Clarification over Ahlfors page 116, 2.1 about winding numbers












3












$begingroup$


Everything on this question is in complex plane.



As the book describes a property of a winding number, it says that:




Outside of the [line segment from $a$ to $b$] the function $(z-a) / (z-b)$ is never real and $leq 0$.




Here, the above statement should be interpreted as "never (real and $leq 0$)".



If anyone could explain why this is true that would be great. I do get why any point on the line segment (other than $b$, in which case the denominator is $0$) has to satisfy the condition that $(z-a) / (z-b)$ is real and $leq 0$, but I am not sure how to prove why any point not on the line has to satisfy the condition also.



Here, $a$ and $b$ are arbitrary complex number in a region determined by a closed curve in the complex plane; both points lie on the same region.










share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Everything on this question is in complex plane.



    As the book describes a property of a winding number, it says that:




    Outside of the [line segment from $a$ to $b$] the function $(z-a) / (z-b)$ is never real and $leq 0$.




    Here, the above statement should be interpreted as "never (real and $leq 0$)".



    If anyone could explain why this is true that would be great. I do get why any point on the line segment (other than $b$, in which case the denominator is $0$) has to satisfy the condition that $(z-a) / (z-b)$ is real and $leq 0$, but I am not sure how to prove why any point not on the line has to satisfy the condition also.



    Here, $a$ and $b$ are arbitrary complex number in a region determined by a closed curve in the complex plane; both points lie on the same region.










    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Everything on this question is in complex plane.



      As the book describes a property of a winding number, it says that:




      Outside of the [line segment from $a$ to $b$] the function $(z-a) / (z-b)$ is never real and $leq 0$.




      Here, the above statement should be interpreted as "never (real and $leq 0$)".



      If anyone could explain why this is true that would be great. I do get why any point on the line segment (other than $b$, in which case the denominator is $0$) has to satisfy the condition that $(z-a) / (z-b)$ is real and $leq 0$, but I am not sure how to prove why any point not on the line has to satisfy the condition also.



      Here, $a$ and $b$ are arbitrary complex number in a region determined by a closed curve in the complex plane; both points lie on the same region.










      share|cite|improve this question











      $endgroup$




      Everything on this question is in complex plane.



      As the book describes a property of a winding number, it says that:




      Outside of the [line segment from $a$ to $b$] the function $(z-a) / (z-b)$ is never real and $leq 0$.




      Here, the above statement should be interpreted as "never (real and $leq 0$)".



      If anyone could explain why this is true that would be great. I do get why any point on the line segment (other than $b$, in which case the denominator is $0$) has to satisfy the condition that $(z-a) / (z-b)$ is real and $leq 0$, but I am not sure how to prove why any point not on the line has to satisfy the condition also.



      Here, $a$ and $b$ are arbitrary complex number in a region determined by a closed curve in the complex plane; both points lie on the same region.







      complex-analysis winding-number






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Oct 27 '18 at 3:35







      Cute Brownie

















      asked Oct 27 '18 at 3:12









      Cute BrownieCute Brownie

      992416




      992416






















          3 Answers
          3






          active

          oldest

          votes


















          0












          $begingroup$

          Note that $frac{z-a}{z-b}$ is unchanged if we add the same number to each of $z$, $a$, and $b$. So, we may translate all of our points by $-a$ to assume that $a=0$. Now let $$t=frac{z}{z-b}.$$ Solving for $z$, we have $$z=frac{t}{t-1}b.$$ If $t$ is real, then we see that $z$ is a real multiple of $b$, so it is on the line between $a=0$ and $b$. More specifically, if $tleq 0$, then $frac{t}{t-1}in [0,1)$, so $z$ is in fact on the line segment between $a=0$ and $b$.



          From a geometric perspective, $frac{z-a}{z-b}$ being negative means that the vector from $a$ to $z$ and the vector from $b$ to $z$ point in opposite directions. It should not be hard to convince yourself with a picture that this only happens when $z$ is on the line segment between $a$ and $b$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Sorry, I misread the question at first; thanks to Eric for pointing it out. Also see his answer for the geometric perspective.



            Anyways, if the statement were true, then there exists $c$ such that $k(c - b) = c - a$ for a real number $k le 0$. That is,
            $$a - kb = (1 - k)c\
            text{Let } l = (1 - k)^{-1}, text{where } 0 lt l le 1.\
            c = {a - kbover 1 - k} = la + (1 - l)b = a + (1 - l)(-a + b)$$



            Yet, $c$ cannot intersect the line segment from $a$ to $b$. That is, $c neq a + m(b - a)$ for any real number $0 le m le 1$. But we have just found $0 le 1 - l lt 1$ above. So the statement can only be false.



            $$ $$



            Original answer:



            I'm not sure if that statement as displayed is actually true. Say f{z} = (z - a)/(z - b), and let a = -1 - i, b = 1 + i. f{c} = 3 for c = 2 + 2i.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              You seem to have missed the "$leq 0$" part of the statement.
              $endgroup$
              – Eric Wofsey
              Oct 27 '18 at 3:32



















            0












            $begingroup$

            The imaginary part of $u=frac{z-a}{z-b}$ is $$frac{u-overline{u}}{2i}=frac{1}{2i}left(frac{z-a}{z-b}-overline{frac{z-a}{z-b}}right)=frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)$$



            So, if we want this to be real, we need



            $$
            begin{eqnarray*}
            frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)&=&0
            \
            (z-a)(overline{z}-overline{b})&=&(z-b)(overline{z}-overline{a}) \
            zoverline{z}-zoverline{b}-aoverline{z}+aoverline{b}&=&zoverline{z}-zoverline{a}-boverline{z}+boverline{a}
            \
            0&=&z(overline{a}-overline{b})+b(overline{z}-overline{a})-a(overline{z}-overline{b})
            \
            0&=&-Re(b)Im(z)+Im(b)Re(z)+Re(a)(Im(z)-Im(b))+Im(a)(Re(b)-Re(z)) \
            0&=&Re(b)Im{a}-Re(a)Im(b)-(Re(a)-Re(b))Im(z)+(Im(b)-Im(a))Re(z) \
            Im(z)&=&frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}
            end{eqnarray*}
            $$

            which is the complex point-slope form of the line connecting $a$ and $b$.



            So, what about whether the real part is positive or negative? We'll use this formula for $z$ to compute $u$.



            $$
            begin{eqnarray*}
            u&=&frac{Re(z)+iIm(z)-left(Re(a)-i Im(a)right)}{Re(z)+iIm(z)-left(Re(b)-i Im(b)right)}\
            &=&
            frac{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(a)+i Im(a))}{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(b)+i Im(b))}\
            &=&
            frac{left(Re(z)-Re(a)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}{left(Re(z)-Re(b)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}\
            &=&
            frac{Re(z)-Re(a)}{Re(z)-Re(b)}
            end{eqnarray*}
            $$

            From there, it's simple algebra to see that $u=frac{Re(z)-Re(a)}{Re(z)-Re(b)}$ is negative only between $Re(a)$ and $Re(b)$, which corresponds to the line segment joining $a$ and $b$.






            share|cite|improve this answer











            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2972950%2fclarification-over-ahlfors-page-116-2-1-about-winding-numbers%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              0












              $begingroup$

              Note that $frac{z-a}{z-b}$ is unchanged if we add the same number to each of $z$, $a$, and $b$. So, we may translate all of our points by $-a$ to assume that $a=0$. Now let $$t=frac{z}{z-b}.$$ Solving for $z$, we have $$z=frac{t}{t-1}b.$$ If $t$ is real, then we see that $z$ is a real multiple of $b$, so it is on the line between $a=0$ and $b$. More specifically, if $tleq 0$, then $frac{t}{t-1}in [0,1)$, so $z$ is in fact on the line segment between $a=0$ and $b$.



              From a geometric perspective, $frac{z-a}{z-b}$ being negative means that the vector from $a$ to $z$ and the vector from $b$ to $z$ point in opposite directions. It should not be hard to convince yourself with a picture that this only happens when $z$ is on the line segment between $a$ and $b$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Note that $frac{z-a}{z-b}$ is unchanged if we add the same number to each of $z$, $a$, and $b$. So, we may translate all of our points by $-a$ to assume that $a=0$. Now let $$t=frac{z}{z-b}.$$ Solving for $z$, we have $$z=frac{t}{t-1}b.$$ If $t$ is real, then we see that $z$ is a real multiple of $b$, so it is on the line between $a=0$ and $b$. More specifically, if $tleq 0$, then $frac{t}{t-1}in [0,1)$, so $z$ is in fact on the line segment between $a=0$ and $b$.



                From a geometric perspective, $frac{z-a}{z-b}$ being negative means that the vector from $a$ to $z$ and the vector from $b$ to $z$ point in opposite directions. It should not be hard to convince yourself with a picture that this only happens when $z$ is on the line segment between $a$ and $b$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Note that $frac{z-a}{z-b}$ is unchanged if we add the same number to each of $z$, $a$, and $b$. So, we may translate all of our points by $-a$ to assume that $a=0$. Now let $$t=frac{z}{z-b}.$$ Solving for $z$, we have $$z=frac{t}{t-1}b.$$ If $t$ is real, then we see that $z$ is a real multiple of $b$, so it is on the line between $a=0$ and $b$. More specifically, if $tleq 0$, then $frac{t}{t-1}in [0,1)$, so $z$ is in fact on the line segment between $a=0$ and $b$.



                  From a geometric perspective, $frac{z-a}{z-b}$ being negative means that the vector from $a$ to $z$ and the vector from $b$ to $z$ point in opposite directions. It should not be hard to convince yourself with a picture that this only happens when $z$ is on the line segment between $a$ and $b$.






                  share|cite|improve this answer









                  $endgroup$



                  Note that $frac{z-a}{z-b}$ is unchanged if we add the same number to each of $z$, $a$, and $b$. So, we may translate all of our points by $-a$ to assume that $a=0$. Now let $$t=frac{z}{z-b}.$$ Solving for $z$, we have $$z=frac{t}{t-1}b.$$ If $t$ is real, then we see that $z$ is a real multiple of $b$, so it is on the line between $a=0$ and $b$. More specifically, if $tleq 0$, then $frac{t}{t-1}in [0,1)$, so $z$ is in fact on the line segment between $a=0$ and $b$.



                  From a geometric perspective, $frac{z-a}{z-b}$ being negative means that the vector from $a$ to $z$ and the vector from $b$ to $z$ point in opposite directions. It should not be hard to convince yourself with a picture that this only happens when $z$ is on the line segment between $a$ and $b$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 27 '18 at 3:44









                  Eric WofseyEric Wofsey

                  187k14215344




                  187k14215344























                      0












                      $begingroup$

                      Sorry, I misread the question at first; thanks to Eric for pointing it out. Also see his answer for the geometric perspective.



                      Anyways, if the statement were true, then there exists $c$ such that $k(c - b) = c - a$ for a real number $k le 0$. That is,
                      $$a - kb = (1 - k)c\
                      text{Let } l = (1 - k)^{-1}, text{where } 0 lt l le 1.\
                      c = {a - kbover 1 - k} = la + (1 - l)b = a + (1 - l)(-a + b)$$



                      Yet, $c$ cannot intersect the line segment from $a$ to $b$. That is, $c neq a + m(b - a)$ for any real number $0 le m le 1$. But we have just found $0 le 1 - l lt 1$ above. So the statement can only be false.



                      $$ $$



                      Original answer:



                      I'm not sure if that statement as displayed is actually true. Say f{z} = (z - a)/(z - b), and let a = -1 - i, b = 1 + i. f{c} = 3 for c = 2 + 2i.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        You seem to have missed the "$leq 0$" part of the statement.
                        $endgroup$
                        – Eric Wofsey
                        Oct 27 '18 at 3:32
















                      0












                      $begingroup$

                      Sorry, I misread the question at first; thanks to Eric for pointing it out. Also see his answer for the geometric perspective.



                      Anyways, if the statement were true, then there exists $c$ such that $k(c - b) = c - a$ for a real number $k le 0$. That is,
                      $$a - kb = (1 - k)c\
                      text{Let } l = (1 - k)^{-1}, text{where } 0 lt l le 1.\
                      c = {a - kbover 1 - k} = la + (1 - l)b = a + (1 - l)(-a + b)$$



                      Yet, $c$ cannot intersect the line segment from $a$ to $b$. That is, $c neq a + m(b - a)$ for any real number $0 le m le 1$. But we have just found $0 le 1 - l lt 1$ above. So the statement can only be false.



                      $$ $$



                      Original answer:



                      I'm not sure if that statement as displayed is actually true. Say f{z} = (z - a)/(z - b), and let a = -1 - i, b = 1 + i. f{c} = 3 for c = 2 + 2i.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        You seem to have missed the "$leq 0$" part of the statement.
                        $endgroup$
                        – Eric Wofsey
                        Oct 27 '18 at 3:32














                      0












                      0








                      0





                      $begingroup$

                      Sorry, I misread the question at first; thanks to Eric for pointing it out. Also see his answer for the geometric perspective.



                      Anyways, if the statement were true, then there exists $c$ such that $k(c - b) = c - a$ for a real number $k le 0$. That is,
                      $$a - kb = (1 - k)c\
                      text{Let } l = (1 - k)^{-1}, text{where } 0 lt l le 1.\
                      c = {a - kbover 1 - k} = la + (1 - l)b = a + (1 - l)(-a + b)$$



                      Yet, $c$ cannot intersect the line segment from $a$ to $b$. That is, $c neq a + m(b - a)$ for any real number $0 le m le 1$. But we have just found $0 le 1 - l lt 1$ above. So the statement can only be false.



                      $$ $$



                      Original answer:



                      I'm not sure if that statement as displayed is actually true. Say f{z} = (z - a)/(z - b), and let a = -1 - i, b = 1 + i. f{c} = 3 for c = 2 + 2i.






                      share|cite|improve this answer











                      $endgroup$



                      Sorry, I misread the question at first; thanks to Eric for pointing it out. Also see his answer for the geometric perspective.



                      Anyways, if the statement were true, then there exists $c$ such that $k(c - b) = c - a$ for a real number $k le 0$. That is,
                      $$a - kb = (1 - k)c\
                      text{Let } l = (1 - k)^{-1}, text{where } 0 lt l le 1.\
                      c = {a - kbover 1 - k} = la + (1 - l)b = a + (1 - l)(-a + b)$$



                      Yet, $c$ cannot intersect the line segment from $a$ to $b$. That is, $c neq a + m(b - a)$ for any real number $0 le m le 1$. But we have just found $0 le 1 - l lt 1$ above. So the statement can only be false.



                      $$ $$



                      Original answer:



                      I'm not sure if that statement as displayed is actually true. Say f{z} = (z - a)/(z - b), and let a = -1 - i, b = 1 + i. f{c} = 3 for c = 2 + 2i.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Oct 27 '18 at 4:37

























                      answered Oct 27 '18 at 3:31









                      d0SO'Nd0SO'N

                      12




                      12












                      • $begingroup$
                        You seem to have missed the "$leq 0$" part of the statement.
                        $endgroup$
                        – Eric Wofsey
                        Oct 27 '18 at 3:32


















                      • $begingroup$
                        You seem to have missed the "$leq 0$" part of the statement.
                        $endgroup$
                        – Eric Wofsey
                        Oct 27 '18 at 3:32
















                      $begingroup$
                      You seem to have missed the "$leq 0$" part of the statement.
                      $endgroup$
                      – Eric Wofsey
                      Oct 27 '18 at 3:32




                      $begingroup$
                      You seem to have missed the "$leq 0$" part of the statement.
                      $endgroup$
                      – Eric Wofsey
                      Oct 27 '18 at 3:32











                      0












                      $begingroup$

                      The imaginary part of $u=frac{z-a}{z-b}$ is $$frac{u-overline{u}}{2i}=frac{1}{2i}left(frac{z-a}{z-b}-overline{frac{z-a}{z-b}}right)=frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)$$



                      So, if we want this to be real, we need



                      $$
                      begin{eqnarray*}
                      frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)&=&0
                      \
                      (z-a)(overline{z}-overline{b})&=&(z-b)(overline{z}-overline{a}) \
                      zoverline{z}-zoverline{b}-aoverline{z}+aoverline{b}&=&zoverline{z}-zoverline{a}-boverline{z}+boverline{a}
                      \
                      0&=&z(overline{a}-overline{b})+b(overline{z}-overline{a})-a(overline{z}-overline{b})
                      \
                      0&=&-Re(b)Im(z)+Im(b)Re(z)+Re(a)(Im(z)-Im(b))+Im(a)(Re(b)-Re(z)) \
                      0&=&Re(b)Im{a}-Re(a)Im(b)-(Re(a)-Re(b))Im(z)+(Im(b)-Im(a))Re(z) \
                      Im(z)&=&frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}
                      end{eqnarray*}
                      $$

                      which is the complex point-slope form of the line connecting $a$ and $b$.



                      So, what about whether the real part is positive or negative? We'll use this formula for $z$ to compute $u$.



                      $$
                      begin{eqnarray*}
                      u&=&frac{Re(z)+iIm(z)-left(Re(a)-i Im(a)right)}{Re(z)+iIm(z)-left(Re(b)-i Im(b)right)}\
                      &=&
                      frac{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(a)+i Im(a))}{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(b)+i Im(b))}\
                      &=&
                      frac{left(Re(z)-Re(a)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}{left(Re(z)-Re(b)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}\
                      &=&
                      frac{Re(z)-Re(a)}{Re(z)-Re(b)}
                      end{eqnarray*}
                      $$

                      From there, it's simple algebra to see that $u=frac{Re(z)-Re(a)}{Re(z)-Re(b)}$ is negative only between $Re(a)$ and $Re(b)$, which corresponds to the line segment joining $a$ and $b$.






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        The imaginary part of $u=frac{z-a}{z-b}$ is $$frac{u-overline{u}}{2i}=frac{1}{2i}left(frac{z-a}{z-b}-overline{frac{z-a}{z-b}}right)=frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)$$



                        So, if we want this to be real, we need



                        $$
                        begin{eqnarray*}
                        frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)&=&0
                        \
                        (z-a)(overline{z}-overline{b})&=&(z-b)(overline{z}-overline{a}) \
                        zoverline{z}-zoverline{b}-aoverline{z}+aoverline{b}&=&zoverline{z}-zoverline{a}-boverline{z}+boverline{a}
                        \
                        0&=&z(overline{a}-overline{b})+b(overline{z}-overline{a})-a(overline{z}-overline{b})
                        \
                        0&=&-Re(b)Im(z)+Im(b)Re(z)+Re(a)(Im(z)-Im(b))+Im(a)(Re(b)-Re(z)) \
                        0&=&Re(b)Im{a}-Re(a)Im(b)-(Re(a)-Re(b))Im(z)+(Im(b)-Im(a))Re(z) \
                        Im(z)&=&frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}
                        end{eqnarray*}
                        $$

                        which is the complex point-slope form of the line connecting $a$ and $b$.



                        So, what about whether the real part is positive or negative? We'll use this formula for $z$ to compute $u$.



                        $$
                        begin{eqnarray*}
                        u&=&frac{Re(z)+iIm(z)-left(Re(a)-i Im(a)right)}{Re(z)+iIm(z)-left(Re(b)-i Im(b)right)}\
                        &=&
                        frac{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(a)+i Im(a))}{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(b)+i Im(b))}\
                        &=&
                        frac{left(Re(z)-Re(a)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}{left(Re(z)-Re(b)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}\
                        &=&
                        frac{Re(z)-Re(a)}{Re(z)-Re(b)}
                        end{eqnarray*}
                        $$

                        From there, it's simple algebra to see that $u=frac{Re(z)-Re(a)}{Re(z)-Re(b)}$ is negative only between $Re(a)$ and $Re(b)$, which corresponds to the line segment joining $a$ and $b$.






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          The imaginary part of $u=frac{z-a}{z-b}$ is $$frac{u-overline{u}}{2i}=frac{1}{2i}left(frac{z-a}{z-b}-overline{frac{z-a}{z-b}}right)=frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)$$



                          So, if we want this to be real, we need



                          $$
                          begin{eqnarray*}
                          frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)&=&0
                          \
                          (z-a)(overline{z}-overline{b})&=&(z-b)(overline{z}-overline{a}) \
                          zoverline{z}-zoverline{b}-aoverline{z}+aoverline{b}&=&zoverline{z}-zoverline{a}-boverline{z}+boverline{a}
                          \
                          0&=&z(overline{a}-overline{b})+b(overline{z}-overline{a})-a(overline{z}-overline{b})
                          \
                          0&=&-Re(b)Im(z)+Im(b)Re(z)+Re(a)(Im(z)-Im(b))+Im(a)(Re(b)-Re(z)) \
                          0&=&Re(b)Im{a}-Re(a)Im(b)-(Re(a)-Re(b))Im(z)+(Im(b)-Im(a))Re(z) \
                          Im(z)&=&frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}
                          end{eqnarray*}
                          $$

                          which is the complex point-slope form of the line connecting $a$ and $b$.



                          So, what about whether the real part is positive or negative? We'll use this formula for $z$ to compute $u$.



                          $$
                          begin{eqnarray*}
                          u&=&frac{Re(z)+iIm(z)-left(Re(a)-i Im(a)right)}{Re(z)+iIm(z)-left(Re(b)-i Im(b)right)}\
                          &=&
                          frac{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(a)+i Im(a))}{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(b)+i Im(b))}\
                          &=&
                          frac{left(Re(z)-Re(a)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}{left(Re(z)-Re(b)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}\
                          &=&
                          frac{Re(z)-Re(a)}{Re(z)-Re(b)}
                          end{eqnarray*}
                          $$

                          From there, it's simple algebra to see that $u=frac{Re(z)-Re(a)}{Re(z)-Re(b)}$ is negative only between $Re(a)$ and $Re(b)$, which corresponds to the line segment joining $a$ and $b$.






                          share|cite|improve this answer











                          $endgroup$



                          The imaginary part of $u=frac{z-a}{z-b}$ is $$frac{u-overline{u}}{2i}=frac{1}{2i}left(frac{z-a}{z-b}-overline{frac{z-a}{z-b}}right)=frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)$$



                          So, if we want this to be real, we need



                          $$
                          begin{eqnarray*}
                          frac{1}{2i}left(frac{z-a}{z-b}-frac{overline{z}-overline{a}}{overline{z}-overline{b}}right)&=&0
                          \
                          (z-a)(overline{z}-overline{b})&=&(z-b)(overline{z}-overline{a}) \
                          zoverline{z}-zoverline{b}-aoverline{z}+aoverline{b}&=&zoverline{z}-zoverline{a}-boverline{z}+boverline{a}
                          \
                          0&=&z(overline{a}-overline{b})+b(overline{z}-overline{a})-a(overline{z}-overline{b})
                          \
                          0&=&-Re(b)Im(z)+Im(b)Re(z)+Re(a)(Im(z)-Im(b))+Im(a)(Re(b)-Re(z)) \
                          0&=&Re(b)Im{a}-Re(a)Im(b)-(Re(a)-Re(b))Im(z)+(Im(b)-Im(a))Re(z) \
                          Im(z)&=&frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}
                          end{eqnarray*}
                          $$

                          which is the complex point-slope form of the line connecting $a$ and $b$.



                          So, what about whether the real part is positive or negative? We'll use this formula for $z$ to compute $u$.



                          $$
                          begin{eqnarray*}
                          u&=&frac{Re(z)+iIm(z)-left(Re(a)-i Im(a)right)}{Re(z)+iIm(z)-left(Re(b)-i Im(b)right)}\
                          &=&
                          frac{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(a)+i Im(a))}{Re(z)+ileft(frac{Im(b) Re(a) - Im(a) Re(b) + (Im(a) - Im(b)) Re(z)}{Re(a) - Re(b)}right)-(Re(b)+i Im(b))}\
                          &=&
                          frac{left(Re(z)-Re(a)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}{left(Re(z)-Re(b)right)left(frac{Re(a)-Re(b)+i(-Im(a)+Im(b))}{Re(a)-Re(b)}right)}\
                          &=&
                          frac{Re(z)-Re(a)}{Re(z)-Re(b)}
                          end{eqnarray*}
                          $$

                          From there, it's simple algebra to see that $u=frac{Re(z)-Re(a)}{Re(z)-Re(b)}$ is negative only between $Re(a)$ and $Re(b)$, which corresponds to the line segment joining $a$ and $b$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 8 at 8:43

























                          answered Oct 27 '18 at 4:43









                          Alexander GruberAlexander Gruber

                          20.1k25102172




                          20.1k25102172






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2972950%2fclarification-over-ahlfors-page-116-2-1-about-winding-numbers%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              Human spaceflight

                              Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                              張江高科駅