I want to find out, which singualrities $f(z)=\frac{z}{e^z+1}$ have?
$e^z+1=0 \Leftrightarrow z_k=(2k+1)i \pi $
But how can I find out, of which type these singualrities are?
Asked
Active
Viewed 56 times
0
-
3$(2k+1)\pi $ is a zero of order 1 in $e^z+1$ and it is not a root of $z$ . So each of these is a pole of order $1$ – user6 Feb 20 '19 at 13:34
-
I see:) The Laurent- series is too difficult to compute? – Sven Feb 20 '19 at 13:40
-
u dont need to compute it eplicitly – user6 Feb 20 '19 at 13:41
1 Answers
1
The numerator plays no role in this matter. Just compute the Laurent series of $f(z)=\frac{1}{1+e^z}$ and determine if the number of terms with negative index is finite (in this case the singularity is a pole) or infinite (in this case it is an essential singularity). You can search for inspiration in This link
PierreCarre
- 23,092
- 1
- 20
- 36
-
Can you maybe explain, how he got from the first line to the second: $$\frac{1}{e^z-1} = 1/z - P(z)/z + P(z)^2/z - P(z)^3/z + ... \ = 1/z - 1/2 - z/3! + z/(2!)^2 - z^2/4! + 2z^2/(2!\cdot 3!) - z^2/(2!)^3 + O(z^3) \$$ – Sven Feb 20 '19 at 13:52