Finding singularities and residualsLet f ( z ) = 1 ( z − 1 ) ( 2 z − 1 ) . Then f ( z ) = 1 z − 1 − 2 2 z − 1 = − ∑ k = 0 ∞ z k − 1 z ( 1 − 1 2 z ) = − ( 1 + z + z 2 + z 3 + ⋯ + z k + … ) − 1 z ∗ ( 1 + 1 2 z + 1 ( 2 z ) 2 + ⋯ + 1 ( 2 z ) k + … ) = − ( 1 + z + z 2 + z 3 + ⋯ + z k + … ) − 1 z ∗ ( 1 + 1 2 z + 1 ( 2 z ) 2 + ⋯ + 1 ( 2 z ) k + … )Therefore, the origin is an essential singularity of f ( z ) with residue − 1.

Question

Finding singularities and residualsLet   f  (  z  )  =      1          (      z      −      1      )      (      2      z      −      1      )      . Then  f  (  z  )  =      1          z      −      1        −      2          2      z      −      1        =  −      ∑          k      =      0        ∞        z    k    −      1          z      (      1      −              1                  2          z                    )        =  −  (  1  +  z  +      z    2    +      z    3    +  ⋯  +      z    k    +  …  )  −      1    z    ∗  (  1  +      1          2      z        +      1          (      2      z              )        2              +  ⋯  +      1          (      2      z              )        k              +  …  )  =  −  (  1  +  z  +      z    2    +      z    3    +  ⋯  +      z    k    +  …  )  −      1    z    ∗  (  1  +      1          2      z        +      1          (      2      z              )        2              +  ⋯  +      1          (      2      z              )        k              +  …  )Therefore, the origin is an essential singularity of   f  (  z  ) with residue   −  1.

crearti2d4 · Accepted Answer

There are two singularities   1 and       1    2  , so you need to expand around these singularities.Around   1:                    f        (        z        )                            =                  1                      z            −            1                          −                  2                      2            z            −            1                                                            =                  1                      z            −            1                          −                  2                      2            (            z            −            1            )            −            1                                                            =                  1                      z            −            1                          −        2                  ∑                      n            =            0                            ∞                (        −        1                  )          n                          2          n                (        z        −        1                  )          n                                                  =                  1                      z            −            1                          −                  ∑                      n            =            0                            ∞                (        −        1                  )          n                          2                      n            +            1                          (        z        −        1                  )          n                    So   1 is a pole of order   1 with residue   =  1.Around       1    2  :                    f        (        z        )                            =                  1                      z            −            1                          −                  2                      2            z            −            1                                                            =                  1                      z            −                          1              2                        −                          1              2                                      −                  1                      z            −                          1              2                                                                        =                              −            2                                1            −            2            (            z            −                          1              2                        )                          −                  1                      z            −                          1              2                                                                        =        −        2                  ∑                      n            =            0                            ∞                          2          n                (        z        −                  1          2                          )          n                −                  1                      z            −                          1              2                                                                        =                  ∑                      n            =            0                            ∞                −                  2                      n            +            1                          (        z        −                  1          2                          )          n                −                  1                      z            −                          1              2                                          So       1    2   is a pole of order   1 with residue   =  −  1.

Let f(z)=1/((z-1)(2z-1)). Then f(z)=1/(z-1)-2/(2z-1)=-sum_(k=0)^infty z^k-1/(z(1-1/2z))

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