Let f(z) = (e^{z})/z. How do find the Laurent series

kihanja20

kihanja20

Answered question

2022-01-16

Let f(z)=ezz. How do find the Laurent series expansion centered at z=0 in the domain |z|<1. What's the coefficient of z1?

Answer & Explanation

puhnut1m

puhnut1m

Beginner2022-01-17Added 33 answers

f(z)=k=01k!zk1
Which is valid for z0, which includes |z|<1 (with the exception of z=0).
The requested coefficient is thus 1
Melissa Moore

Melissa Moore

Beginner2022-01-18Added 32 answers

Laurent series of a complex function, centered at z=0, involves powers of z, including some negative powers, but it does not involve powers of z1.
The function f(z)=ezz is analytic in the punctured plane 0<|z| and therefore, it has Laurent series expansion in that domain, centered at the puncture z=0 which we get by dividing the classic exponential series by z:
1z+n=0zn(n+1)!
alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

The series is about z=0 so it is written in powers of z rather than powers of (z1). The series is [(zk1/k!,{k,0,}] so the coefficient of z is 1

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