- #1

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**Okay so the partial fraction decomposition theorem is that if f(z) is a rational function, f(z)=sum of the principal parts of a laurent expansion of f(z) about each root.**

I'm working through an example in my book, I am fine to follow it. (method 1 below)

But instinctively , I would have thought you would have started with a laurent expansion rather than a taylor expansion - i.e- a expansion of z^-1 terms, as we are after the principal part.(method 2 below)

f(z)=[itex]\frac{1}{z^{2}+1}[/itex]=[itex]\frac{1}{z+i}[/itex][itex]\frac{1}{z-i}[/itex]

I am attaining the pricipal part corresponding to the root z=i :

**Method 1**

[itex]\frac{1}{z+i}[/itex][itex]\frac{1}{z-i}[/itex]=[itex]\frac{1}{z-i}[/itex][itex]\frac{1}{z-i+2i}[/itex]=[itex]\frac{1}{z-i}[/itex][itex]\frac{1}{2i}[/itex][itex]\frac{1}{\frac{z-i}{2i}+1}[/itex]

=[itex]\frac{1}{z-i}[/itex][itex]\frac{1}{2i}[/itex][itex]^{\infty}_{0}[/itex][itex]\sum[/itex]((-1)([itex]\frac{z-i}{2i}[/itex]))[itex]^{n}[/itex] [1]

From which I can observe the principal part is given by n=0: [itex]\frac{1}{2i(z-i)}[/itex]

**Method 2**

f(z)=[itex]\frac{1}{z-i}[/itex][itex]\frac{1}{z-i+2i}[/itex]=[itex]\frac{1}{z-i^{2}}[/itex][itex]\frac{1}{\frac{2i}{z-i}+1}[/itex]= [itex]\frac{1}{z-i^{2}}[/itex][itex]^{\infty}_{0}[/itex][itex]\sum[/itex]((-1)[itex]\frac{2i}{z-i}[/itex])[itex]^{n}[/itex] [2]

**From which I can see that to attain a (z-i)[itex]^{-1}[/itex] we require n=-3, which by equality [2], is not valid as n runs from n=0. (it would also yield the answer to method 1 multiplied by -(2i)[itex]^{-3}[/itex], so they don't agree anyway.**

I am not entirely sure, but I think which summation includes the zero n power may be causing some confusion. But I can't see how this would be a strict rule - are my equivalent GP summation expressions [1] and [2] okay?

*(I can see that conventionally the ≥ 0 is for positive powers and ≥ 1 for negative powers. But doing this I still get an incorrect answer with method 2*)

So I don't understand why you would not attain the same answer expanding in terms of negative z powers.

**Many thanks for any assistance, greatly appreciated !**