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I have to find the Laurent series expansion of $$f(z) = \frac{1}{(z + 1)(z +3)}$$ for the region $$0 \lt |{z + 1}| \lt 2$$

Using partial fration f(z) can written as:

$$f(z) = \frac{1}{2} \frac{1}{z + 1} - \frac{1}{2} \frac{1}{z + 3}$$

since we have from Maclurin's series,

$$\sum_0^\infty \frac{1}{1+z} = (-z)^n for |z| <1$$

so ,

$$f(z) = \frac{1}{2} [\sum (-1)^n z^n - \frac{1}{1 + (z + 2)}]$$

How to proceed from here?

Abhilekh Gautam
  • 285

3 Answers3

1

It should rather be $$f(z)=\frac{1}{2}\frac{1}{z+1}-\frac{1}{4} \frac{1}{1+\frac{(z+1)}{2}}$$ $$\implies f(z)=\frac{1}{2}\frac{1}{z+1}-\frac{1}{4} \sum_{k=0}^{\infty} (-1)^k \left(\frac{z+1}{2}\right)^k.$$

Dharmendra Singh
  • 987
1

For the first term, it's already done: $$\dfrac 12\dfrac 1{z+1}$$ on $\mid z+1\mid\gt0$.

For the other one, $$\dfrac 1{z+3}=\dfrac 1{(z+1)+2}=\dfrac 1{2--(z+1)}=\dfrac 12\dfrac 1{1--\dfrac {z+1}2}=\dfrac 12\sum_{n\ge0}(-\dfrac {z+1}2)^n$$ on $\mid z+1\mid\lt2$.

So to finish we have $$\dfrac 12\dfrac 1{z+1}-\dfrac 14\sum_{n\ge0}(-\dfrac{z+1}2)^n$$, on the overlap $0\lt\mid z+1\mid\lt2$.

calc ll
  • 8,427
0

we have,

$$0 < | z + 1| < 2 \implies |z| \lt 1 , \frac{|z + 1|}{2} \lt 1$$

and Using partial fraction we have,

$$f(z) = \frac{1}{2}\frac{1}{z+1} - \frac{1}{2}\frac{1}{z+3}$$

$$\implies \frac{1}{2} \sum(-1)^n z ^n - \frac{1}{4}\frac{1}{1 + \frac{z + 1}{2}}$$

$$\implies \frac{1}{2} \sum(-1)^n z ^n - \frac{1}{4} \sum(-1)^n \frac{(z + 1)^n}{2^n}$$

Abhilekh Gautam
  • 285