Let r 1 , r 2 , … , r n be distinct complex numbers. Show that a rational function of the form f ( z ) = b 0 + b 1 z + ⋯ + b n − 2 z n − 2 + b n − 1 z n − 1 ( z − r 1 ) ( z − r 2 ) ⋯ ( z − r n ) can be written as a sum f ( z ) = A 1 z − r 1 + A 2 z − r 2 + ⋯ + A n z − r n for some constant A 1 , … A n .Solution:Defining g ( z ) = f ( z ) − A 1 z − r 1 + A 2 z − r 2 + ⋯ + A n z − r n where A 1 , … A n are the residues of f at each of the points r 1 , … r n .I suppose I want to show that g ( z ) is 0, but not sure how to do this. I know that g ( z ) tends to 0 as | z | → ∞....

Question

Let       r    1    ,      r    2    ,  …  ,      r    n   be distinct complex numbers. Show that a rational function of the form  f  (  z  )  =                    b        0            +              b        1            z      +      ⋯      +              b                  n          −          2                            z                  n          −          2                    +              b                  n          −          1                            z                  n          −          1                            (      z      −              r        1            )      (      z      −              r        2            )      ⋯      (      z      −              r        n            )      can be written as a sum  f  (  z  )  =            A      1              z      −              r        1              +            A      2              z      −              r        2              +  ⋯  +            A      n              z      −              r        n            for some constant       A    1    ,  …      A    n  .Solution:Defining  g  (  z  )  =  f  (  z  )  −            A      1              z      −              r        1              +            A      2              z      −              r        2              +  ⋯  +            A      n              z      −              r        n             where       A    1    ,  …      A    n   are the residues of   f at each of the points       r    1    ,  …      r    n  .I suppose I want to show that   g  (  z  ) is 0, but not sure how to do this. I know that   g  (  z  ) tends to 0 as       |    z      |    →  ∞....

relientaaho2 · Accepted Answer

Each       r    j   is a pole of order   ≤  1 of   f, hence there is a polynomial       p    j   such that  f  (  z  )  =            A      j              z      −              r        j              +      p    j    (  z  )(Laurent expansion around       r    j  ).Then it follows that   g is a polynomial. Since   g  (  z  )  →  0 for       |    z      |    →  ∞, we get   g  (  z  )  =  0 for all   z.

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