$ on $partial D(P,r) implies du = f'(z)dz$



                  3.Rewrite: $$f^{-1}(w) = frac{1}{2pi i }oint_{partial D(P,r)}frac{zf'(z)}{f(z)-w}dz$$






                  share|cite|improve this answer









                  $endgroup$



                  I had a similar problem:




                  1. $f(z)$ has local inverse $f^{-1}(w) = z$. Write down an integral formula that gives $f^{-1}(w)$ in terms of $f(z)$.


                  My solution:




                  1. Apply Cauchy Integral formula: Since $f^{-1}(w)$ is analytic by the inverse function theorem, we can say:


                  $$f^{-1}(w) = frac{1}{2pi i }oint_{f(partial D(P,r))}frac{f^{-1}(u)}{u-w}du$$
                  2. Using Substitution: $u=f(z)$ for $z$ on $partial D(P,r) implies du = f'(z)dz$



                  3.Rewrite: $$f^{-1}(w) = frac{1}{2pi i }oint_{partial D(P,r)}frac{zf'(z)}{f(z)-w}dz$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 25 '14 at 21:53









                  J_Lopez8J_Lopez8

                  555




                  555























                      1












                      $begingroup$

                      After some thought, it makes sense after applying the Cauchy Integral Formula to the inverse function and then making the substitution $ f^{-1}(s) = t $ where t is taken along the path given by $f^{-1}(s)$ where s is along $partial D(P,r)$.



                      The problem is being able to deform that curve into a circle which goes around $f^{-1}(w)$. This can be done if we can show the path only goes around $f^{-1}(w)$ once.



                      I still have no idea how the argument principle is involved






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        It's just something very similar to the argument principle (in particular the proof), eg, see the answer. Or just look at the special case $w = 0$, and compare the statements and the proofs of this and argument principle.
                        $endgroup$
                        – user27126
                        Nov 15 '12 at 17:59










                      • $begingroup$
                        I would but I still don't have a proof of this, can you direct me to one?
                        $endgroup$
                        – Mathstudent
                        Nov 15 '12 at 18:15










                      • $begingroup$
                        What's wrong with the other answer?
                        $endgroup$
                        – user27126
                        Nov 15 '12 at 18:57










                      • $begingroup$
                        The other answer is a hint, not a proof of the the above, which I am still confused on.
                        $endgroup$
                        – Mathstudent
                        Nov 19 '12 at 2:17










                      • $begingroup$
                        OK. Use Cauchy's integral formula. There is only one pole to be picked up on the right hand side, which is at $z_0 = f^{-1}(w)$. So we only need to compute the residue. Near $z_0$, write $f(z)$ as $w + (z-z_0)^k h(z)$, where $h(z_0) neq 0$. Then $f'(z) = k(z-z_0)^{k-1} (h(z) + (z-z_0)h'(z)/k)$, where the thing inside the bracket does not vanish at $z_0$. Can you see what the residue at $z_0$ is now?
                        $endgroup$
                        – user27126
                        Nov 19 '12 at 6:53


















                      1












                      $begingroup$

                      After some thought, it makes sense after applying the Cauchy Integral Formula to the inverse function and then making the substitution $ f^{-1}(s) = t $ where t is taken along the path given by $f^{-1}(s)$ where s is along $partial D(P,r)$.



                      The problem is being able to deform that curve into a circle which goes around $f^{-1}(w)$. This can be done if we can show the path only goes around $f^{-1}(w)$ once.



                      I still have no idea how the argument principle is involved






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        It's just something very similar to the argument principle (in particular the proof), eg, see the answer. Or just look at the special case $w = 0$, and compare the statements and the proofs of this and argument principle.
                        $endgroup$
                        – user27126
                        Nov 15 '12 at 17:59










                      • $begingroup$
                        I would but I still don't have a proof of this, can you direct me to one?
                        $endgroup$
                        – Mathstudent
                        Nov 15 '12 at 18:15










                      • $begingroup$
                        What's wrong with the other answer?
                        $endgroup$
                        – user27126
                        Nov 15 '12 at 18:57










                      • $begingroup$
                        The other answer is a hint, not a proof of the the above, which I am still confused on.
                        $endgroup$
                        – Mathstudent
                        Nov 19 '12 at 2:17










                      • $begingroup$
                        OK. Use Cauchy's integral formula. There is only one pole to be picked up on the right hand side, which is at $z_0 = f^{-1}(w)$. So we only need to compute the residue. Near $z_0$, write $f(z)$ as $w + (z-z_0)^k h(z)$, where $h(z_0) neq 0$. Then $f'(z) = k(z-z_0)^{k-1} (h(z) + (z-z_0)h'(z)/k)$, where the thing inside the bracket does not vanish at $z_0$. Can you see what the residue at $z_0$ is now?
                        $endgroup$
                        – user27126
                        Nov 19 '12 at 6:53
















                      1












                      1








                      1





                      $begingroup$

                      After some thought, it makes sense after applying the Cauchy Integral Formula to the inverse function and then making the substitution $ f^{-1}(s) = t $ where t is taken along the path given by $f^{-1}(s)$ where s is along $partial D(P,r)$.



                      The problem is being able to deform that curve into a circle which goes around $f^{-1}(w)$. This can be done if we can show the path only goes around $f^{-1}(w)$ once.



                      I still have no idea how the argument principle is involved






                      share|cite|improve this answer











                      $endgroup$



                      After some thought, it makes sense after applying the Cauchy Integral Formula to the inverse function and then making the substitution $ f^{-1}(s) = t $ where t is taken along the path given by $f^{-1}(s)$ where s is along $partial D(P,r)$.



                      The problem is being able to deform that curve into a circle which goes around $f^{-1}(w)$. This can be done if we can show the path only goes around $f^{-1}(w)$ once.



                      I still have no idea how the argument principle is involved







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Nov 15 '12 at 18:05









                      Martin Argerami

                      125k1181180




                      125k1181180










                      answered Nov 15 '12 at 17:41









                      MathstudentMathstudent

                      512




                      512












                      • $begingroup$
                        It's just something very similar to the argument principle (in particular the proof), eg, see the answer. Or just look at the special case $w = 0$, and compare the statements and the proofs of this and argument principle.
                        $endgroup$
                        – user27126
                        Nov 15 '12 at 17:59










                      • $begingroup$
                        I would but I still don't have a proof of this, can you direct me to one?
                        $endgroup$
                        – Mathstudent
                        Nov 15 '12 at 18:15










                      • $begingroup$
                        What's wrong with the other answer?
                        $endgroup$
                        – user27126
                        Nov 15 '12 at 18:57










                      • $begingroup$
                        The other answer is a hint, not a proof of the the above, which I am still confused on.
                        $endgroup$
                        – Mathstudent
                        Nov 19 '12 at 2:17










                      • $begingroup$
                        OK. Use Cauchy's integral formula. There is only one pole to be picked up on the right hand side, which is at $z_0 = f^{-1}(w)$. So we only need to compute the residue. Near $z_0$, write $f(z)$ as $w + (z-z_0)^k h(z)$, where $h(z_0) neq 0$. Then $f'(z) = k(z-z_0)^{k-1} (h(z) + (z-z_0)h'(z)/k)$, where the thing inside the bracket does not vanish at $z_0$. Can you see what the residue at $z_0$ is now?
                        $endgroup$
                        – user27126
                        Nov 19 '12 at 6:53




















                      • $begingroup$
                        It's just something very similar to the argument principle (in particular the proof), eg, see the answer. Or just look at the special case $w = 0$, and compare the statements and the proofs of this and argument principle.
                        $endgroup$
                        – user27126
                        Nov 15 '12 at 17:59










                      • $begingroup$
                        I would but I still don't have a proof of this, can you direct me to one?
                        $endgroup$
                        – Mathstudent
                        Nov 15 '12 at 18:15










                      • $begingroup$
                        What's wrong with the other answer?
                        $endgroup$
                        – user27126
                        Nov 15 '12 at 18:57










                      • $begingroup$
                        The other answer is a hint, not a proof of the the above, which I am still confused on.
                        $endgroup$
                        – Mathstudent
                        Nov 19 '12 at 2:17










                      • $begingroup$
                        OK. Use Cauchy's integral formula. There is only one pole to be picked up on the right hand side, which is at $z_0 = f^{-1}(w)$. So we only need to compute the residue. Near $z_0$, write $f(z)$ as $w + (z-z_0)^k h(z)$, where $h(z_0) neq 0$. Then $f'(z) = k(z-z_0)^{k-1} (h(z) + (z-z_0)h'(z)/k)$, where the thing inside the bracket does not vanish at $z_0$. Can you see what the residue at $z_0$ is now?
                        $endgroup$
                        – user27126
                        Nov 19 '12 at 6:53


















                      $begingroup$
                      It's just something very similar to the argument principle (in particular the proof), eg, see the answer. Or just look at the special case $w = 0$, and compare the statements and the proofs of this and argument principle.
                      $endgroup$
                      – user27126
                      Nov 15 '12 at 17:59




                      $begingroup$
                      It's just something very similar to the argument principle (in particular the proof), eg, see the answer. Or just look at the special case $w = 0$, and compare the statements and the proofs of this and argument principle.
                      $endgroup$
                      – user27126
                      Nov 15 '12 at 17:59












                      $begingroup$
                      I would but I still don't have a proof of this, can you direct me to one?
                      $endgroup$
                      – Mathstudent
                      Nov 15 '12 at 18:15




                      $begingroup$
                      I would but I still don't have a proof of this, can you direct me to one?
                      $endgroup$
                      – Mathstudent
                      Nov 15 '12 at 18:15












                      $begingroup$
                      What's wrong with the other answer?
                      $endgroup$
                      – user27126
                      Nov 15 '12 at 18:57




                      $begingroup$
                      What's wrong with the other answer?
                      $endgroup$
                      – user27126
                      Nov 15 '12 at 18:57












                      $begingroup$
                      The other answer is a hint, not a proof of the the above, which I am still confused on.
                      $endgroup$
                      – Mathstudent
                      Nov 19 '12 at 2:17




                      $begingroup$
                      The other answer is a hint, not a proof of the the above, which I am still confused on.
                      $endgroup$
                      – Mathstudent
                      Nov 19 '12 at 2:17












                      $begingroup$
                      OK. Use Cauchy's integral formula. There is only one pole to be picked up on the right hand side, which is at $z_0 = f^{-1}(w)$. So we only need to compute the residue. Near $z_0$, write $f(z)$ as $w + (z-z_0)^k h(z)$, where $h(z_0) neq 0$. Then $f'(z) = k(z-z_0)^{k-1} (h(z) + (z-z_0)h'(z)/k)$, where the thing inside the bracket does not vanish at $z_0$. Can you see what the residue at $z_0$ is now?
                      $endgroup$
                      – user27126
                      Nov 19 '12 at 6:53






                      $begingroup$
                      OK. Use Cauchy's integral formula. There is only one pole to be picked up on the right hand side, which is at $z_0 = f^{-1}(w)$. So we only need to compute the residue. Near $z_0$, write $f(z)$ as $w + (z-z_0)^k h(z)$, where $h(z_0) neq 0$. Then $f'(z) = k(z-z_0)^{k-1} (h(z) + (z-z_0)h'(z)/k)$, where the thing inside the bracket does not vanish at $z_0$. Can you see what the residue at $z_0$ is now?
                      $endgroup$
                      – user27126
                      Nov 19 '12 at 6:53




















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                      Human spaceflight

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                      Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
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                      張江高科駅

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