Types of Singularities for singularities of the complex function 1 − e z 1 + e z I have two questions:Question 1For Functions with a holomorphic numerator and polynomial denominator, can I simply find the Taylor Series of the numerator in the neighbourhood of the first Singularity, and determine the type in the neighbourhood? And do so for each singularity?ExampleGiven f ( z ) = e z z 2 + 1 with Singularities at z = ± i. Let us check the type of z=i first. Since e z ∈ H ( C ), we can use the taylor expansion in the neighbourhood of i, ie. B ϵ ( i ) where 0 &lt; ϵ &lt; 2 so that − i ∉ B ϵ ( i ). Then f ( z ) = 1 ( z − i ) ( z + i ) ∑ k = 0 ∞ e i k ! ( z − i ) k = 1 z + i ∑ k = 0 ∞ e i k ! ( z − i ) k − 1 = 1 z + i ( e i z − i + ∑ k = 1 ∞ e i k ! ( z − i ) k − 1 )So i has a Pole of first order. Similarily we can show that −i also is a Pole of first order.Is that right?Question 2 How do you go about functions like f ( z ) = 1 − e z 1 + e z , | z | &lt; 4?I've found the Singularities, namely z = ± i π, but have no idea how to proceed, because I can't find a way to use a similar approach to the above example...!

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Types of Singularities for singularities of the complex function              1      −              e        z                    1      +              e        z            I have two questions:Question 1For Functions with a holomorphic numerator and polynomial denominator, can I simply find the Taylor Series of the numerator in the neighbourhood of the first Singularity, and determine the type in the neighbourhood? And do so for each singularity?ExampleGiven   f  (  z  )  =            e      z                      z        2            +      1       with Singularities at   z  =  ±  i. Let us check the type of z=i first. Since       e    z    ∈  H  (      C    ), we can use the taylor expansion in the neighbourhood of i, ie.       B          ϵ        (  i  ) where   0  &amp;lt;  ϵ  &amp;lt;  2 so that   −  i  ∉      B          ϵ        (  i  ). Then  f  (  z  )  =      1          (      z      −      i      )      (      z      +      i      )            ∑          k      =      0              ∞                  e      i              k      !        (  z  −  i      )    k    =      1          z      +      i            ∑          k      =      0              ∞                  e      i              k      !        (  z  −  i      )          k      −      1        =      1          z      +      i        (            e      i              z      −      i        +      ∑          k      =      1              ∞                  e      i              k      !        (  z  −  i      )          k      −      1        )So i has a Pole of first order. Similarily we can show that −i also is a Pole of first order.Is that right?Question 2 How do you go about functions like   f  (  z  )  =            1      −              e        z                    1      +              e        z              ,  |      z    |  &amp;lt;  4?I&#039;ve found the Singularities, namely   z  =  ±  i  π, but have no idea how to proceed, because I can&#039;t find a way to use a similar approach to the above example...!

Alisson Frazier · Accepted Answer

I think your approach to the first one is correctFor the second question I found an approach that can also become an alternative approach for the first questionLet&#039;s say the function   f  (  z  )  =            g      (      z      )              h      (      z      )       has an isolated singularity at   z  =      z    0   because   h  (      z    0    )  =  0. Assume that g and h are analytic in a neighborhood of       z    0  . First expand both g and h into power series about z0, say   g  (  z  )  =      α    k    (  z  −      z    0        )    k    +      α          k      +      1        (  z  −      z    0        )          k      +      1        +  … and   h  (  z  )  =      β    m    (  z  −      z    0        )    m    +  β      m    +    1    (  z  −      z    0        )          m      +      1        +  …If   k  ≥  m, this is a removable singularity which is removed by setting   f  (      z    0    )  =  0 in case k&amp;gt;m and   f  (      z    0    )  =            α      k              β      k       in case   k  =  m (just like in L&#039;Hopital&#039;s rule).If   k  &amp;lt;  m, this is a pole of order   m  −  k.

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