For each $ a $ with $ |a| > 1 $, $f^{-1}(a)$ contains exactly one point












2












$begingroup$



Let $ G = { z in mathbb{C} : |z-2| < 1} $ and let $ f $ be
analytic in $ overline{G} $ except for one simple pole $ z_0 $
inside. Suppose that $ |f(z)| = 1$ for all $ z in partial G $. Show
that for all $ a $ with $ |a| > 1 $, $ f^{-1}(a) $ contains exactly
one point.




So $f(z) $ can be written as $ sum a_n (z-2)^n + frac{b}{z - z_0} $, but I don't know how to use the fact that $ |f(z)| = 1 $ for all $ z in partial G $. How to solve this question?






complex-analysis







share|cite|improve this question









$endgroup$

















    2












    $begingroup$



    Let $ G = { z in mathbb{C} : |z-2| < 1} $ and let $ f $ be
    analytic in $ overline{G} $ except for one simple pole $ z_0 $
    inside. Suppose that $ |f(z)| = 1$ for all $ z in partial G $. Show
    that for all $ a $ with $ |a| > 1 $, $ f^{-1}(a) $ contains exactly
    one point.




    So $f(z) $ can be written as $ sum a_n (z-2)^n + frac{b}{z - z_0} $, but I don't know how to use the fact that $ |f(z)| = 1 $ for all $ z in partial G $. How to solve this question?






    complex-analysis







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$



      Let $ G = { z in mathbb{C} : |z-2| < 1} $ and let $ f $ be
      analytic in $ overline{G} $ except for one simple pole $ z_0 $
      inside. Suppose that $ |f(z)| = 1$ for all $ z in partial G $. Show
      that for all $ a $ with $ |a| > 1 $, $ f^{-1}(a) $ contains exactly
      one point.




      So $f(z) $ can be written as $ sum a_n (z-2)^n + frac{b}{z - z_0} $, but I don't know how to use the fact that $ |f(z)| = 1 $ for all $ z in partial G $. How to solve this question?






      complex-analysis







      share|cite|improve this question









      $endgroup$





      Let $ G = { z in mathbb{C} : |z-2| < 1} $ and let $ f $ be
      analytic in $ overline{G} $ except for one simple pole $ z_0 $
      inside. Suppose that $ |f(z)| = 1$ for all $ z in partial G $. Show
      that for all $ a $ with $ |a| > 1 $, $ f^{-1}(a) $ contains exactly
      one point.




      So $f(z) $ can be written as $ sum a_n (z-2)^n + frac{b}{z - z_0} $, but I don't know how to use the fact that $ |f(z)| = 1 $ for all $ z in partial G $. How to solve this question?






      complex-analysis




      complex-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 10:01









      calmcalm

      1387




      1387






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Since this is complex analysis, $ |f(z)| = 1 $ on $ partial G $ should be interpreted in one of two different ways :




          • For the maximum modulus principle, however this result doesn't help at all with finding only one preimage of $ a $.


          • For the indices of points relative to some curves : here, we get by setting the usual curve $ gamma : t mapsto e^{2i pi t} + 2 $ that $ n(f circ gamma, a) = 0 $ because $ operatorname{Im} f circ gamma subset D_c(0, 1)$ and $ a not in D_c(0, 1)$. Thus , we can use the argument principle :



          $$ int_gamma frac{f'(z)}{f(z) - a}dz = 2 i pi (sum n(gamma, text{zero of } f(z) - a) - sum n(gamma, text{pole of } f(z) - a))$$



          and notice that $ int_gamma frac{f'(z)}{f(z) - a}dz = n(fcirc gamma, a) = 0$. The result then follows.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So for this problem, $ |f(z) | leq 1 $ on $ partial G $ is enough since the winding number is still $ 0 $ in this case, correct?
            $endgroup$
            – calm
            Jan 14 at 2:24










          • $begingroup$
            @lifeishard911 I do think that is the case.
            $endgroup$
            – FreeSalad
            Jan 14 at 9:00











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Since this is complex analysis, $ |f(z)| = 1 $ on $ partial G $ should be interpreted in one of two different ways :




          • For the maximum modulus principle, however this result doesn't help at all with finding only one preimage of $ a $.


          • For the indices of points relative to some curves : here, we get by setting the usual curve $ gamma : t mapsto e^{2i pi t} + 2 $ that $ n(f circ gamma, a) = 0 $ because $ operatorname{Im} f circ gamma subset D_c(0, 1)$ and $ a not in D_c(0, 1)$. Thus , we can use the argument principle :



          $$ int_gamma frac{f'(z)}{f(z) - a}dz = 2 i pi (sum n(gamma, text{zero of } f(z) - a) - sum n(gamma, text{pole of } f(z) - a))$$



          and notice that $ int_gamma frac{f'(z)}{f(z) - a}dz = n(fcirc gamma, a) = 0$. The result then follows.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So for this problem, $ |f(z) | leq 1 $ on $ partial G $ is enough since the winding number is still $ 0 $ in this case, correct?
            $endgroup$
            – calm
            Jan 14 at 2:24










          • $begingroup$
            @lifeishard911 I do think that is the case.
            $endgroup$
            – FreeSalad
            Jan 14 at 9:00
















          3












          $begingroup$

          Since this is complex analysis, $ |f(z)| = 1 $ on $ partial G $ should be interpreted in one of two different ways :




          • For the maximum modulus principle, however this result doesn't help at all with finding only one preimage of $ a $.


          • For the indices of points relative to some curves : here, we get by setting the usual curve $ gamma : t mapsto e^{2i pi t} + 2 $ that $ n(f circ gamma, a) = 0 $ because $ operatorname{Im} f circ gamma subset D_c(0, 1)$ and $ a not in D_c(0, 1)$. Thus , we can use the argument principle :



          $$ int_gamma frac{f'(z)}{f(z) - a}dz = 2 i pi (sum n(gamma, text{zero of } f(z) - a) - sum n(gamma, text{pole of } f(z) - a))$$



          and notice that $ int_gamma frac{f'(z)}{f(z) - a}dz = n(fcirc gamma, a) = 0$. The result then follows.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So for this problem, $ |f(z) | leq 1 $ on $ partial G $ is enough since the winding number is still $ 0 $ in this case, correct?
            $endgroup$
            – calm
            Jan 14 at 2:24










          • $begingroup$
            @lifeishard911 I do think that is the case.
            $endgroup$
            – FreeSalad
            Jan 14 at 9:00














          3












          3








          3





          $begingroup$

          Since this is complex analysis, $ |f(z)| = 1 $ on $ partial G $ should be interpreted in one of two different ways :




          • For the maximum modulus principle, however this result doesn't help at all with finding only one preimage of $ a $.


          • For the indices of points relative to some curves : here, we get by setting the usual curve $ gamma : t mapsto e^{2i pi t} + 2 $ that $ n(f circ gamma, a) = 0 $ because $ operatorname{Im} f circ gamma subset D_c(0, 1)$ and $ a not in D_c(0, 1)$. Thus , we can use the argument principle :



          $$ int_gamma frac{f'(z)}{f(z) - a}dz = 2 i pi (sum n(gamma, text{zero of } f(z) - a) - sum n(gamma, text{pole of } f(z) - a))$$



          and notice that $ int_gamma frac{f'(z)}{f(z) - a}dz = n(fcirc gamma, a) = 0$. The result then follows.






          share|cite|improve this answer











          $endgroup$



          Since this is complex analysis, $ |f(z)| = 1 $ on $ partial G $ should be interpreted in one of two different ways :




          • For the maximum modulus principle, however this result doesn't help at all with finding only one preimage of $ a $.


          • For the indices of points relative to some curves : here, we get by setting the usual curve $ gamma : t mapsto e^{2i pi t} + 2 $ that $ n(f circ gamma, a) = 0 $ because $ operatorname{Im} f circ gamma subset D_c(0, 1)$ and $ a not in D_c(0, 1)$. Thus , we can use the argument principle :



          $$ int_gamma frac{f'(z)}{f(z) - a}dz = 2 i pi (sum n(gamma, text{zero of } f(z) - a) - sum n(gamma, text{pole of } f(z) - a))$$



          and notice that $ int_gamma frac{f'(z)}{f(z) - a}dz = n(fcirc gamma, a) = 0$. The result then follows.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 7 at 13:51

























          answered Jan 7 at 10:53









          FreeSaladFreeSalad

          479211




          479211












          • $begingroup$
            So for this problem, $ |f(z) | leq 1 $ on $ partial G $ is enough since the winding number is still $ 0 $ in this case, correct?
            $endgroup$
            – calm
            Jan 14 at 2:24










          • $begingroup$
            @lifeishard911 I do think that is the case.
            $endgroup$
            – FreeSalad
            Jan 14 at 9:00


















          • $begingroup$
            So for this problem, $ |f(z) | leq 1 $ on $ partial G $ is enough since the winding number is still $ 0 $ in this case, correct?
            $endgroup$
            – calm
            Jan 14 at 2:24










          • $begingroup$
            @lifeishard911 I do think that is the case.
            $endgroup$
            – FreeSalad
            Jan 14 at 9:00
















          $begingroup$
          So for this problem, $ |f(z) | leq 1 $ on $ partial G $ is enough since the winding number is still $ 0 $ in this case, correct?
          $endgroup$
          – calm
          Jan 14 at 2:24




          $begingroup$
          So for this problem, $ |f(z) | leq 1 $ on $ partial G $ is enough since the winding number is still $ 0 $ in this case, correct?
          $endgroup$
          – calm
          Jan 14 at 2:24












          $begingroup$
          @lifeishard911 I do think that is the case.
          $endgroup$
          – FreeSalad
          Jan 14 at 9:00




          $begingroup$
          @lifeishard911 I do think that is the case.
          $endgroup$
          – FreeSalad
          Jan 14 at 9:00


















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