I am trying to prove that ∫ C P ( z ) Q ( z ) d z = 0If polynomial order of Q is 2 or more than that of P, using the theorem stating that if a function has a finite number of singular points all interior to a contour C, then ∫ C f ( z ) d z = 2 π i Res [ 1 z 2 f ( 1 z ) ] I received the hint that, under the conditions described above, the rational function in the integrand can be written as a McLauran Series with no negative powers of z. This would imply that the residue is zero...My problem is, I can't seem to wrap my head around how the rational function given to me can be written into the form of a power series with only positive exponents...So, as the title says, When can a rational function be represented as a power series? I'm not looking for a full proof, but a few details would be nice, just so I feel more comfortable running with the hint.

Question

I am trying to prove that      ∫    C              P      (      z      )              Q      (      z      )        d  z  =  0If polynomial order of   Q is 2 or more than that of   P, using the theorem stating that if a function has a finite number of singular points all interior to a contour   C, then      ∫    C    f  (  z  )  d  z  =  2  π  i  Res            [            1          z      2        f  (      1    z    )            ]      I received the hint that, under the conditions described above, the rational function in the integrand can be written as a McLauran Series with no negative powers of z. This would imply that the residue is zero...My problem is, I can&#039;t seem to wrap my head around how the rational function given to me can be written into the form of a power series with only positive exponents...So, as the title says, When can a rational function be represented as a power series? I&#039;m not looking for a full proof, but a few details would be nice, just so I feel more comfortable running with the hint.

kovilovop2 · Accepted Answer

The answer is that whenever you center the expansion at a point where the function is analytic, it can be represented as a Taylor series. In this specific case, we will see that the function in question is analytic at   z  =  0 and therefore has a Maclaurin expansion.To see why we can write       1          z      2                  P      (              1        z            )              Q      (              1        z            )       as a Taylor series centered at   z  =  0, consider distributing the       1          z      2       onto the rational function as shown below:  g  (  z  )  =      1          z      2                  P      (              1        z            )              Q      (              1        z            )        =                    a        0            +              a        1                    1        z            +      ⋯      +              a        n                    1                  z          n                                    b        0                    z        2            +              b        1            z      +              b        2            +      ⋯      +              b        m                    1                  z                      m            −            2                                ×            z              m        −        2                    z              m        −        2              =                    a        0                    z                  m          −          2                    +              a        1                    z                  m          −          3                    +      ⋯      +              a        n                    z                  m          −          n          −          2                                    b        0                    z        m            +              b        1                    z                  m          −          1                    +      ⋯      +              b        m            Now, take note of the far right hand side of the above equation and consider what happens when   z  =  0. Because we know       b    m   is non zero, the denominator is non-zero at   z  =  0. This means that   g  (  z  ) is analytic at   z  =  0We know then that if a function is analytic within some neighborhood of   z  =  0 that it can be written as a Maclaurin power series:  g  (  z  )  =      ∑          n      =      0        ∞        d    n        z    n  Because there are no negative powers of z in this expansion, its residue is zero at   z  =  0.Using the above information and the fact that             P      (      z      )              Q      (      z      )       is analytic except for a finite number   n of singular points (the zeros of   Q) contained in a large enough contour   C, then:      ∫    C    f  (  z  )  d  z  =  2  π  i      ∑          j      =      1        n        Res          z      =              z        j              [  f  (  x  )  ]  =  2  π  i            Res              z        =        0                            [                    1              z        2              f                  (                    1      z                      )                            ]              =  0

I am trying to prove that ∫ C </msub> P

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