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Complex Analysis and Differential Geometry




                    Notes                                          d r    2  k
                                                                    2
                                                                   dt 2   r 3    r 2 .

                                   Although this now involves only the one unknown function r, as it stands it is tough to solve.
                                   Let’s change variables and think of r as a function of . Let’s also write things in terms of the
                                             1
                                   function  s   .  Then,
                                             r

                                                                  d  d  d    d  .
                                                                            2
                                                                 dt    dt d    r d
                                   Hence,
                                                                 dr  a dr     ds  ,
                                                                      2
                                                                 dt    r d     d
                                   and our differential equation looks like

                                                                          2
                                                            2
                                                           d r   2   2 2 d s  2 3  ks ,
                                                                       s
                                                                                     2
                                                           dt 2    r 3     d 2     s  
                                   or,
                                                                    d s     k  .
                                                                     2
                                                                    d 2    s    2
                                   This one is easy. From high school differential equations class, we remember that

                                                                  1             k
                                                                             )
                                                               s     A cos(     ,
                                                                  r              2
                                   where A and  are constants which depend on the initial conditions. At long last,

                                                                         2 /k
                                                                 r             ,
                                                                    1   cos(   )
                                   where we  have set   =  A /k. The  graph of  this equation  is, of  course, a  conic section  of
                                                         2
                                   eccentricity .
                                   2.2 Functions of a Complex Variable

                                   The real excitement begins when we consider function f : D  C in which the domain D is a
                                   subset of the complex numbers. In some sense, these too are familiar to us from elementary
                                   calculus—they are simply functions from a subset of the plane into the plane:

                                                      f(z) = f(x, y) = u(x, y) + iv(x, y) = (u(x, y), v(x, y))
                                   Thus, f(z) = z  looks like f(z) = z  = (x + iy)  = x  – y  + 2xyi. In other words, u(x, y) = x  – y  and
                                             2
                                                                           2
                                                                        2
                                                                                                           2
                                                            2
                                                                    2
                                                                                                       2
                                   v(x, y) = 2xy. The  complex perspective,  as we  shall see, generally provides richer and more
                                   profitable insights into these functions.
                                   The definition of the limit of a function f at a point z = z  is essentially the same as that which we
                                                                              0
                                   learned in elementary calculus:
                                                                    lim f(z) L
                                                                           
                                                                    z  0 z

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