Given \(\displaystyle{f{{\left({z}\right)}}}={\frac{{{1}}}{{{z}^{{2}}{\left({1}-{z}\right)}}}}\) I am to find two

markush35q

markush35q

Answered question

2022-03-25

Given f(z)=1z2(1z) I am to find two Laurent series expansions. There are two singularities, z=0 and z=1. So for the first expansion, I used the region 0<|z|<1 and I got n=0zn+1z+1z2. The second expansion is for the region 1<|z|<. I don't know how to approach this, the explanation in my book is confusing. Any help?

Answer & Explanation

Pubephenedsjq

Pubephenedsjq

Beginner2022-03-26Added 11 answers

The problem is that for |z|>1 the absolute value of z is greater than 1 (ok that sound ridicolus but it is really the problem) as the geometric sequence wont work in the given form.
We just change it a bit.
1z2(1z)=1z211z=1z21zz(1z)=1z21z(1z1)
This is equivalent to
1z2(1z)=1z3111z
note that |1z|<1. hence
1z2(1z)=1z3k=0(1z)k
This is the same as
k=0(1z)k+3=n=3(1z)n=n=3zn=m=3zm

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