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Complex Variables

                              g ( ) z                                                              − 1
                   (ii) Then         has a pole at z = 0 of              (c) cos (1)           (d)    sin ( ) 1
                              zf  ( ) z                                                            2
                   order                                          48.    Consider the function


                   (a) 1                 (b) 2                            f  ( ) z =  e iz
                                                                                      2
                                                                                    ( z z +  ) 1
                   (c) 3                 (d) greater than 3
            45.    For the function                                      (i) The residue of  f  at the isolated
                                     1                                 singular point in the upper half plane
                    f z =                       , the point z       z = +          : y    0  is
                      ( ) sin
                                                                               x iy C
                                 cos (1/ z ) 
                   = 0 is                                                    − 1                   − 1
                                                                         (a)                   (b)
                   (a) a removable singularity                               2e                    e

                   (b) a pole                                            (c)   e               (d) 1
                                                                             2
                   (c) an essential singularity
                                                                         (ii) The Cauchy Principal Value of
                   (d) a non-isolated singularity                                        sin xdx
                                                                         the integral       2      is
                                15
                                            
                                    n
                        f
            46.    Let  ( ) z =   z  for  z C . If                                   −  ( x x +  ) 1
                                n= 0                                                     −                −
                                                                                                  
                                                                                    +
                                                                                                       −
                                                                             −
                   C  : z i −  =  2  then     f  ( ) z dz  =            (a)  2  (1 2e   1 )   (b)  (1 e  1 )
                                        C  (z i −  ) 15
                                                                                      )
                                                                                   +
                                                                                                   −
                                                                                                         +
                                                                                                    
                                                                         (c) 2 (1 e           (d)  (1 e    − 1 )
                                   )
                              +
                   (a) 2 i  (1 15i
                                                                                                 )
                                                                                              −
                                                                             u
                                                                  49.    Let  ( , x y =  ) 2x (1 y  for all real x
                                                                                                        )
                   (b) 2 i (1 15i−  )                                   and y. then a function  ( , x y , so
                                                                                                
                                                                                                         )
                   (c) 4 i (1 15i+  )                                   that  ( ) z =  f  u ( , x y +  ) i ( , x y  is
                                                                         analytic, is
                   (d) 2 i
                                                                                         2
                                                                                                               2
                                                                         (a)  x −  2  ( y −  ) 1    (b) ( x −  ) 1 −  2  y
                         
                                     n
            47.    Let    a n (z +  ) 1  be the Laurent
                                                                                                               2
                       n=−                                              (c) ( x −  ) 1 +  2  y    (d)  x +  2  ( y −  ) 1
                                                                                         2
                   series expansion of
                                  z  
                                                                              f
                    f  ( ) sinz =             . Then a =    50.    Let  ( ) z  be analytic o n
                                                 −
                                 z + 1          2                      D =   z C   : z − 1 1  such that
                                                                                            
                                                                                
                   (a) 1                 (b) 0                            f  ( ) 1 = 1. If  ( ) z =  f z 2
                                                                                       f
                                                                                                 ( )  for all

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