Suppose we are given a rational function f ( s ) = 25 s s 4 + 18 s 3 + 134 s 2 + 472 s + 680 and we need to find the zeros and poles of the function.Suppose f ( s ) = a ( s ) b ( s ) , then a ( s ) = 25 s and s = 0 is a root of a ( s ) = 0 and hence 0 is a zero of f(s). Again b(s)=0 has roots at s = − 5 ± 3 i and s = − 4 ± 2 i, hence − 5 ± 3 i and − 4 ± 2 i are poles of f(s).Is the solution correct?

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Suppose we are given a rational function  f  (  s  )  =            25      s                      s        4            +      18              s        3            +      134              s        2            +      472      s      +      680      and we need to find the zeros and poles of the function.Suppose   f  (  s  )  =            a      (      s      )              b      (      s      )      , then   a  (  s  )  =  25  s and   s  =  0 is a root of   a  (  s  )  =  0 and hence 0 is a zero of f(s). Again b(s)=0 has roots at   s  =  −  5  ±  3  i and   s  =  −  4  ±  2  i, hence   −  5  ±  3  i and   −  4  ±  2  i are poles of f(s).Is the solution correct?

Lucille Melton · Accepted Answer

Just because s=0 is a zero of a(s) does not mean s=0 is a zero of f(s). To be a zero, the numerator must be zero and the denominator not (otherwise, apply l&#039;Hopital&#039;s rule to agonizingly cancel the common factor in a and b). Happily,   b  (  0  )  ≠  0 and once you observe this fact, you have found all the zeroes of f (and shown that it is a genuine zero of f). (Example: z/z has a removable singularity at z=0 and upon removal takes the value 1.)f  has a pole at s if b(s)=0 and   a  (  s  )  ≠  0. With the above modification, you would have already disposed of the possibility of common zeroes in a and b, so every zero of b is a pole, as you write.

Suppose we are given a rational function f ( s ) = 25 s </mrow>

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