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Complex Analysis and Differential Geometry




                    Notes
                                          Example 1: Let (t) = t + it , –1  t  1. One easily sees that this function describes that part
                                                             2
                                   of the curve y = x  between x = –1 and x = 1:
                                                2













                                   Another example. Suppose there is a body of mass M  “fixed” at the origin–perhaps the sun–and
                                   there is a body of mass m which is free to move–perhaps a planet. Let the location of this second
                                   body at time t be given by the complex-valued function z(t). We assume the only force on this
                                   mass is the gravitational force of the fixed body. This force f is thus,

                                                                    GMm    z(t) 
                                                                 f =    2     
                                                                     z(t)     z(t)   
                                   where G is the universal gravitational constant. Sir Isaac Newton tells us that


                                                                     f
                                                                           2 
                                                              mz"(t)   GMm    z(t)  
                                                                        z(t)     z(t)   
                                   Hence,
                                                                         GM
                                                                    z” =    3  z
                                                                          z
                                   Next, let’s write this in polar form, z = re :
                                                                    i
                                                                  d 2  i   k  e  i
                                                                  dt 2  (re )   r 2
                                   where we have written GM = k. Now, let’s see what we have.

                                                               d   i  r  d  i  d  e i
                                                               dt  (re )   dt (e )  dt
                                   Now,

                                              d  (re )  =  d   isin
                                                  i
                                              dt       dt  cos    

                                                     = ( sin    i cos )  d
                                                                 
                                                                   dt
                                                     = i(cos  isin ) d
                                                                
                                                                  dt
                                                     = i  d e .
                                                           i
                                                        dt
                                   (Additional evidence that our notation e  = cos  + i sin  is reasonable.)
                                                                   i


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