Question

Find series expansion of 1/ z^2-3z+2 in region |z| <2

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Answer 1

Answer:

My attempt:

f(z)=1z2−3z+2

=1(z−2)(z−1)=A(z−2)+B(z−1)

After finding common denominator, equating the numerators, and letting z=0, we get:

=1(z−2)(z−1)=1(z−2)+−1(z−1)

Letting w=z−2, we get:

f(w)=11−(−w+1)−11−(−w)

But this is the part where I get confused. I believe the next step is to find the sums about the new singularities, right? The singularity for the term on the left is w=0 and for the term on the right it is w=−1. The question is asking to find the series in the given region, where the w=−1 does not appear.

The rest of my attempt:

For |w|<0:

−∑n=0∞(−w+1)n=−∑n=0∞(−z+3)n

For |w|>0:

∑n=0∞1(−w+1)n+1=∑n=0∞1(−z+3)n+1

For |w|<−1:

−∑n=0∞(−w)n=−∑n=0∞(−z+2)n

For |w|>−1:

∑n=0∞1(−w)n+1=∑n=0∞1(−z+2)n+1

So what do I do about the w=−1 singularity

Step-by-step explanation:

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