Page 18 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY
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Unit 2: Complex Functions




          Thus,                                                                                 Notes

                      d (re )  = r  d  i  d  e i
                          i
                      dt       dt (e )  dt

                               
                            = r i  d e i     dr e i
                               
                                 dt    dt
                            =    dr   ir d   e .
                                         i
                               dt   dt 
          Now,

           d 2  i   d r  dr d  d     i   dr  d   d  i
                      2
                                   2
                                    2 
           dt 2  (re )  =    dt 2   i dt dt   ir  dt  e     dt   ir  dt   i  dt  e
                              2
             d r    d  2      d   dr d  
               2
                                      
          =     2   r         i r  2   2  e    i
                           
               dt    dt       dt  dt dt  
                           d 2      k
          Now, the equation   (re )    e  becomes
                                       i
                               i
                          dt 2      r 2
                                                 2
                                  2
                                 d r    d  2      d   dr d   k
                                                         
                                   2   r          i r   2   2       2 .
                                 dt    dt       dt  dt dt   r
          This gives us the two equations
                                         d r  r   d  2  k  ,
                                          2
                                         dt 2       dt      r 2
          and,
                                           2
                                         r d    2  dr d   0.
                                          dt 2  dt dt
          Multiply by r and this second equation becomes


                                           d  r 2 d    0
                                           dt    dt  
          This tells us that


                                               r 2 d
                                                  dt
          is a constant. (This constant  is called the angular momentum.) This result allows us to get rid
             d
          of    in the first of the two differential equations above:
             dt
                                         d r   r    2     k
                                          2
                                         dt 2     r    r 2
          or




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